AMUSEMENTS IN MATHEMATICS

by

HENRY ERNEST DUDENEY (1857-1930)

In Mathematicks he was greater Than Tycho Brahe or Erra Pater: For he, by geometrick scale, Could take the size of pots of ale; Resolve, by sines and tangents, straight, If bread or butter wanted weight; And wisely tell what hour o' th' day The clock does strike by algebra.

BUTLER'S _Hudibras_.

1917

PREFACE

In issuing this volume of my Mathematical Puzzles, of which some haveappeared in periodicals and others are given here for the first time, Imust acknowledge the encouragement that I have received from manyunknown correspondents, at home and abroad, who have expressed a desireto have the problems in a collected form, with some of the solutionsgiven at greater length than is possible in magazines and newspapers.Though I have included a few old puzzles that have interested the worldfor generations, where I felt that there was something new to be saidabout them, the problems are in the main original. It is true that someof these have become widely known through the press, and it is possiblethat the reader may be glad to know their source.

On the question of Mathematical Puzzles in general there is, perhaps,little more to be said than I have written elsewhere. The history of thesubject entails nothing short of the actual story of the beginnings anddevelopment of exact thinking in man. The historian must start from thetime when man first succeeded in counting his ten fingers and individing an apple into two approximately equal parts. Every puzzle thatis worthy of consideration can be referred to mathematics and logic.Every man, woman, and child who tries to "reason out" the answer to thesimplest puzzle is working, though not of necessity consciously, onmathematical lines. Even those puzzles that we have no way of attackingexcept by haphazard attempts can be brought under a method of what hasbeen called "glorified trial"--a system of shortening our labours byavoiding or eliminating what our reason tells us is useless. It is, infact, not easy to say sometimes where the "empirical" begins and whereit ends.

When a man says, "I have never solved a puzzle in my life," it isdifficult to know exactly what he means, for every intelligentindividual is doing it every day. The unfortunate inmates of our lunaticasylums are sent there expressly because they cannot solvepuzzles--because they have lost their powers of reason. If there were nopuzzles to solve, there would be no questions to ask; and if there wereno questions to be asked, what a world it would be! We should all beequally omniscient, and conversation would be useless and idle.

It is possible that some few exceedingly sober-minded mathematicians,who are impatient of any terminology in their favourite science but theacademic, and who object to the elusive x and y appearing under anyother names, will have wished that various problems had been presentedin a less popular dress and introduced with a less flippant phraseology.I can only refer them to the first word of my title and remind them thatwe are primarily out to be amused--not, it is true, without some hope ofpicking up morsels of knowledge by the way. If the manner is light, Ican only say, in the words of Touchstone, that it is "an ill-favouredthing, sir, but my own; a poor humour of mine, sir."

As for the question of difficulty, some of the puzzles, especially inthe Arithmetical and Algebraical category, are quite easy. Yet some ofthose examples that look the simplest should not be passed over withouta little consideration, for now and again it will be found that there issome more or less subtle pitfall or trap into which the reader may beapt to fall. It is good exercise to cultivate the habit of being verywary over the exact wording of a puzzle. It teaches exactitude andcaution. But some of the problems are very hard nuts indeed, and notunworthy of the attention of the advanced mathematician. Readers willdoubtless select according to their individual tastes.

In many cases only the mere answers are given. This leaves the beginnersomething to do on his own behalf in working out the method of solution,and saves space that would be wasted from the point of view of theadvanced student. On the other hand, in particular cases where it seemedlikely to interest, I have given rather extensive solutions and treatedproblems in a general manner. It will often be found that the notes onone problem will serve to elucidate a good many others in the book; sothat the reader's difficulties will sometimes be found cleared up as headvances. Where it is possible to say a thing in a manner that may be"understanded of the people" generally, I prefer to use this simplephraseology, and so engage the attention and interest of a largerpublic. The mathematician will in such cases have no difficulty inexpressing the matter under consideration in terms of his familiarsymbols.

I have taken the greatest care in reading the proofs, and trust that anyerrors that may have crept in are very few. If any such should occur, Ican only plead, in the words of Horace, that "good Homer sometimesnods," or, as the bishop put it, "Not even the youngest curate in mydiocese is infallible."

I have to express my thanks in particular to the proprietors of _TheStrand Magazine_, _Cassell's Magazine_, _The Queen_, _Tit-Bits_, and_The Weekly Dispatch_ for their courtesy in allowing me to reprint someof the puzzles that have appeared in their pages.

THE AUTHORS' CLUB _March_ 25, 1917

CONTENTS

PREFACE v ARITHMETICAL AND ALGEBRAICAL PROBLEMS 1 Money Puzzles 1 Age and Kinship Puzzles 6 Clock Puzzles 9 Locomotion and Speed Puzzles 11 Digital Puzzles 13 Various Arithmetical and Algebraical Problems 17 GEOMETRICAL PROBLEMS 27 Dissection Puzzles 27 Greek Cross Puzzles 28 Various Dissection Puzzles 35 Patchwork Puzzles 46 Various Geometrical Puzzles 49 POINTS AND LINES PROBLEMS 56 MOVING COUNTER PROBLEMS 58 UNICURSAL AND ROUTE PROBLEMS 68 COMBINATION AND GROUP PROBLEMS 76 CHESSBOARD PROBLEMS 85 The Chessboard 85 Statical Chess Puzzles 88 The Guarded Chessboard 95 Dynamical Chess Puzzles 96 Various Chess Puzzles 105 MEASURING, WEIGHING, AND PACKING PUZZLES 109 CROSSING RIVER PROBLEMS 112 PROBLEMS CONCERNING GAMES 114 PUZZLE GAMES 117 MAGIC SQUARE PROBLEMS 119 Subtracting, Multiplying, and Dividing Magics 124 Magic Squares of Primes 125 MAZES AND HOW TO THREAD THEM 127 THE PARADOX PARTY 137 UNCLASSIFIED PROBLEMS 142 SOLUTIONS 148 INDEX 253

AMUSEMENTS IN MATHEMATICS.

ARITHMETICAL AND ALGEBRAICAL PROBLEMS.

"And what was he? Forsooth, a great arithmetician." _Othello_, I. i.

The puzzles in this department are roughly thrown together in classesfor the convenience of the reader. Some are very easy, others quitedifficult. But they are not arranged in any order of difficulty--andthis is intentional, for it is well that the solver should not be warnedthat a puzzle is just what it seems to be. It may, therefore, prove tobe quite as simple as it looks, or it may contain some pitfall intowhich, through want of care or over-confidence, we may stumble.

Also, the arithmetical and algebraical puzzles are not separated in themanner adopted by some authors, who arbitrarily require certain problemsto be solved by one method or the other. The reader is left to make hisown choice and determine which puzzles are capable of being solved byhim on purely arithmetical lines.

MONEY PUZZLES.

"Put not your trust in money, but put your money in trust."

OLIVER WENDELL HOLMES.

1.--A POST-OFFICE PERPLEXITY.

In every business of life we are occasionally perplexed by some chancequestion that for the moment staggers us. I quite pitied a young lady ina branch post-office when a gentleman entered and deposited a crown onthe counter with this request: "Please give me some twopenny stamps, sixtimes as many penny stamps, and make up the rest of the money intwopence-halfpenny stamps." For a moment she seemed bewildered, then herbrain cleared, and with a smile she handed over stamps in exactfulfilment of the order. How long would it have taken you to think itout?

2.--YOUTHFUL PRECOCITY.

The precocity of some youths is surprising. One is disposed to say onoccasion, "That boy of yours is a genius, and he is certain to do greatthings when he grows up;" but past experience has taught us that heinvariably becomes quite an ordinary citizen. It is so often the case,on the contrary, that the dull boy becomes a great man. You never cantell. Nature loves to present to us these queer paradoxes. It is wellknown that those wonderful "lightning calculators," who now and againsurprise the world by their feats, lose all their mysterious powersdirectly they are taught the elementary rules of arithmetic.

A boy who was demolishing a choice banana was approached by a youngfriend, who, regarding him with envious eyes, asked, "How much did youpay for that banana, Fred?" The prompt answer was quite remarkable inits way: "The man what I bought it of receives just half as manysixpences for sixteen dozen dozen bananas as he gives bananas for afiver."

Now, how long will it take the reader to say correctly just how muchFred paid for his rare and refreshing fruit?

3.--AT A CATTLE MARKET.

Three countrymen met at a cattle market. "Look here," said Hodge toJakes, "I'll give you six of my pigs for one of your horses, and thenyou'll have twice as many animals here as I've got." "If that's yourway of doing business," said Durrant to Hodge, "I'll give you fourteenof my sheep for a horse, and then you'll have three times as manyanimals as I." "Well, I'll go better than that," said Jakes to Durrant;"I'll give you four cows for a horse, and then you'll have six times asmany animals as I've got here."

No doubt this was a very primitive way of bartering animals, but it isan interesting little puzzle to discover just how many animals Jakes,Hodge, and Durrant must have taken to the cattle market.

4.--THE BEANFEAST PUZZLE.

A number of men went out together on a bean-feast. There were fourparties invited--namely, 25 cobblers, 20 tailors, 18 hatters, and 12glovers. They spent altogether L6, 13s. It was found that five cobblersspent as much as four tailors; that twelve tailors spent as much as ninehatters; and that six hatters spent as much as eight glovers. The puzzleis to find out how much each of the four parties spent.

5.--A QUEER COINCIDENCE.

Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards,Francis, and Gudgeon, were recently engaged in play. The name of theparticular game is of no consequence. They had agreed that whenever aplayer won a game he should double the money of each of the otherplayers--that is, he was to give the players just as much money as theyhad already in their pockets. They played seven games, and, strange tosay, each won a game in turn, in the order in which their names aregiven. But a more curious coincidence is this--that when they hadfinished play each of the seven men had exactly the same amount--twoshillings and eightpence--in his pocket. The puzzle is to find out howmuch money each man had with him before he sat down to play.

6.--A CHARITABLE BEQUEST.

A man left instructions to his executors to distribute once a yearexactly fifty-five shillings among the poor of his parish; but they wereonly to continue the gift so long as they could make it in differentways, always giving eighteenpence each to a number of women and half acrown each to men. During how many years could the charity beadministered? Of course, by "different ways" is meant a different numberof men and women every time.

7.--THE WIDOW'S LEGACY.

A gentleman who recently died left the sum of L8,000 to be divided amonghis widow, five sons, and four daughters. He directed that every sonshould receive three times as much as a daughter, and that everydaughter should have twice as much as their mother. What was the widow'sshare?

8.--INDISCRIMINATE CHARITY.

A charitable gentleman, on his way home one night, was appealed to bythree needy persons in succession for assistance. To the first person hegave one penny more than half the money he had in his pocket; to thesecond person he gave twopence more than half the money he then had inhis pocket; and to the third person he handed over threepence more thanhalf of what he had left. On entering his house he had only one penny inhis pocket. Now, can you say exactly how much money that gentleman hadon him when he started for home?

9.--THE TWO AEROPLANES.

A man recently bought two aeroplanes, but afterwards found that theywould not answer the purpose for which he wanted them. So he sold themfor L600 each, making a loss of 20 per cent. on one machine and a profitof 20 per cent. on the other. Did he make a profit on the wholetransaction, or a loss? And how much?

10.--BUYING PRESENTS.

"Whom do you think I met in town last week, Brother William?" said UncleBenjamin. "That old skinflint Jorkins. His family had been taking himaround buying Christmas presents. He said to me, 'Why cannot thegovernment abolish Christmas, and make the giving of presents punishableby law? I came out this morning with a certain amount of money in mypocket, and I find I have spent just half of it. In fact, if you willbelieve me, I take home just as many shillings as I had pounds, and halfas many pounds as I had shillings. It is monstrous!'" Can you sayexactly how much money Jorkins had spent on those presents?

11.--THE CYCLISTS' FEAST.

'Twas last Bank Holiday, so I've been told, Some cyclists rode abroad in glorious weather. Resting at noon within a tavern old, They all agreed to have a feast together. "Put it all in one bill, mine host," they said, "For every man an equal share will pay." The bill was promptly on the table laid, And four pounds was the reckoning that day. But, sad to state, when they prepared to square, 'Twas found that two had sneaked outside and fled. So, for two shillings more than his due share Each honest man who had remained was bled. They settled later with those rogues, no doubt. How many were they when they first set out?

12.--A QUEER THING IN MONEY.

It will be found that L66, 6s. 6d. equals 15,918 pence. Now, the four6's added together make 24, and the figures in 15,918 also add to 24. Itis a curious fact that there is only one other sum of money, in pounds,shillings, and pence (all similarly repetitions of one figure), of whichthe digits shall add up the same as the digits of the amount in pence.What is the other sum of money?

13.--A NEW MONEY PUZZLE.

The largest sum of money that can be written in pounds, shillings,pence, and farthings, using each of the nine digits once and only once,is L98,765, 4s. 31/2d. Now, try to discover the smallest sum of moneythat can be written down under precisely the same conditions. There mustbe some value given for each denomination--pounds, shillings, pence,and farthings--and the nought may not be used. It requires just a littlejudgment and thought.

14.--SQUARE MONEY.

"This is queer," said McCrank to his friend. "Twopence added to twopenceis fourpence, and twopence multiplied by twopence is also fourpence." Ofcourse, he was wrong in thinking you can multiply money by money. Themultiplier must be regarded as an abstract number. It is true that twofeet multiplied by two feet will make four square feet. Similarly, twopence multiplied by two pence will produce four square pence! And itwill perplex the reader to say what a "square penny" is. But we willassume for the purposes of our puzzle that twopence multiplied bytwopence is fourpence. Now, what two amounts of money will produce thenext smallest possible result, the same in both cases, when added ormultiplied in this manner? The two amounts need not be alike, but theymust be those that can be paid in current coins of the realm.

15.--POCKET MONEY.

What is the largest sum of money--all in current silver coins and nofour-shilling piece--that I could have in my pocket without being ableto give change for a half-sovereign?

16.--THE MILLIONAIRE'S PERPLEXITY.

Mr. Morgan G. Bloomgarten, the millionaire, known in the States as theClam King, had, for his sins, more money than he knew what to do with.It bored him. So he determined to persecute some of his poor but happyfriends with it. They had never done him any harm, but he resolved toinoculate them with the "source of all evil." He therefore proposed todistribute a million dollars among them and watch them go rapidly to thebad. But he was a man of strange fancies and superstitions, and it wasan inviolable rule with him never to make a gift that was not either onedollar or some power of seven--such as 7, 49, 343, 2,401, which numbersof dollars are produced by simply multiplying sevens together. Anotherrule of his was that he would never give more than six persons exactlythe same sum. Now, how was he to distribute the 1,000,000 dollars? Youmay distribute the money among as many people as you like, under theconditions given.

17.--THE PUZZLING MONEY-BOXES.

Four brothers--named John, William, Charles, and Thomas--had each amoney-box. The boxes were all given to them on the same day, and they atonce put what money they had into them; only, as the boxes were not verylarge, they first changed the money into as few coins as possible. Afterthey had done this, they told one another how much money they had saved,and it was found that if John had had 2s. more in his box than atpresent, if William had had 2s. less, if Charles had had twice as much,and if Thomas had had half as much, they would all have had exactly thesame amount.

Now, when I add that all four boxes together contained 45s., and thatthere were only six coins in all in them, it becomes an entertainingpuzzle to discover just what coins were in each box.

18.--THE MARKET WOMEN.

A number of market women sold their various products at a certain priceper pound (different in every case), and each received the sameamount--2s. 21/2d. What is the greatest number of women there couldhave been? The price per pound in every case must be such as could bepaid in current money.

19.--THE NEW YEAR'S EVE SUPPERS.

The proprietor of a small London cafe has given me some interestingfigures. He says that the ladies who come alone to his place forrefreshment spend each on an average eighteenpence, that theunaccompanied men spend half a crown each, and that when a gentlemanbrings in a lady he spends half a guinea. On New Year's Eve he suppliedsuppers to twenty-five persons, and took five pounds in all. Now,assuming his averages to have held good in every case, how was hiscompany made up on that occasion? Of course, only single gentlemen,single ladies, and pairs (a lady and gentleman) can be supposed to havebeen present, as we are not considering larger parties.

20.--BEEF AND SAUSAGES.

"A neighbour of mine," said Aunt Jane, "bought a certain quantity ofbeef at two shillings a pound, and the same quantity of sausages ateighteenpence a pound. I pointed out to her that if she had divided thesame money equally between beef and sausages she would have gained twopounds in the total weight. Can you tell me exactly how much she spent?"

"Of course, it is no business of mine," said Mrs. Sunniborne; "but alady who could pay such prices must be somewhat inexperienced indomestic economy."

"I quite agree, my dear," Aunt Jane replied, "but you see that is notthe precise point under discussion, any more than the name and morals ofthe tradesman."

21.--A DEAL IN APPLES.

I paid a man a shilling for some apples, but they were so small that Imade him throw in two extra apples. I find that made them cost just apenny a dozen less than the first price he asked. How many apples did Iget for my shilling?

22.--A DEAL IN EGGS.

A man went recently into a dairyman's shop to buy eggs. He wanted themof various qualities. The salesman had new-laid eggs at the high priceof fivepence each, fresh eggs at one penny each, eggs at a halfpennyeach, and eggs for electioneering purposes at a greatly reduced figure,but as there was no election on at the time the buyer had no use for thelast. However, he bought some of each of the three other kinds andobtained exactly one hundred eggs for eight and fourpence. Now, as hebrought away exactly the same number of eggs of two of the threequalities, it is an interesting puzzle to determine just how many hebought at each price.

23.--THE CHRISTMAS-BOXES.

Some years ago a man told me he had spent one hundred English silvercoins in Christmas-boxes, giving every person the same amount, and itcost him exactly L1, 10s. 1d. Can you tell just how many personsreceived the present, and how he could have managed the distribution?That odd penny looks queer, but it is all right.

24.--A SHOPPING PERPLEXITY.

Two ladies went into a shop where, through some curious eccentricity, nochange was given, and made purchases amounting together to less thanfive shillings. "Do you know," said one lady, "I find I shall require nofewer than six current coins of the realm to pay for what I havebought." The other lady considered a moment, and then exclaimed: "By apeculiar coincidence, I am exactly in the same dilemma." "Then we willpay the two bills together." But, to their astonishment, they stillrequired six coins. What is the smallest possible amount of theirpurchases--both different?

25.--CHINESE MONEY.

The Chinese are a curious people, and have strange inverted ways ofdoing things. It is said that they use a saw with an upward pressureinstead of a downward one, that they plane a deal board by pulling thetool toward them instead of pushing it, and that in building a housethey first construct the roof and, having raised that into position,proceed to work downwards. In money the currency of the country consistsof taels of fluctuating value. The tael became thinner and thinner until2,000 of them piled together made less than three inches in height. Thecommon cash consists of brass coins of varying thicknesses, with around, square, or triangular hole in the centre, as in our illustration.

[Illustration]

These are strung on wires like buttons. Supposing that eleven coins withround holes are worth fifteen ching-changs, that eleven with squareholes are worth sixteen ching-changs, and that eleven with triangularholes are worth seventeen ching-changs, how can a Chinaman give mechange for half a crown, using no coins other than the three mentioned?A ching-chang is worth exactly twopence and four-fifteenths of aching-chang.

26.--THE JUNIOR CLERK'S PUZZLE.

Two youths, bearing the pleasant names of Moggs and Snoggs, wereemployed as junior clerks by a merchant in Mincing Lane. They were bothengaged at the same salary--that is, commencing at the rate of L50 ayear, payable half-yearly. Moggs had a yearly rise of L10, and Snoggswas offered the same, only he asked, for reasons that do not concern ourpuzzle, that he might take his rise at L2, 10s. half-yearly, to whichhis employer (not, perhaps, unnaturally!) had no objection.

Now we come to the real point of the puzzle. Moggs put regularly intothe Post Office Savings Bank a certain proportion of his salary, whileSnoggs saved twice as great a proportion of his, and at the end of fiveyears they had together saved L268, 15s. How much had each saved? Thequestion of interest can be ignored.

27.--GIVING CHANGE.

Every one is familiar with the difficulties that frequently arise overthe giving of change, and how the assistance of a third person with afew coins in his pocket will sometimes help us to set the matter right.Here is an example. An Englishman went into a shop in New York andbought goods at a cost of thirty-four cents. The only money he had was adollar, a three-cent piece, and a two-cent piece. The tradesman had onlya half-dollar and a quarter-dollar. But another customer happened to bepresent, and when asked to help produced two dimes, a five-cent piece, atwo-cent piece, and a one-cent piece. How did the tradesman manage togive change? For the benefit of those readers who are not familiar withthe American coinage, it is only necessary to say that a dollar is ahundred cents and a dime ten cents. A puzzle of this kind should rarelycause any difficulty if attacked in a proper manner.

28.--DEFECTIVE OBSERVATION.

Our observation of little things is frequently defective, and ourmemories very liable to lapse. A certain judge recently remarked in acase that he had no recollection whatever of putting the wedding-ring onhis wife's finger. Can you correctly answer these questions withouthaving the coins in sight? On which side of a penny is the date given?Some people are so unobservant that, although they are handling the coinnearly every day of their lives, they are at a loss to answer thissimple question. If I lay a penny flat on the table, how many otherpennies can I place around it, every one also lying flat on the table,so that they all touch the first one? The geometrician will, of course,give the answer at once, and not need to make any experiment. He willalso know that, since all circles are similar, the same answer willnecessarily apply to any coin. The next question is a most interestingone to ask a company, each person writing down his answer on a slip ofpaper, so that no one shall be helped by the answers of others. What isthe greatest number of three-penny-pieces that may be laid flat on thesurface of a half-crown, so that no piece lies on another or overlapsthe surface of the half-crown? It is amazing what a variety of differentanswers one gets to this question. Very few people will be found to givethe correct number. Of course the answer must be given without lookingat the coins.

29.--THE BROKEN COINS.

A man had three coins--a sovereign, a shilling, and a penny--and hefound that exactly the same fraction of each coin had been broken away.Now, assuming that the original intrinsic value of these coins was thesame as their nominal value--that is, that the sovereign was worth apound, the shilling worth a shilling, and the penny worth a penny--whatproportion of each coin has been lost if the value of the threeremaining fragments is exactly one pound?

30.--TWO QUESTIONS IN PROBABILITIES.

There is perhaps no class of puzzle over which people so frequentlyblunder as that which involves what is called the theory ofprobabilities. I will give two simple examples of the sort of puzzle Imean. They are really quite easy, and yet many persons are tripped up bythem. A friend recently produced five pennies and said to me: "Inthrowing these five pennies at the same time, what are the chances thatat least four of the coins will turn up either all heads or all tails?"His own solution was quite wrong, but the correct answer ought not to behard to discover. Another person got a wrong answer to the followinglittle puzzle which I heard him propound: "A man placed three sovereignsand one shilling in a bag. How much should be paid for permission todraw one coin from it?" It is, of course, understood that you are aslikely to draw any one of the four coins as another.

31.--DOMESTIC ECONOMY.

Young Mrs. Perkins, of Putney, writes to me as follows: "I should bevery glad if you could give me the answer to a little sum that has beenworrying me a good deal lately. Here it is: We have only been married ashort time, and now, at the end of two years from the time when we setup housekeeping, my husband tells me that he finds we have spent a thirdof his yearly income in rent, rates, and taxes, one-half in domesticexpenses, and one-ninth in other ways. He has a balance of L190remaining in the bank. I know this last, because he accidentally leftout his pass-book the other day, and I peeped into it. Don't you thinkthat a husband ought to give his wife his entire confidence in his moneymatters? Well, I do; and--will you believe it?--he has never told mewhat his income really is, and I want, very naturally, to find out. Canyou tell me what it is from the figures I have given you?"

Yes; the answer can certainly be given from the figures contained inMrs. Perkins's letter. And my readers, if not warned, will bepractically unanimous in declaring the income to be--something absurdlyin excess of the correct answer!

32.--THE EXCURSION TICKET PUZZLE.

When the big flaming placards were exhibited at the little provincialrailway station, announcing that the Great ---- Company would run cheapexcursion trains to London for the Christmas holidays, the inhabitantsof Mudley-cum-Turmits were in quite a flutter of excitement. Half anhour before the train came in the little booking office was crowded withcountry passengers, all bent on visiting their friends in the greatMetropolis. The booking clerk was unaccustomed to dealing with crowds ofsuch a dimension, and he told me afterwards, while wiping his manlybrow, that what caused him so much trouble was the fact that theserustics paid their fares in such a lot of small money.

He said that he had enough farthings to supply a West End draper withchange for a week, and a sufficient number of threepenny pieces for thecongregations of three parish churches. "That excursion fare," said he,"is nineteen shillings and ninepence, and I should like to know in justhow many different ways it is possible for such an amount to be paid inthe current coin of this realm."

Here, then, is a puzzle: In how many different ways may nineteenshillings and ninepence be paid in our current coin? Remember that thefourpenny-piece is not now current.

33.--PUZZLE IN REVERSALS.

Most people know that if you take any sum of money in pounds, shillings,and pence, in which the number of pounds (less than L12) exceeds that ofthe pence, reverse it (calling the pounds pence and the pence pounds),find the difference, then reverse and add this difference, the result isalways L12, 18s. 11d. But if we omit the condition, "less than L12," andallow nought to represent shillings or pence--(1) What is the lowestamount to which the rule will not apply? (2) What is the highest amountto which it will apply? Of course, when reversing such a sum as L14,15s. 3d. it may be written L3, 16s. 2d., which is the same as L3, 15s.14d.

34.--THE GROCER AND DRAPER.

A country "grocer and draper" had two rival assistants, who pridedthemselves on their rapidity in serving customers. The young man on thegrocery side could weigh up two one-pound parcels of sugar per minute,while the drapery assistant could cut three one-yard lengths of cloth inthe same time. Their employer, one slack day, set them a race, givingthe grocer a barrel of sugar and telling him to weigh up forty-eightone-pound parcels of sugar While the draper divided a roll offorty-eight yards of cloth into yard pieces. The two men wereinterrupted together by customers for nine minutes, but the draper wasdisturbed seventeen times as long as the grocer. What was the result ofthe race?

35.--JUDKINS'S CATTLE.

Hiram B. Judkins, a cattle-dealer of Texas, had five droves of animals,consisting of oxen, pigs, and sheep, with the same number of animals ineach drove. One morning he sold all that he had to eight dealers. Eachdealer bought the same number of animals, paying seventeen dollars foreach ox, four dollars for each pig, and two dollars for each sheep; andHiram received in all three hundred and one dollars. What is thegreatest number of animals he could have had? And how many would therebe of each kind?

36.--BUYING APPLES.

As the purchase of apples in small quantities has always presentedconsiderable difficulties, I think it well to offer a few remarks onthis subject. We all know the story of the smart boy who, on being toldby the old woman that she was selling her apples at four for threepence,said: "Let me see! Four for threepence; that's three for twopence, twofor a penny, one for nothing--I'll take _one_!"

There are similar cases of perplexity. For example, a boy once picked upa penny apple from a stall, but when he learnt that the woman's pearswere the same price he exchanged it, and was about to walk off. "Stop!"said the woman. "You haven't paid me for the pear!" "No," said the boy,"of course not. I gave you the apple for it." "But you didn't pay forthe apple!" "Bless the woman! You don't expect me to pay for the appleand the pear too!" And before the poor creature could get out of thetangle the boy had disappeared.

Then, again, we have the case of the man who gave a boy sixpence andpromised to repeat the gift as soon as the youngster had made it intoninepence. Five minutes later the boy returned. "I have made it intoninepence," he said, at the same time handing his benefactor threepence."How do you make that out?" he was asked. "I bought threepennyworth ofapples." "But that does not make it into ninepence!" "I should ratherthink it did," was the boy's reply. "The apple woman has threepence,hasn't she? Very well, I have threepennyworth of apples, and I have justgiven you the other threepence. What's that but ninepence?"

I cite these cases just to show that the small boy really stands in needof a little instruction in the art of buying apples. So I will give asimple poser dealing with this branch of commerce.

An old woman had apples of three sizes for sale--one a penny, two apenny, and three a penny. Of course two of the second size and three ofthe third size were respectively equal to one apple of the largest size.Now, a gentleman who had an equal number of boys and girls gave hischildren sevenpence to be spent amongst them all on these apples. Thepuzzle is to give each child an equal distribution of apples. How wasthe sevenpence spent, and how many children were there?

37.--BUYING CHESTNUTS.

Though the following little puzzle deals with the purchase of chestnuts,it is not itself of the "chestnut" type. It is quite new. At first sightit has certainly the appearance of being of the "nonsense puzzle"character, but it is all right when properly considered.

A man went to a shop to buy chestnuts. He said he wanted a pennyworth,and was given five chestnuts. "It is not enough; I ought to have asixth," he remarked! "But if I give you one chestnut more." the shopmanreplied, "you will have five too many." Now, strange to say, they wereboth right. How many chestnuts should the buyer receive for half acrown?

38.--THE BICYCLE THIEF.

Here is a little tangle that is perpetually cropping up in variousguises. A cyclist bought a bicycle for L15 and gave in payment a chequefor L25. The seller went to a neighbouring shopkeeper and got him tochange the cheque for him, and the cyclist, having received his L10change, mounted the machine and disappeared. The cheque proved to bevalueless, and the salesman was requested by his neighbour to refund theamount he had received. To do this, he was compelled to borrow the L25from a friend, as the cyclist forgot to leave his address, and could notbe found. Now, as the bicycle cost the salesman L11, how much money didhe lose altogether?

39.--THE COSTERMONGER'S PUZZLE.

"How much did yer pay for them oranges, Bill?"

"I ain't a-goin' to tell yer, Jim. But I beat the old cove downfourpence a hundred."

"What good did that do yer?"

"Well, it meant five more oranges on every ten shillin's-worth."

Now, what price did Bill actually pay for the oranges? There is only onerate that will fit in with his statements.

AGE AND KINSHIP PUZZLES.

"The days of our years are threescore years and ten."

--_Psalm_ xc. 10.

For centuries it has been a favourite method of propounding arithmeticalpuzzles to pose them in the form of questions as to the age of anindividual. They generally lend themselves to very easy solution by theuse of algebra, though often the difficulty lies in stating themcorrectly. They may be made very complex and may demand considerableingenuity, but no general laws can well be laid down for their solution.The solver must use his own sagacity. As for puzzles in relationship orkinship, it is quite curious how bewildering many people find thesethings. Even in ordinary conversation, some statement as torelationship, which is quite clear in the mind of the speaker, willimmediately tie the brains of other people into knots. Such expressionsas "He is my uncle's son-in-law's sister" convey absolutely nothing tosome people without a detailed and laboured explanation. In such casesthe best course is to sketch a brief genealogical table, when the eyecomes immediately to the assistance of the brain. In these days, when wehave a growing lack of respect for pedigrees, most people have got outof the habit of rapidly drawing such tables, which is to be regretted,as they would save a lot of time and brain racking on occasions.

40.--MAMMA'S AGE.

Tommy: "How old are you, mamma?"

Mamma: "Let me think, Tommy. Well, our three ages add up to exactlyseventy years."

Tommy: "That's a lot, isn't it? And how old are you, papa?"

Papa: "Just six times as old as you, my son."

Tommy: "Shall I ever be half as old as you, papa?"

Papa: "Yes, Tommy; and when that happens our three ages will add up toexactly twice as much as to-day."

Tommy: "And supposing I was born before you, papa; and supposing mammahad forgot all about it, and hadn't been at home when I came; andsupposing--"

Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'llhave a headache."

Now, if Tommy had been some years older he might have calculated theexact ages of his parents from the information they had given him. Canyou find out the exact age of mamma?

41.--THEIR AGES.

"My husband's age," remarked a lady the other day, "is represented bythe figures of my own age reversed. He is my senior, and the differencebetween our ages is one-eleventh of their sum."

42.--THE FAMILY AGES.

When the Smileys recently received a visit from the favourite uncle, thefond parents had all the five children brought into his presence. Firstcame Billie and little Gertrude, and the uncle was informed that the boywas exactly twice as old as the girl. Then Henrietta arrived, and it waspointed out that the combined ages of herself and Gertrude equalledtwice the age of Billie. Then Charlie came running in, and somebodyremarked that now the combined ages of the two boys were exactly twicethe combined ages of the two girls. The uncle was expressing hisastonishment at these coincidences when Janet came in. "Ah! uncle," sheexclaimed, "you have actually arrived on my twenty-first birthday!" Tothis Mr. Smiley added the final staggerer: "Yes, and now the combinedages of the three girls are exactly equal to twice the combined ages ofthe two boys." Can you give the age of each child?

43.--MRS. TIMPKINS'S AGE.

Edwin: "Do you know, when the Timpkinses married eighteen years agoTimpkins was three times as old as his wife, and to-day he is just twiceas old as she?"

Angelina: "Then how old was Mrs. Timpkins on the wedding day?"

Can you answer Angelina's question?

44--A CENSUS PUZZLE.

Mr. and Mrs. Jorkins have fifteen children, all born at intervals of oneyear and a half. Miss Ada Jorkins, the eldest, had an objection to stateher age to the census man, but she admitted that she was just seventimes older than little Johnnie, the youngest of all. What was Ada'sage? Do not too hastily assume that you have solved this little poser.You may find that you have made a bad blunder!

45.--MOTHER AND DAUGHTER.

"Mother, I wish you would give me a bicycle," said a girl of twelve theother day.

"I do not think you are old enough yet, my dear," was the reply. "When Iam only three times as old as you are you shall have one."

Now, the mother's age is forty-five years. When may the young ladyexpect to receive her present?

46.--MARY AND MARMADUKE.

Marmaduke: "Do you know, dear, that in seven years' time our combinedages will be sixty-three years?"

Mary: "Is that really so? And yet it is a fact that when you were mypresent age you were twice as old as I was then. I worked it out lastnight."

Now, what are the ages of Mary and Marmaduke?

47--ROVER'S AGE.

"Now, then, Tommy, how old is Rover?" Mildred's young man asked herbrother.

"Well, five years ago," was the youngster's reply, "sister was fourtimes older than the dog, but now she is only three times as old."

Can you tell Rover's age?

48.--CONCERNING TOMMY'S AGE.

Tommy Smart was recently sent to a new school. On the first day of hisarrival the teacher asked him his age, and this was his curious reply:"Well, you see, it is like this. At the time I was born--I forget theyear--my only sister, Ann, happened to be just one-quarter the age ofmother, and she is now one-third the age of father." "That's all verywell," said the teacher, "but what I want is not the age of your sisterAnn, but your own age." "I was just coming to that," Tommy answered; "Iam just a quarter of mother's present age, and in four years' time Ishall be a quarter the age of father. Isn't that funny?"

This was all the information that the teacher could get out of TommySmart. Could you have told, from these facts, what was his precise age?It is certainly a little puzzling.

49.--NEXT-DOOR NEIGHBOURS.

There were two families living next door to one another at TootingBec--the Jupps and the Simkins. The united ages of the four Juppsamounted to one hundred years, and the united ages of the Simkins alsoamounted to the same. It was found in the case of each family that thesum obtained by adding the squares of each of the children's ages to thesquare of the mother's age equalled the square of the father's age. Inthe case of the Jupps, however, Julia was one year older than herbrother Joe, whereas Sophy Simkin was two years older than her brotherSammy. What was the age of each of the eight individuals?

50.--THE BAG OF NUTS.

Three boys were given a bag of nuts as a Christmas present, and it wasagreed that they should be divided in proportion to their ages, whichtogether amounted to 171/2 years. Now the bag contained 770 nuts, andas often as Herbert took four Robert took three, and as often as Herberttook six Christopher took seven. The puzzle is to find out how many nutseach had, and what were the boys' respective ages.

51.--HOW OLD WAS MARY?

Here is a funny little age problem, by the late Sam Loyd, which has beenvery popular in the United States. Can you unravel the mystery?

The combined ages of Mary and Ann are forty-four years, and Mary istwice as old as Ann was when Mary was half as old as Ann will be whenAnn is three times as old as Mary was when Mary was three times as oldas Ann. How old is Mary? That is all, but can you work it out? If not,ask your friends to help you, and watch the shadow of bewilderment creepover their faces as they attempt to grip the intricacies of thequestion.

52.--QUEER RELATIONSHIPS.

"Speaking of relationships," said the Parson at a certain dinner-party,"our legislators are getting the marriage law into a frightful tangle,Here, for example, is a puzzling case that has come under my notice. Twobrothers married two sisters. One man died and the other man's wife alsodied. Then the survivors married."

"The man married his deceased wife's sister under the recent Act?" putin the Lawyer.

"Exactly. And therefore, under the civil law, he is legally married andhis child is legitimate. But, you see, the man is the woman's deceasedhusband's brother, and therefore, also under the civil law, she is notmarried to him and her child is illegitimate."

"He is married to her and she is not married to him!" said the Doctor.

"Quite so. And the child is the legitimate son of his father, but theillegitimate son of his mother."

"Undoubtedly 'the law is a hass,'" the Artist exclaimed, "if I may bepermitted to say so," he added, with a bow to the Lawyer.

"Certainly," was the reply. "We lawyers try our best to break in thebeast to the service of man. Our legislators are responsible for thebreed."

"And this reminds me," went on the Parson, "of a man in my parish whomarried the sister of his widow. This man--"

"Stop a moment, sir," said the Professor. "Married the sister of hiswidow? Do you marry dead men in your parish?"

"No; but I will explain that later. Well, this man has a sister of hisown. Their names are Stephen Brown and Jane Brown. Last week a youngfellow turned up whom Stephen introduced to me as his nephew. Naturally,I spoke of Jane as his aunt, but, to my astonishment, the youthcorrected me, assuring me that, though he was the nephew of Stephen, hewas not the nephew of Jane, the sister of Stephen. This perplexed me agood deal, but it is quite correct."

The Lawyer was the first to get at the heart of the mystery. What washis solution?

53.--HEARD ON THE TUBE RAILWAY.

First Lady: "And was he related to you, dear?"

Second Lady: "Oh, yes. You see, that gentleman's mother was my mother'smother-in-law, but he is not on speaking terms with my papa."

First Lady: "Oh, indeed!" (But you could see that she was not muchwiser.)

How was the gentleman related to the Second Lady?

54.--A FAMILY PARTY.

A certain family party consisted of 1 grandfather, 1 grandmother, 2fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2sons, 2 daughters, 1 father-in-law, 1 mother-in-law, and 1daughter-in-law. Twenty-three people, you will say. No; there were onlyseven persons present. Can you show how this might be?

55.--A MIXED PEDIGREE.

Joseph Bloggs: "I can't follow it, my dear boy. It makes me dizzy!"

John Snoggs: "It's very simple. Listen again! You happen to be myfather's brother-in-law, my brother's father-in-law, and also myfather-in-law's brother. You see, my father was--"

But Mr. Bloggs refused to hear any more. Can the reader show how thisextraordinary triple relationship might have come about?

56.--WILSON'S POSER.

"Speaking of perplexities--" said Mr. Wilson, throwing down a magazineon the table in the commercial room of the Railway Hotel.

"Who was speaking of perplexities?" inquired Mr. Stubbs.

"Well, then, reading about them, if you want to be exact--it justoccurred to me that perhaps you three men may be interested in a littlematter connected with myself."

It was Christmas Eve, and the four commercial travellers were spendingthe holiday at Grassminster. Probably each suspected that the others hadno homes, and perhaps each was conscious of the fact that he was in thatpredicament himself. In any case they seemed to be perfectlycomfortable, and as they drew round the cheerful fire the conversationbecame general.

"What is the difficulty?" asked Mr. Packhurst.

"There's no difficulty in the matter, when you rightly understand it. Itis like this. A man named Parker had a flying-machine that would carrytwo. He was a venturesome sort of chap--reckless, I should call him--andhe had some bother in finding a man willing to risk his life in makingan ascent with him. However, an uncle of mine thought he would chanceit, and one fine morning he took his seat in the machine and she startedoff well. When they were up about a thousand feet, my nephewsuddenly--"

"Here, stop, Wilson! What was your nephew doing there? You said youruncle," interrupted Mr. Stubbs.

"Did I? Well, it does not matter. My nephew suddenly turned to Parkerand said that the engine wasn't running well, so Parker called out to myuncle--"

"Look here," broke in Mr. Waterson, "we are getting mixed. Was it youruncle or your nephew? Let's have it one way or the other."

"What I said is quite right. Parker called out to my uncle to dosomething or other, when my nephew--"

"There you are again, Wilson," cried Mr. Stubbs; "once for all, are weto understand that both your uncle and your nephew were on the machine?"

"Certainly. I thought I made that clear. Where was I? Well, my nephewshouted back to Parker--"

"Phew! I'm sorry to interrupt you again, Wilson, but we can't get onlike this. Is it true that the machine would only carry two?"

"Of course. I said at the start that it only carried two."

"Then what in the name of aerostation do you mean by saying that therewere three persons on board?" shouted Mr. Stubbs.

"Who said there were three?"

"You have told us that Parker, your uncle, and your nephew went up onthis blessed flying-machine."

"That's right."

"And the thing would only carry two!"

"Right again."

"Wilson, I have known you for some time as a truthful man and atemperate man," said Mr. Stubbs, solemnly. "But I am afraid since youtook up that new line of goods you have overworked yourself."

"Half a minute, Stubbs," interposed Mr. Waterson. "I see clearly wherewe all slipped a cog. Of course, Wilson, you meant us to understand thatParker is either your uncle or your nephew. Now we shall be all right ifyou will just tell us whether Parker is your uncle or nephew."

"He is no relation to me whatever."

The three men sighed and looked anxiously at one another. Mr. Stubbs gotup from his chair to reach the matches, Mr. Packhurst proceeded to windup his watch, and Mr. Waterson took up the poker to attend to the fire.It was an awkward moment, for at the season of goodwill nobody wished totell Mr. Wilson exactly what was in his mind.

"It's curious," said Mr. Wilson, very deliberately, "and it's rathersad, how thick-headed some people are. You don't seem to grip the facts.It never seems to have occurred to either of you that my uncle and mynephew are one and the same man."

"What!" exclaimed all three together.

"Yes; David George Linklater is my uncle, and he is also my nephew.Consequently, I am both his uncle and nephew. Queer, isn't it? I'llexplain how it comes about."

Mr. Wilson put the case so very simply that the three men saw how itmight happen without any marriage within the prohibited degrees. Perhapsthe reader can work it out for himself.

CLOCK PUZZLES.

"Look at the clock!"

_Ingoldsby Legends_.

In considering a few puzzles concerning clocks and watches, and thetimes recorded by their hands under given conditions, it is well that aparticular convention should always be kept in mind. It is frequentlythe case that a solution requires the assumption that the hands canactually record a time involving a minute fraction of a second. Such atime, of course, cannot be really indicated. Is the puzzle, therefore,impossible of solution? The conclusion deduced from a logical syllogismdepends for its truth on the two premises assumed, and it is the same inmathematics. Certain things are antecedently assumed, and the answerdepends entirely on the truth of those assumptions.

"If two horses," says Lagrange, "can pull a load of a certain weight, itis natural to suppose that four horses could pull a load of double thatweight, six horses a load of three times that weight. Yet, strictlyspeaking, such is not the case. For the inference is based on theassumption that the four horses pull alike in amount and direction,which in practice can scarcely ever be the case. It so happens that weare frequently led in our reckonings to results which diverge widelyfrom reality. But the fault is not the fault of mathematics; formathematics always gives back to us exactly what we have put into it.The ratio was constant according to that supposition. The result isfounded upon that supposition. If the supposition is false the result isnecessarily false."

If one man can reap a field in six days, we say two men will reap it inthree days, and three men will do the work in two days. We here assume,as in the case of Lagrange's horses, that all the men are exactlyequally capable of work. But we assume even more than this. For whenthree men get together they may waste time in gossip or play; or, on theother hand, a spirit of rivalry may spur them on to greater diligence.We may assume any conditions we like in a problem, provided they beclearly expressed and understood, and the answer will be in accordancewith those conditions.

57.--WHAT WAS THE TIME?

"I say, Rackbrane, what is the time?" an acquaintance asked our friendthe professor the other day. The answer was certainly curious.

"If you add one quarter of the time from noon till now to half the timefrom now till noon to-morrow, you will get the time exactly."

What was the time of day when the professor spoke?

58.--A TIME PUZZLE.

How many minutes is it until six o'clock if fifty minutes ago it wasfour times as many minutes past three o'clock?

59.--A PUZZLING WATCH.

A friend pulled out his watch and said, "This watch of mine does notkeep perfect time; I must have it seen to. I have noticed that theminute hand and the hour hand are exactly together every sixty-fiveminutes." Does that watch gain or lose, and how much per hour?

60.--THE WAPSHAW'S WHARF MYSTERY.

There was a great commotion in Lower Thames Street on the morning ofJanuary 12, 1887. When the early members of the staff arrived atWapshaw's Wharf they found that the safe had been broken open, aconsiderable sum of money removed, and the offices left in greatdisorder. The night watchman was nowhere to be found, but nobody who hadbeen acquainted with him for one moment suspected him to be guilty ofthe robbery. In this belief the proprietors were confirmed when, laterin the day, they were informed that the poor fellow's body had beenpicked up by the River Police. Certain marks of violence pointed to thefact that he had been brutally attacked and thrown into the river. Awatch found in his pocket had stopped, as is invariably the case in suchcircumstances, and this was a valuable clue to the time of the outrage.But a very stupid officer (and we invariably find one or two stupidindividuals in the most intelligent bodies of men) had actually amusedhimself by turning the hands round and round, trying to set the watchgoing again. After he had been severely reprimanded for this seriousindiscretion, he was asked whether he could remember the time that wasindicated by the watch when found. He replied that he could not, but herecollected that the hour hand and minute hand were exactly together,one above the other, and the second hand had just passed the forty-ninthsecond. More than this he could not remember.

What was the exact time at which the watchman's watch stopped? The watchis, of course, assumed to have been an accurate one.

61.--CHANGING PLACES.

[Illustration]

The above clock face indicates a little before 42 minutes past 4. Thehands will again point at exactly the same spots a little after 23minutes past 8. In fact, the hands will have changed places. How manytimes do the hands of a clock change places between three o'clock p.m.and midnight? And out of all the pairs of times indicated by thesechanges, what is the exact time when the minute hand will be nearest tothe point IX?

62.--THE CLUB CLOCK.

One of the big clocks in the Cogitators' Club was found the other nightto have stopped just when, as will be seen in the illustration, thesecond hand was exactly midway between the other two hands. One of themembers proposed to some of his friends that they should tell him theexact time when (if the clock had not stopped) the second hand wouldnext again have been midway between the minute hand and the hour hand.Can you find the correct time that it would happen?

[Illustration]

63.--THE STOP-WATCH.

[Illustration]

We have here a stop-watch with three hands. The second hand, whichtravels once round the face in a minute, is the one with the little ringat its end near the centre. Our dial indicates the exact time when itsowner stopped the watch. You will notice that the three hands are nearlyequidistant. The hour and minute hands point to spots that are exactly athird of the circumference apart, but the second hand is a little tooadvanced. An exact equidistance for the three hands is not possible.Now, we want to know what the time will be when the three hands are nextat exactly the same distances as shown from one another. Can you statethe time?

64.--THE THREE CLOCKS.

On Friday, April 1, 1898, three new clocks were all set going preciselyat the same time--twelve noon. At noon on the following day it was foundthat clock A had kept perfect time, that clock B had gained exactly oneminute, and that clock C had lost exactly one minute. Now, supposingthat the clocks B and C had not been regulated, but all three allowed togo on as they had begun, and that they maintained the same rates ofprogress without stopping, on what date and at what time of day wouldall three pairs of hands again point at the same moment at twelveo'clock?

65.--THE RAILWAY STATION CLOCK.

A clock hangs on the wall of a railway station, 71 ft. 9 in. long and 10ft. 4 in. high. Those are the dimensions of the wall, not of the clock!While waiting for a train we noticed that the hands of the clock werepointing in opposite directions, and were parallel to one of thediagonals of the wall. What was the exact time?

66.--THE VILLAGE SIMPLETON.

A facetious individual who was taking a long walk in the country cameupon a yokel sitting on a stile. As the gentleman was not quite sure ofhis road, he thought he would make inquiries of the local inhabitant;but at the first glance he jumped too hastily to the conclusion that hehad dropped on the village idiot. He therefore decided to test thefellow's intelligence by first putting to him the simplest question hecould think of, which was, "What day of the week is this, my good man?"The following is the smart answer that he received:--

"When the day after to-morrow is yesterday, to-day will be as far fromSunday as to-day was from Sunday when the day before yesterday wasto-morrow."

Can the reader say what day of the week it was? It is pretty evidentthat the countryman was not such a fool as he looked. The gentleman wenton his road a puzzled but a wiser man.

LOCOMOTION AND SPEED PUZZLES.

"The race is not to the swift."--_Ecclesiastes_ ix. II.

67.--AVERAGE SPEED.

In a recent motor ride it was found that we had gone at the rate of tenmiles an hour, but we did the return journey over the same route, owingto the roads being more clear of traffic, at fifteen miles an hour. Whatwas our average speed? Do not be too hasty in your answer to this simplelittle question, or it is pretty certain that you will be wrong.

68.--THE TWO TRAINS.

I put this little question to a stationmaster, and his correct answerwas so prompt that I am convinced there is no necessity to seek talentedrailway officials in America or elsewhere.

Two trains start at the same time, one from London to Liverpool, theother from Liverpool to London. If they arrive at their destinations onehour and four hours respectively after passing one another, how muchfaster is one train running than the other?

69.--THE THREE VILLAGES.

I set out the other day to ride in a motor-car from Acrefield toButterford, but by mistake I took the road going _via_ Cheesebury, whichis nearer Acrefield than Butterford, and is twelve miles to the left ofthe direct road I should have travelled. After arriving at Butterford Ifound that I had gone thirty-five miles. What are the three distancesbetween these villages, each being a whole number of miles? I maymention that the three roads are quite straight.

70.--DRAWING HER PENSION.

"Speaking of odd figures," said a gentleman who occupies some post in aGovernment office, "one of the queerest characters I know is an old lamewidow who climbs up a hill every week to draw her pension at the villagepost office. She crawls up at the rate of a mile and a half an hour andcomes down at the rate of four and a half miles an hour, so that ittakes her just six hours to make the double journey. Can any of you tellme how far it is from the bottom of the hill to the top?"

[Illustration]

71.--SIR EDWYN DE TUDOR.

In the illustration we have a sketch of Sir Edwyn de Tudor going torescue his lady-love, the fair Isabella, who was held a captive by aneighbouring wicked baron. Sir Edwyn calculated that if he rode fifteenmiles an hour he would arrive at the castle an hour too soon, while ifhe rode ten miles an hour he would get there just an hour too late. Now,it was of the first importance that he should arrive at the exact timeappointed, in order that the rescue that he had planned should be asuccess, and the time of the tryst was five o'clock, when the captivelady would be taking her afternoon tea. The puzzle is to discoverexactly how far Sir Edwyn de Tudor had to ride.

72.--THE HYDROPLANE QUESTION.

The inhabitants of Slocomb-on-Sea were greatly excited over the visit ofa certain flying man. All the town turned out to see the flight of thewonderful hydroplane, and, of course, Dobson and his family were there.Master Tommy was in good form, and informed his father that Englishmenmade better airmen than Scotsmen and Irishmen because they are not soheavy. "How do you make that out?" asked Mr. Dobson. "Well, you see,"Tommy replied, "it is true that in Ireland there are men of Cork and inScotland men of Ayr, which is better still, but in England there arelightermen." Unfortunately it had to be explained to Mrs. Dobson, andthis took the edge off the thing. The hydroplane flight was from Slocombto the neighbouring watering-place Poodleville--five miles distant. Butthere was a strong wind, which so helped the airman that he made theoutward journey in the short time of ten minutes, though it took him anhour to get back to the starting point at Slocomb, with the wind deadagainst him. Now, how long would the ten miles have taken him if therehad been a perfect calm? Of course, the hydroplane's engine workeduniformly throughout.

73.--DONKEY RIDING.

During a visit to the seaside Tommy and Evangeline insisted on having adonkey race over the mile course on the sands. Mr. Dobson and some ofhis friends whom he had met on the beach acted as judges, but, as thedonkeys were familiar acquaintances and declined to part company thewhole way, a dead heat was unavoidable. However, the judges, beingstationed at different points on the course, which was marked off inquarter-miles, noted the following results:--The first three-quarterswere run in six and three-quarter minutes, the first half-mile took thesame time as the second half, and the third quarter was run in exactlythe same time as the last quarter. From these results Mr. Dobson amusedhimself in discovering just how long it took those two donkeys to runthe whole mile. Can you give the answer?

74.--THE BASKET OF POTATOES.

A man had a basket containing fifty potatoes. He proposed to his son, asa little recreation, that he should place these potatoes on the groundin a straight line. The distance between the first and second potatoeswas to be one yard, between the second and third three yards, betweenthe third and fourth five yards, between the fourth and fifth sevenyards, and so on--an increase of two yards for every successive potatolaid down. Then the boy was to pick them up and put them in the basketone at a time, the basket being placed beside the first potato. How farwould the boy have to travel to accomplish the feat of picking them allup? We will not consider the journey involved in placing the potatoes,so that he starts from the basket with them all laid out.

75.--THE PASSENGER'S FARE.

At first sight you would hardly think there was matter for dispute inthe question involved in the following little incident, yet it took thetwo persons concerned some little time to come to an agreement. Mr.Smithers hired a motor-car to take him from Addleford to Clinkervilleand back again for L3. At Bakenham, just midway, he picked up anacquaintance, Mr. Tompkins, and agreed to take him on to Clinkervilleand bring him back to Bakenham on the return journey. How much should hehave charged the passenger? That is the question. What was a reasonablefare for Mr. Tompkins?

DIGITAL PUZZLES.

"Nine worthies were they called." DRYDEN: _The Flower and the Leaf._

I give these puzzles, dealing with the nine digits, a class tothemselves, because I have always thought that they deserve moreconsideration than they usually receive. Beyond the mere trick of"casting out nines," very little seems to be generally known of the lawsinvolved in these problems, and yet an acquaintance with the propertiesof the digits often supplies, among other uses, a certain number ofarithmetical checks that are of real value in the saving of labour. Letme give just one example--the first that occurs to me.

If the reader were required to determine whether or not15,763,530,163,289 is a square number, how would he proceed? If thenumber had ended with a 2, 3, 7, or 8 in the digits place, of course hewould know that it could not be a square, but there is nothing in itsapparent form to prevent its being one. I suspect that in such a case hewould set to work, with a sigh or a groan, at the laborious task ofextracting the square root. Yet if he had given a little attention tothe study of the digital properties of numbers, he would settle thequestion in this simple way. The sum of the digits is 59, the sum ofwhich is 14, the sum of which is 5 (which I call the "digital root"),and therefore I know that the number cannot be a square, and for thisreason. The digital root of successive square numbers from 1 upwards isalways 1, 4, 7, or 9, and can never be anything else. In fact, theseries, 1, 4, 9, 7, 7, 9, 4, 1, 9, is repeated into infinity. Theanalogous series for triangular numbers is 1, 3, 6, 1, 6, 3, 1, 9, 9. Sohere we have a similar negative check, for a number cannot be triangular(that is, (n squared + n)/2) if its digital root be 2, 4, 5, 7, or 8.

76.--THE BARREL OF BEER.

A man bought an odd lot of wine in barrels and one barrel containingbeer. These are shown in the illustration, marked with the number ofgallons that each barrel contained. He sold a quantity of the wine toone man and twice the quantity to another, but kept the beer to himself.The puzzle is to point out which barrel contains beer. Can you say whichone it is? Of course, the man sold the barrels just as he bought them,without manipulating in any way the contents.

[Illustration:

( 15 Gals )

(31 Gals) (19 Gals)

(20 Gals) (16 Gals) (18 Gals)

]

77.--DIGITS AND SQUARES.

[Illustration:

+---+---+---+ | 1 | 9 | 2 | +---+---+---+ | 3 | 8 | 4 | +---+---+---+ | 5 | 7 | 6 | +---+---+---+

]

It will be seen in the diagram that we have so arranged the nine digitsin a square that the number in the second row is twice that in the firstrow, and the number in the bottom row three times that in the top row.There are three other ways of arranging the digits so as to produce thesame result. Can you find them?

78.--ODD AND EVEN DIGITS.

The odd digits, 1, 3, 5, 7, and 9, add up 25, while the even figures, 2,4, 6, and 8, only add up 20. Arrange these figures so that the odd onesand the even ones add up alike. Complex and improper fractions andrecurring decimals are not allowed.

79.--THE LOCKERS PUZZLE.

[Illustration:

A B C ================== ================== ================== | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | ================== ================== ================== | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | ------------------ ------------------ ------------------

]

A man had in his office three cupboards, each containing nine lockers,as shown in the diagram. He told his clerk to place a differentone-figure number on each locker of cupboard A, and to do the same inthe case of B, and of C. As we are here allowed to call nought a digit,and he was not prohibited from using nought as a number, he clearly hadthe option of omitting any one of ten digits from each cupboard.

Now, the employer did not say the lockers were to be numbered in anynumerical order, and he was surprised to find, when the work was done,that the figures had apparently been mixed up indiscriminately. Callingupon his clerk for an explanation, the eccentric lad stated that thenotion had occurred to him so to arrange the figures that in each casethey formed a simple addition sum, the two upper rows of figuresproducing the sum in the lowest row. But the most surprising point wasthis: that he had so arranged them that the addition in A gave thesmallest possible sum, that the addition in C gave the largest possiblesum, and that all the nine digits in the three totals were different.The puzzle is to show how this could be done. No decimals are allowedand the nought may not appear in the hundreds place.

80.--THE THREE GROUPS.

There appeared in "Nouvelles Annales de Mathematiques" the followingpuzzle as a modification of one of my "Canterbury Puzzles." Arrange thenine digits in three groups of two, three, and four digits, so that thefirst two numbers when multiplied together make the third. Thus, 12 x483 = 5,796. I now also propose to include the cases where there areone, four, and four digits, such as 4 x 1,738 = 6,952. Can you find allthe possible solutions in both cases?

81.--THE NINE COUNTERS.

[Illustration:

(1)(5)(8) (7)(9) (2)(3) (4)(6)

]

I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4,5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shownin the illustration, so as to form two multiplication sums, and foundthat both sums gave the same product. You will find that 158 multipliedby 23 is 3,634, and that 79 multiplied by 46 is also 3,634. Now, thepuzzle I propose is to rearrange the counters so as to get as large aproduct as possible. What is the best way of placing them? Remember bothgroups must multiply to the same amount, and there must be threecounters multiplied by two in one case, and two multiplied by twocounters in the other, just as at present.

82.--THE TEN COUNTERS.

In this case we use the nought in addition to the 1, 2, 3, 4, 5, 6, 7,8, 9. The puzzle is, as in the last case, so to arrange the ten countersthat the products of the two multiplications shall be the same, and youmay here have one or more figures in the multiplier, as you choose. Theabove is a very easy feat; but it is also required to find the twoarrangements giving pairs of the highest and lowest products possible.Of course every counter must be used, and the cipher may not be placedto the left of a row of figures where it would have no effect. Vulgarfractions or decimals are not allowed.

83.--DIGITAL MULTIPLICATION.

Here is another entertaining problem with the nine digits, the noughtbeing excluded. Using each figure once, and only once, we can form twomultiplication sums that have the same product, and this may be done inmany ways. For example, 7 x 658 and 14 x 329 contain all the digitsonce, and the product in each case is the same--4,606. Now, it will beseen that the sum of the digits in the product is 16, which is neitherthe highest nor the lowest sum so obtainable. Can you find the solutionof the problem that gives the lowest possible sum of digits in thecommon product? Also that which gives the highest possible sum?

84.--THE PIERROT'S PUZZLE.

[Illustration]

The Pierrot in the illustration is standing in a posture that representsthe sign of multiplication. He is indicating the peculiar fact that 15multiplied by 93 produces exactly the same figures (1,395), differentlyarranged. The puzzle is to take any four digits you like (all different)and similarly arrange them so that the number formed on one side of thePierrot when multiplied by the number on the other side shall producethe same figures. There are very few ways of doing it, and I shall giveall the cases possible. Can you find them all? You are allowed to puttwo figures on each side of the Pierrot as in the example shown, or toplace a single figure on one side and three figures on the other. If weonly used three digits instead of four, the only possible ways arethese: 3 multiplied by 51 equals 153, and 6 multiplied by 21 equals 126.

85.--THE CAB NUMBERS.

A London policeman one night saw two cabs drive off in oppositedirections under suspicious circumstances. This officer was aparticularly careful and wide-awake man, and he took out his pocket-bookto make an entry of the numbers of the cabs, but discovered that he hadlost his pencil. Luckily, however, he found a small piece of chalk, withwhich he marked the two numbers on the gateway of a wharf close by. Whenhe returned to the same spot on his beat he stood and looked again atthe numbers, and noticed this peculiarity, that all the nine digits (nonought) were used and that no figure was repeated, but that if hemultiplied the two numbers together they again produced the nine digits,all once, and once only. When one of the clerks arrived at the wharf inthe early morning, he observed the chalk marks and carefully rubbed themout. As the policeman could not remember them, certain mathematicianswere then consulted as to whether there was any known method fordiscovering all the pairs of numbers that have the peculiarity that theofficer had noticed; but they knew of none. The investigation, however,was interesting, and the following question out of many was proposed:What two numbers, containing together all the nine digits, will, whenmultiplied together, produce another number (the _highest possible_)containing also all the nine digits? The nought is not allowed anywhere.

86.--QUEER MULTIPLICATION.

If I multiply 51,249,876 by 3 (thus using all the nine digits once, andonce only), I get 153,749,628 (which again contains all the nine digitsonce). Similarly, if I multiply 16,583,742 by 9 the result is149,253,678, where in each case all the nine digits are used. Now, take6 as your multiplier and try to arrange the remaining eight digits so asto produce by multiplication a number containing all nine once, and onceonly. You will find it far from easy, but it can be done.

87.--THE NUMBER-CHECKS PUZZLE.

[Illustration]

Where a large number of workmen are employed on a building it iscustomary to provide every man with a little disc bearing his number.These are hung on a board by the men as they arrive, and serve as acheck on punctuality. Now, I once noticed a foreman remove a number ofthese checks from his board and place them on a split-ring which hecarried in his pocket. This at once gave me the idea for a good puzzle.In fact, I will confide to my readers that this is just how ideas forpuzzles arise. You cannot really create an idea: it happens--and youhave to be on the alert to seize it when it does so happen.

It will be seen from the illustration that there are ten of thesechecks on a ring, numbered 1 to 9 and 0. The puzzle is to divide theminto three groups without taking any off the ring, so that the firstgroup multiplied by the second makes the third group. For example, wecan divide them into the three groups, 2--8 9 0 7--1 5 4 6 3, bybringing the 6 and the 3 round to the 4, but unfortunately the firsttwo when multiplied together do not make the third. Can you separatethem correctly? Of course you may have as many of the checks as youlike in any group. The puzzle calls for some ingenuity, unless youhave the luck to hit on the answer by chance.

88.--DIGITAL DIVISION.

It is another good puzzle so to arrange the nine digits (the noughtexcluded) into two groups so that one group when divided by the otherproduces a given number without remainder. For example, 1 3 4 5 8divided by 6 7 2 9 gives 2. Can the reader find similar arrangementsproducing 3, 4, 5, 6, 7, 8, and 9 respectively? Also, can he find thepairs of smallest possible numbers in each case? Thus, 1 4 6 5 8 dividedby 7 3 2 9 is just as correct for 2 as the other example we have given,but the numbers are higher.

89.--ADDING THE DIGITS.

If I write the sum of money, L987, 5s. 41/2d., and add up the digits,they sum to 36. No digit has thus been used a second time in the amountor addition. This is the largest amount possible under the conditions.Now find the smallest possible amount, pounds, shillings, pence, andfarthings being all represented. You need not use more of the ninedigits than you choose, but no digit may be repeated throughout. Thenought is not allowed.

90.--THE CENTURY PUZZLE.

Can you write 100 in the form of a mixed number, using all the ninedigits once, and only once? The late distinguished French mathematician,Edouard Lucas, found seven different ways of doing it, and expressed hisdoubts as to there being any other ways. As a matter of fact there arejust eleven ways and no more. Here is one of them, 91+5742/638. Nine ofthe other ways have similarly two figures in the integral part of thenumber, but the eleventh expression has only one figure there. Can thereader find this last form?

91.--MORE MIXED FRACTIONS.

When I first published my solution to the last puzzle, I was led toattempt the expression of all numbers in turn up to 100 by a mixedfraction containing all the nine digits. Here are twelve numbers for thereader to try his hand at: 13, 14, 15, 16, 18, 20, 27, 36, 40, 69, 72,94. Use every one of the nine digits once, and only once, in every case.

92.--DIGITAL SQUARE NUMBERS.

Here are the nine digits so arranged that they form four square numbers:9, 81, 324, 576. Now, can you put them all together so as to form asingle square number--(I) the smallest possible, and (II) the largestpossible?

93.--THE MYSTIC ELEVEN.

Can you find the largest possible number containing any nine of the tendigits (calling nought a digit) that can be divided by 11 without aremainder? Can you also find the smallest possible number produced inthe same way that is divisible by 11? Here is an example, where thedigit 5 has been omitted: 896743012. This number contains nine of thedigits and is divisible by 11, but it is neither the largest nor thesmallest number that will work.

94.--THE DIGITAL CENTURY.

1 2 3 4 5 6 7 8 9 = 100.

It is required to place arithmetical signs between the nine figures sothat they shall equal 100. Of course, you must not alter the presentnumerical arrangement of the figures. Can you give a correct solutionthat employs (1) the fewest possible signs, and (2) the fewest possibleseparate strokes or dots of the pen? That is, it is necessary to use asfew signs as possible, and those signs should be of the simplest form.The signs of addition and multiplication (+ and x) will thus count astwo strokes, the sign of subtraction (-) as one stroke, the sign ofdivision (/) as three, and so on.

95.--THE FOUR SEVENS.

[Illustration]

In the illustration Professor Rackbrane is seen demonstrating one of thelittle posers with which he is accustomed to entertain his class. Hebelieves that by taking his pupils off the beaten tracks he is thebetter able to secure their attention, and to induce original andingenious methods of thought. He has, it will be seen, just shown howfour 5's may be written with simple arithmetical signs so as torepresent 100. Every juvenile reader will see at a glance that hisexample is quite correct. Now, what he wants you to do is this: Arrangefour 7's (neither more nor less) with arithmetical signs so that theyshall represent 100. If he had said we were to use four 9's we might atonce have written 99+9/9, but the four 7's call for rather moreingenuity. Can you discover the little trick?

96.--THE DICE NUMBERS.

[Illustration]

I have a set of four dice, not marked with spots in the ordinary way,but with Arabic figures, as shown in the illustration. Each die, ofcourse, bears the numbers 1 to 6. When put together they will form agood many, different numbers. As represented they make the number 1246.Now, if I make all the different four-figure numbers that are possiblewith these dice (never putting the same figure more than once in anynumber), what will they all add up to? You are allowed to turn the 6upside down, so as to represent a 9. I do not ask, or expect, the readerto go to all the labour of writing out the full list of numbers and thenadding them up. Life is not long enough for such wasted energy. Can youget at the answer in any other way?

VARIOUS ARITHMETICAL AND ALGEBRAICAL PROBLEMS.

"Variety's the very spice of life, That gives it all its flavour."

COWPER: _The Task._

97.--THE SPOT ON THE TABLE.

A boy, recently home from school, wished to give his father anexhibition of his precocity. He pushed a large circular table into thecorner of the room, as shown in the illustration, so that it touchedboth walls, and he then pointed to a spot of ink on the extreme edge.

[Illustration]

"Here is a little puzzle for you, pater," said the youth. "That spot isexactly eight inches from one wall and nine inches from the other. Canyou tell me the diameter of the table without measuring it?"

The boy was overheard to tell a friend, "It fairly beat the guv'nor;"but his father is known to have remarked to a City acquaintance that hesolved the thing in his head in a minute. I often wonder which spoke thetruth.

98.--ACADEMIC COURTESIES.

In a certain mixed school, where a special feature was made of theinculcation of good manners, they had a curious rule on assembling everymorning. There were twice as many girls as boys. Every girl made a bowto every other girl, to every boy, and to the teacher. Every boy made abow to every other boy, to every girl, and to the teacher. In all therewere nine hundred bows made in that model academy every morning. Now,can you say exactly how many boys there were in the school? If you arenot very careful, you are likely to get a good deal out in yourcalculation.

99.--THE THIRTY-THREE PEARLS.

[Illustration]

"A man I know," said Teddy Nicholson at a certain family party,"possesses a string of thirty-three pearls. The middle pearl is thelargest and best of all, and the others are so selected and arrangedthat, starting from one end, each successive pearl is worth L100 morethan the preceding one, right up to the big pearl. From the other endthe pearls increase in value by L150 up to the large pearl. The wholestring is worth L65,000. What is the value of that large pearl?"

"Pearls and other articles of clothing," said Uncle Walter, when theprice of the precious gem had been discovered, "remind me of Adam andEve. Authorities, you may not know, differ as to the number of applesthat were eaten by Adam and Eve. It is the opinion of some that Eve 8(ate) and Adam 2 (too), a total of 10 only. But certain mathematicianshave figured it out differently, and hold that Eve 8 and Adam a total of16. Yet the most recent investigators think the above figures entirelywrong, for if Eve 8 and Adam 8 2, the total must be 90."

"Well," said Harry, "it seems to me that if there were giants in thosedays, probably Eve 8 1 and Adam 8 2, which would give a total of 163."

"I am not at all satisfied," said Maud. "It seems to me that if Eve 8 1and Adam 8 1 2, they together consumed 893."

"I am sure you are all wrong," insisted Mr. Wilson, "for I consider thatEve 8 1 4 Adam, and Adam 8 1 2 4 Eve, so we get a total of 8,938."

"But, look here," broke in Herbert. "If Eve 8 1 4 Adam and Adam 8 1 2 42 oblige Eve, surely the total must have been 82,056!"

At this point Uncle Walter suggested that they might let the matterrest. He declared it to be clearly what mathematicians call anindeterminate problem.

100.--THE LABOURER'S PUZZLE.

Professor Rackbrane, during one of his rambles, chanced to come upon aman digging a deep hole.

"Good morning," he said. "How deep is that hole?"

"Guess," replied the labourer. "My height is exactly five feet teninches."

"How much deeper are you going?" said the professor.

"I am going twice as deep," was the answer, "and then my head will betwice as far below ground as it is now above ground."

Rackbrane now asks if you could tell how deep that hole would be whenfinished.

101.--THE TRUSSES OF HAY.

Farmer Tompkins had five trusses of hay, which he told his man Hodge toweigh before delivering them to a customer. The stupid fellow weighedthem two at a time in all possible ways, and informed his master thatthe weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120,and 121. Now, how was Farmer Tompkins to find out from these figures howmuch every one of the five trusses weighed singly? The reader may atfirst think that he ought to be told "which pair is which pair," orsomething of that sort, but it is quite unnecessary. Can you give thefive correct weights?

102.--MR. GUBBINS IN A FOG.

Mr. Gubbins, a diligent man of business, was much inconvenienced by aLondon fog. The electric light happened to be out of order and he had tomanage as best he could with two candles. His clerk assured him thatthough both were of the same length one candle would burn for four hoursand the other for five hours. After he had been working some time he putthe candles out as the fog had lifted, and he then noticed that whatremained of one candle was exactly four times the length of what wasleft of the other.

When he got home that night Mr. Gubbins, who liked a good puzzle, saidto himself, "Of course it is possible to work out just how long thosetwo candles were burning to-day. I'll have a shot at it." But he soonfound himself in a worse fog than the atmospheric one. Could you haveassisted him in his dilemma? How long were the candles burning?

103.--PAINTING THE LAMP-POSTS.

Tim Murphy and Pat Donovan were engaged by the local authorities topaint the lamp-posts in a certain street. Tim, who was an early riser,arrived first on the job, and had painted three on the south side whenPat turned up and pointed out that Tim's contract was for the northside. So Tim started afresh on the north side and Pat continued on thesouth. When Pat had finished his side he went across the street andpainted six posts for Tim, and then the job was finished. As there wasan equal number of lamp-posts on each side of the street, the simplequestion is: Which man painted the more lamp-posts, and just how manymore?

104.--CATCHING THE THIEF.

"Now, constable," said the defendant's counsel in cross-examination,"you say that the prisoner was exactly twenty-seven steps ahead of youwhen you started to run after him?"

"Yes, sir."

"And you swear that he takes eight steps to your five?"

"That is so."

"Then I ask you, constable, as an intelligent man, to explain how youever caught him, if that is the case?"

"Well, you see, I have got a longer stride. In fact, two of my steps areequal in length to five of the prisoner's. If you work it out, you willfind that the number of steps I required would bring me exactly to thespot where I captured him."

Here the foreman of the jury asked for a few minutes to figure out thenumber of steps the constable must have taken. Can you also say how manysteps the officer needed to catch the thief?

105.--THE PARISH COUNCIL ELECTION.

Here is an easy problem for the novice. At the last election of theparish council of Tittlebury-in-the-Marsh there were twenty-threecandidates for nine seats. Each voter was qualified to vote for nine ofthese candidates or for any less number. One of the electors wants toknow in just how many different ways it was possible for him to vote.

106.--THE MUDDLETOWN ELECTION.

At the last Parliamentary election at Muddletown 5,473 votes werepolled. The Liberal was elected by a majority of 18 over theConservative, by 146 over the Independent, and by 575 over theSocialist. Can you give a simple rule for figuring out how many voteswere polled for each candidate?

107.--THE SUFFRAGISTS' MEETING.

At a recent secret meeting of Suffragists a serious difference ofopinion arose. This led to a split, and a certain number left themeeting. "I had half a mind to go myself," said the chair-woman, "and ifI had done so, two-thirds of us would have retired." "True," saidanother member; "but if I had persuaded my friends Mrs. Wild andChristine Armstrong to remain we should only have lost half our number."Can you tell how many were present at the meeting at the start?

108.--THE LEAP-YEAR LADIES.

Last leap-year ladies lost no time in exercising the privilege of makingproposals of marriage. If the figures that reached me from an occultsource are correct, the following represents the state of affairs inthis country.

A number of women proposed once each, of whom one-eighth were widows. Inconsequence, a number of men were to be married of whom one-eleventhwere widowers. Of the proposals made to widowers, one-fifth weredeclined. All the widows were accepted. Thirty-five forty-fourths of thewidows married bachelors. One thousand two hundred and twenty-onespinsters were declined by bachelors. The number of spinsters acceptedby bachelors was seven times the number of widows accepted by bachelors.Those are all the particulars that I was able to obtain. Now, how manywomen proposed?

109.--THE GREAT SCRAMBLE.

After dinner, the five boys of a household happened to find a parcel ofsugar-plums. It was quite unexpected loot, and an exciting scrambleensued, the full details of which I will recount with accuracy, as itforms an interesting puzzle.

You see, Andrew managed to get possession of just two-thirds of theparcel of sugar-plums. Bob at once grabbed three-eighths of these, andCharlie managed to seize three-tenths also. Then young David dashed uponthe scene, and captured all that Andrew had left, except one-seventh,which Edgar artfully secured for himself by a cunning trick. Now the funbegan in real earnest, for Andrew and Charlie jointly set upon Bob, whostumbled against the fender and dropped half of all that he had, whichwere equally picked up by David and Edgar, who had crawled under a tableand were waiting. Next, Bob sprang on Charlie from a chair, and upsetall the latter's collection on to the floor. Of this prize Andrew gotjust a quarter, Bob gathered up one-third, David got two-sevenths, whileCharlie and Edgar divided equally what was left of that stock.

[Illustration]

They were just thinking the fray was over when David suddenly struck outin two directions at once, upsetting three-quarters of what Bob andAndrew had last acquired. The two latter, with the greatest difficulty,recovered five-eighths of it in equal shares, but the three others eachcarried off one-fifth of the same. Every sugar-plum was now accountedfor, and they called a truce, and divided equally amongst them theremainder of the parcel. What is the smallest number of sugar-plumsthere could have been at the start, and what proportion did each boyobtain?

110.--THE ABBOT'S PUZZLE.

The first English puzzlist whose name has come down to us was aYorkshireman--no other than Alcuin, Abbot of Canterbury (A.D. 735-804).Here is a little puzzle from his works, which is at least interesting onaccount of its antiquity. "If 100 bushels of corn were distributed among100 people in such a manner that each man received three bushels, eachwoman two, and each child half a bushel, how many men, women, andchildren were there?"

Now, there are six different correct answers, if we exclude a case wherethere would be no women. But let us say that there were just five timesas many women as men, then what is the correct solution?

111.--REAPING THE CORN.

A farmer had a square cornfield. The corn was all ripe for reaping, and,as he was short of men, it was arranged that he and his son should sharethe work between them. The farmer first cut one rod wide all round thesquare, thus leaving a smaller square of standing corn in the middle ofthe field. "Now," he said to his son, "I have cut my half of the field,and you can do your share." The son was not quite satisfied as to theproposed division of labour, and as the village schoolmaster happened tobe passing, he appealed to that person to decide the matter. He foundthe farmer was quite correct, provided there was no dispute as to thesize of the field, and on this point they were agreed. Can you tell thearea of the field, as that ingenious schoolmaster succeeded in doing?

112.--A PUZZLING LEGACY.

A man left a hundred acres of land to be divided among his threesons--Alfred, Benjamin, and Charles--in the proportion of one-third,one-fourth, and one-fifth respectively. But Charles died. How was theland to be divided fairly between Alfred and Benjamin?

113.--THE TORN NUMBER.

[Illustration]

I had the other day in my possession a label bearing the number 3 0 2 5in large figures. This got accidentally torn in half, so that 3 0 was onone piece and 2 5 on the other, as shown on the illustration. On lookingat these pieces I began to make a calculation, scarcely conscious ofwhat I was doing, when I discovered this little peculiarity. If we addthe 3 0 and the 2 5 together and square the sum we get as the result thecomplete original number on the label! Thus, 30 added to 25 is 55, and55 multiplied by 55 is 3025. Curious, is it not? Now, the puzzle is tofind another number, composed of four figures, all different, which maybe divided in the middle and produce the same result.

114.--CURIOUS NUMBERS.

The number 48 has this peculiarity, that if you add 1 to it the resultis a square number (49, the square of 7), and if you add 1 to its half,you also get a square number (25, the square of 5). Now, there is nolimit to the numbers that have this peculiarity, and it is aninteresting puzzle to find three more of them--the smallest possiblenumbers. What are they?

115.--A PRINTER'S ERROR.

In a certain article a printer had to set up the figures 5^4x2^3, which,of course, means that the fourth power of 5 (625) is to be multiplied bythe cube of 2 (8), the product of which is 5,000. But he printed 5^4x2^3as 5 4 2 3, which is not correct. Can you place four digits in themanner shown, so that it will be equally correct if the printer sets itup aright or makes the same blunder?

116.--THE CONVERTED MISER.

Mr. Jasper Bullyon was one of the very few misers who have ever beenconverted to a sense of their duty towards their less fortunatefellow-men. One eventful night he counted out his accumulated wealth,and resolved to distribute it amongst the deserving poor.

He found that if he gave away the same number of pounds every day in theyear, he could exactly spread it over a twelvemonth without there beinganything left over; but if he rested on the Sundays, and only gave awaya fixed number of pounds every weekday, there would be one sovereignleft over on New Year's Eve. Now, putting it at the lowest possible,what was the exact number of pounds that he had to distribute?

Could any question be simpler? A sum of pounds divided by one number ofdays leaves no remainder, but divided by another number of days leaves asovereign over. That is all; and yet, when you come to tackle thislittle question, you will be surprised that it can become so puzzling.

117.--A FENCE PROBLEM.

[Illustration]

The practical usefulness of puzzles is a point that we are liable tooverlook. Yet, as a matter of fact, I have from time to time receivedquite a large number of letters from individuals who have found that themastering of some little principle upon which a puzzle was built hasproved of considerable value to them in a most unexpected way. Indeed,it may be accepted as a good maxim that a puzzle is of little real valueunless, as well as being amusing and perplexing, it conceals someinstructive and possibly useful feature. It is, however, very curioushow these little bits of acquired knowledge dovetail into theoccasional requirements of everyday life, and equally curious to whatstrange and mysterious uses some of our readers seem to apply them.What, for example, can be the object of Mr. Wm. Oxley, who writes to meall the way from Iowa, in wishing to ascertain the dimensions of a fieldthat he proposes to enclose, containing just as many acres as thereshall be rails in the fence?

The man wishes to fence in a perfectly square field which is to containjust as many acres as there are rails in the required fence. Eachhurdle, or portion of fence, is seven rails high, and two lengths wouldextend one pole (161/2 ft.): that is to say, there are fourteen railsto the pole, lineal measure. Now, what must be the size of the field?

118.--CIRCLING THE SQUARES.

[Illustration]

The puzzle is to place a different number in each of the ten squares sothat the sum of the squares of any two adjacent numbers shall be equalto the sum of the squares of the two numbers diametrically opposite tothem. The four numbers placed, as examples, must stand as they are. Thesquare of 16 is 256, and the square of 2 is 4. Add these together, andthe result is 260. Also--the square of 14 is 196, and the square of 8 is64. These together also make 260. Now, in precisely the same way, B andC should be equal to G and H (the sum will not necessarily be 260), Aand K to F and E, H and I to C and D, and so on, with any two adjoiningsquares in the circle.

All you have to do is to fill in the remaining six numbers. Fractionsare not allowed, and I shall show that no number need contain more thantwo figures.

119.--RACKBRANE'S LITTLE LOSS.

Professor Rackbrane was spending an evening with his old friends, Mr.and Mrs. Potts, and they engaged in some game (he does not say whatgame) of cards. The professor lost the first game, which resulted indoubling the money that both Mr. and Mrs. Potts had laid on the table.The second game was lost by Mrs. Potts, which doubled the money thenheld by her husband and the professor. Curiously enough, the third gamewas lost by Mr. Potts, and had the effect of doubling the money thenheld by his wife and the professor. It was then found that each personhad exactly the same money, but the professor had lost five shillings inthe course of play. Now, the professor asks, what was the sum of moneywith which he sat down at the table? Can you tell him?

120.--THE FARMER AND HIS SHEEP.

[Illustration]

Farmer Longmore had a curious aptitude for arithmetic, and was known inhis district as the "mathematical farmer." The new vicar was not awareof this fact when, meeting his worthy parishioner one day in the lane,he asked him in the course of a short conversation, "Now, how many sheephave you altogether?" He was therefore rather surprised at Longmore'sanswer, which was as follows: "You can divide my sheep into twodifferent parts, so that the difference between the two numbers is thesame as the difference between their squares. Maybe, Mr. Parson, youwill like to work out the little sum for yourself."

Can the reader say just how many sheep the farmer had? Supposing he hadpossessed only twenty sheep, and he divided them into the two parts 12and 8. Now, the difference between their squares, 144 and 64, is 80. Sothat will not do, for 4 and 80 are certainly not the same. If you canfind numbers that work out correctly, you will know exactly how manysheep Farmer Longmore owned.

121.--HEADS OR TAILS.

Crooks, an inveterate gambler, at Goodwood recently said to a friend,"I'll bet you half the money in my pocket on the toss of a coin--heads Iwin, tails I lose." The coin was tossed and the money handed over. Herepeated the offer again and again, each time betting half the moneythen in his possession. We are not told how long the game went on, orhow many times the coin was tossed, but this we know, that the number oftimes that Crooks lost was exactly equal to the number of times that hewon. Now, did he gain or lose by this little venture?

122.--THE SEE-SAW PUZZLE.

Necessity is, indeed, the mother of invention. I was amused the otherday in watching a boy who wanted to play see-saw and, in his failure tofind another child to share the sport with him, had been driven backupon the ingenious resort of tying a number of bricks to one end of theplank to balance his weight at the other.

As a matter of fact, he just balanced against sixteen bricks, when thesewere fixed to the short end of plank, but if he fixed them to the longend of plank he only needed eleven as balance.

Now, what was that boy's weight, if a brick weighs equal to athree-quarter brick and three-quarters of a pound?

123.--A LEGAL DIFFICULTY.

"A client of mine," said a lawyer, "was on the point of death when hiswife was about to present him with a child. I drew up his will, in whichhe settled two-thirds of his estate upon his son (if it should happen tobe a boy) and one-third on the mother. But if the child should be agirl, then two-thirds of the estate should go to the mother andone-third to the daughter. As a matter of fact, after his death twinswere born--a boy and a girl. A very nice point then arose. How was theestate to be equitably divided among the three in the closest possibleaccordance with the spirit of the dead man's will?"

124.--A QUESTION OF DEFINITION.

"My property is exactly a mile square," said one landowner to another.

"Curiously enough, mine is a square mile," was the reply.

"Then there is no difference?"

Is this last statement correct?

125.--THE MINERS' HOLIDAY.

Seven coal-miners took a holiday at the seaside during a big strike. Sixof the party spent exactly half a sovereign each, but Bill Harris wasmore extravagant. Bill spent three shillings more than the average ofthe party. What was the actual amount of Bill's expenditure?

126.--SIMPLE MULTIPLICATION.

If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on thetable in this order:--

1 4 2 8 5 7

We can demonstrate that in order to multiply by 3 all that is necessaryis to remove the 1 to the other end of the row, and the thing is done.The answer is 428571. Can you find a number that, when multiplied by 3and divided by 2, the answer will be the same as if we removed the firstcard (which in this case is to be a 3) From the beginning of the row tothe end?

127.--SIMPLE DIVISION.

Sometimes a very simple question in elementary arithmetic will cause agood deal of perplexity. For example, I want to divide the four numbers,701, 1,059, 1,417, and 2,312, by the largest number possible that willleave the same remainder in every case. How am I to set to work Ofcourse, by a laborious system of trial one can in time discover theanswer, but there is quite a simple method of doing it if you can onlyfind it.

128.--A PROBLEM IN SQUARES.

We possess three square boards. The surface of the first contains fivesquare feet more than the second, and the second contains five squarefeet more than the third. Can you give exact measurements for the sidesof the boards? If you can solve this little puzzle, then try to findthree squares in arithmetical progression, with a common difference of 7and also of 13.

129.--THE BATTLE OF HASTINGS.

All historians know that there is a great deal of mystery anduncertainty concerning the details of the ever-memorable battle on thatfatal day, October 14, 1066. My puzzle deals with a curious passage inan ancient monkish chronicle that may never receive the attention thatit deserves, and if I am unable to vouch for the authenticity of thedocument it will none the less serve to furnish us with a problem thatcan hardly fail to interest those of my readers who have arithmeticalpredilections. Here is the passage in question.

"The men of Harold stood well together, as their wont was, and formedsixty and one squares, with a like number of men in every squarethereof, and woe to the hardy Norman who ventured to enter theirredoubts; for a single blow of a Saxon war-hatchet would break his lanceand cut through his coat of mail.... When Harold threw himself into thefray the Saxons were one mighty square of men, shouting thebattle-cries, 'Ut!' 'Olicrosse!' 'Godemite!'"

Now, I find that all the contemporary authorities agree that the Saxonsdid actually fight in this solid order. For example, in the "Carmen deBello Hastingensi," a poem attributed to Guy, Bishop of Amiens, livingat the time of the battle, we are told that "the Saxons stood fixed in adense mass," and Henry of Huntingdon records that "they were like unto acastle, impenetrable to the Normans;" while Robert Wace, a centuryafter, tells us the same thing. So in this respect my newly-discoveredchronicle may not be greatly in error. But I have reason to believe thatthere is something wrong with the actual figures. Let the reader seewhat he can make of them.

The number of men would be sixty-one times a square number; but whenHarold himself joined in the fray they were then able to form one largesquare. What is the smallest possible number of men there could havebeen?

In order to make clear to the reader the simplicity of the question, Iwill give the lowest solutions in the case of 60 and 62, the numbersimmediately preceding and following 61. They are 60 x 4 squared + 1 = 31 squared,and 62 x 8 squared + 1 = 63 squared. That is, 60 squares of 16 men each would be 960men, and when Harold joined them they would be 961 in number, and soform a square with 31 men on every side. Similarly in the case of thefigures I have given for 62. Now, find the lowest answer for 61.

130.--THE SCULPTOR'S PROBLEM.

An ancient sculptor was commissioned to supply two statues, each on acubical pedestal. It is with these pedestals that we are concerned. Theywere of unequal sizes, as will be seen in the illustration, and when thetime arrived for payment a dispute arose as to whether the agreement wasbased on lineal or cubical measurement. But as soon as they came tomeasure the two pedestals the matter was at once settled, because,curiously enough, the number of lineal feet was exactly the same as thenumber of cubical feet. The puzzle is to find the dimensions for twopedestals having this peculiarity, in the smallest possible figures. Yousee, if the two pedestals, for example, measure respectively 3 ft. and 1ft. on every side, then the lineal measurement would be 4 ft. and thecubical contents 28 ft., which are not the same, so these measurementswill not do.

[Illustration]

131.--THE SPANISH MISER.

There once lived in a small town in New Castile a noted miser named DonManuel Rodriguez. His love of money was only equalled by a strongpassion for arithmetical problems. These puzzles usually dealt in someway or other with his accumulated treasure, and were propounded by himsolely in order that he might have the pleasure of solving them himself.Unfortunately very few of them have survived, and when travellingthrough Spain, collecting material for a proposed work on "The SpanishOnion as a Cause of National Decadence," I only discovered a very few.One of these concerns the three boxes that appear in the accompanyingauthentic portrait.

[Illustration]

Each box contained a different number of golden doubloons. Thedifference between the number of doubloons in the upper box and thenumber in the middle box was the same as the difference between thenumber in the middle box and the number in the bottom box. And if thecontents of any two of the boxes were united they would form a squarenumber. What is the smallest number of doubloons that there could havebeen in any one of the boxes?

132.--THE NINE TREASURE BOXES.

The following puzzle will illustrate the importance on occasions ofbeing able to fix the minimum and maximum limits of a required number.This can very frequently be done. For example, it has not yet beenascertained in how many different ways the knight's tour can beperformed on the chess board; but we know that it is fewer than thenumber of combinations of 168 things taken 63 at a time and is greaterthan 31,054,144--for the latter is the number of routes of a particulartype. Or, to take a more familiar case, if you ask a man how many coinshe has in his pocket, he may tell you that he has not the slightestidea. But on further questioning you will get out of him some suchstatement as the following: "Yes, I am positive that I have more thanthree coins, and equally certain that there are not so many astwenty-five." Now, the knowledge that a certain number lies between 2and 12 in my puzzle will enable the solver to find the exact answer;without that information there would be an infinite number of answers,from which it would be impossible to select the correct one.

This is another puzzle received from my friend Don Manuel Rodriguez, thecranky miser of New Castile. On New Year's Eve in 1879 he showed me ninetreasure boxes, and after informing me that every box contained a squarenumber of golden doubloons, and that the difference between the contentsof A and B was the same as between B and C, D and E, E and F, G and H,or H and I, he requested me to tell him the number of coins in every oneof the boxes. At first I thought this was impossible, as there would bean infinite number of different answers, but on consideration I foundthat this was not the case. I discovered that while every box containedcoins, the contents of A, B, C increased in weight in alphabeticalorder; so did D, E, F; and so did G, H, I; but D or E need not beheavier than C, nor G or H heavier than F. It was also perfectly certainthat box A could not contain more than a dozen coins at the outside;there might not be half that number, but I was positive that there werenot more than twelve. With this knowledge I was able to arrive at thecorrect answer.

In short, we have to discover nine square numbers such that A, B, C; andD, E, F; and G, H, I are three groups in arithmetical progression, thecommon difference being the same in each group, and A being less than12. How many doubloons were there in every one of the nine boxes?

133.--THE FIVE BRIGANDS.

The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and Esteban,were counting their spoils after a raid, when it was found that they hadcaptured altogether exactly 200 doubloons. One of the band pointed outthat if Alfonso had twelve times as much, Benito three times as much,Carlos the same amount, Diego half as much, and Esteban one-third asmuch, they would still have altogether just 200 doubloons. How manydoubloons had each?

There are a good many equally correct answers to this question. Here isone of them:

A 6 x 12 = 72 B 12 x 3 = 36 C 17 x 1 = 17 D 120 x 1/2 = 60 E 45 x 1/3 = 15 ___ ___ 200 200

The puzzle is to discover exactly how many different answers there are,it being understood that every man had something and that there is to beno fractional money--only doubloons in every case.

This problem, worded somewhat differently, was propounded by Tartaglia(died 1559), and he flattered himself that he had found one solution;but a French mathematician of note (M.A. Labosne), in a recent work,says that his readers will be astonished when he assures them that thereare 6,639 different correct answers to the question. Is this so? Howmany answers are there?

134.--THE BANKER'S PUZZLE.

A banker had a sporting customer who was always anxious to wager onanything. Hoping to cure him of his bad habit, he proposed as a wagerthat the customer would not be able to divide up the contents of a boxcontaining only sixpences into an exact number of equal piles ofsixpences. The banker was first to put in one or more sixpences (as manyas he liked); then the customer was to put in one or more (but in hiscase not more than a pound in value), neither knowing what the other putin. Lastly, the customer was to transfer from the banker's counter tothe box as many sixpences as the banker desired him to put in. Thepuzzle is to find how many sixpences the banker should first put in andhow many he should ask the customer to transfer, so that he may have thebest chance of winning.

135.--THE STONEMASON'S PROBLEM.

A stonemason once had a large number of cubic blocks of stone in hisyard, all of exactly the same size. He had some very fanciful littleways, and one of his queer notions was to keep these blocks piled incubical heaps, no two heaps containing the same number of blocks. He haddiscovered for himself (a fact that is well known to mathematicians)that if he took all the blocks contained in any number of heaps inregular order, beginning with the single cube, he could always arrangethose on the ground so as to form a perfect square. This will be clearto the reader, because one block is a square, 1 + 8 = 9 is a square, 1 +8 + 27 = 36 is a square, 1 + 8 + 27 + 64 = 100 is a square, and so on.In fact, the sum of any number of consecutive cubes, beginning alwayswith 1, is in every case a square number.

One day a gentleman entered the mason's yard and offered him a certainprice if he would supply him with a consecutive number of these cubicalheaps which should contain altogether a number of blocks that could belaid out to form a square, but the buyer insisted on more than threeheaps and _declined to take the single block_ because it contained aflaw. What was the smallest possible number of blocks of stone that themason had to supply?

136.--THE SULTAN'S ARMY.

A certain Sultan wished to send into battle an army that could be formedinto two perfect squares in twelve different ways. What is the smallestnumber of men of which that army could be composed? To make it clear tothe novice, I will explain that if there were 130 men, they could beformed into two squares in only two different ways--81 and 49, or 121and 9. Of course, all the men must be used on every occasion.

137.--A STUDY IN THRIFT.

Certain numbers are called triangular, because if they are taken torepresent counters or coins they may be laid out on the table so as toform triangles. The number 1 is always regarded as triangular, just as 1is a square and a cube number. Place one counter on the table--that is,the first triangular number. Now place two more counters beneath it, andyou have a triangle of three counters; therefore 3 is triangular. Nextplace a row of three more counters, and you have a triangle of sixcounters; therefore 6 is triangular. We see that every row of countersthat we add, containing just one more counter than the row above it,makes a larger triangle.

Now, half the sum of any number and its square is always a triangularnumber. Thus half of 2 + 2 squared = 3; half of 3 + 3 squared = 6; half of 4 +4 squared = 10; half of 5 + 5 squared= 15; and so on. So if we want to form atriangle with 8 counters on each side we shall require half of 8 +8 squared, or 36 counters. This is a pretty little property of numbers.Before going further, I will here say that if the reader refers to the"Stonemason's Problem" (No. 135) he will remember that the sum of anynumber of consecutive cubes beginning with 1 is always a square, andthese form the series 1 squared, 3 squared, 6 squared, 10 squared, etc. It will now be understoodwhen I say that one of the keys to the puzzle was the fact that theseare always the squares of triangular numbers--that is, the squares of 1,3, 6, 10, 15, 21, 28, etc., any of which numbers we have seen will forma triangle.

Every whole number is either triangular, or the sum of two triangularnumbers or the sum of three triangular numbers. That is, if we take anynumber we choose we can always form one, two, or three triangles withthem. The number 1 will obviously, and uniquely, only form one triangle;some numbers will only form two triangles (as 2, 4, 11, etc.); somenumbers will only form three triangles (as 5, 8, 14, etc.). Then, again,some numbers will form both one and two triangles (as 6), others bothone and three triangles (as 3 and 10), others both two and threetriangles (as 7 and 9), while some numbers (like 21) will form one, two,or three triangles, as we desire. Now for a little puzzle in triangularnumbers.

Sandy McAllister, of Aberdeen, practised strict domestic economy, andwas anxious to train his good wife in his own habits of thrift. He toldher last New Year's Eve that when she had saved so many sovereigns thatshe could lay them all out on the table so as to form a perfect square,or a perfect triangle, or two triangles, or three triangles, just as hemight choose to ask he would add five pounds to her treasure. Soon shewent to her husband with a little bag of L36 in sovereigns and claimedher reward. It will be found that the thirty-six coins will form asquare (with side 6), that they will form a single triangle (with side8), that they will form two triangles (with sides 5 and 6), and thatthey will form three triangles (with sides 3, 5, and 5). In each of thefour cases all the thirty-six coins are used, as required, and Sandytherefore made his wife the promised present like an honest man.

The Scotsman then undertook to extend his promise for five more years,so that if next year the increased number of sovereigns that she hassaved can be laid out in the same four different ways she will receive asecond present; if she succeeds in the following year she will get athird present, and so on until she has earned six presents in all. Now,how many sovereigns must she put together before she can win the sixthpresent?

What you have to do is to find five numbers, the smallest possible,higher than 36, that can be displayed in the four ways--to form asquare, to form a triangle, to form two triangles, and to form threetriangles. The highest of your five numbers will be your answer.

138.--THE ARTILLERYMEN'S DILEMMA.

[Illustration: [Pyramid of cannon-balls]]

MMMMMMMr MM MM: M 0 rWZX M : MWM aX ,BM M 0M M aMMMM2MW 02 MMWMMr ZM. M@M 8MM 7 XM2 MS2 M.MMMWMMMM MM M MX iMM M7W 8 . M r W M@ Z;M M 0r ; M M M W 22 W M @ M M M.M2WMMMMZ ;MM@X:7MMMB; MMM ZM M:MM0;8: ,MS Ma 8 MMMMMMMi rM 2MMMMMM MB M 7 XM, ,: BMM: r7S .,MM MM MB M i ,M , 2 ; aMMMMMMMMM XM; MZM M . M 7 M . Z M M M8 M M S M .0 M 8MM aMi: MMMM7M ,7 .iM X M @ aZ M M 8 ,@MMMMBMMMa SMW 7M,XZ@MM M 8M M .M MMMM@X MMr Ma MMMMMMMMM@ M .WM M @WM7WMM .WX MZS M 8M :MMMWMMMM 8X MMMBMMM7 7aM 2MM r, 8r ZM2 Mr2 aMM; Mai :MS :iM ZiM @MX M M . M Wr.MMMaBMMMB M M MZ. ,M MMZ Mr M M B0 Z 2S iM S XM 7 WMM MM @.M M M W M. M M 0;M2M;MMMM: WW8aMM M S@ M M M : MaMMMMMM MM0W;MZM: M i M M MM MMMZMBZa0ar B20rMMM Si i BW MMM02 7MM0 2MMM MMMMMMMM . M SM@aiMM20BWM XM 0ZMMM:MMMMW; r. 0WMBM2XrB: .

"All cannon-balls are to be piled in square pyramids," was the orderissued to the regiment. This was done. Then came the further order, "Allpyramids are to contain a square number of balls." Whereupon the troublearose. "It can't be done," said the major. "Look at this pyramid, forexample; there are sixteen balls at the base, then nine, then four, thenone at the top, making thirty balls in all. But there must be six moreballs, or five fewer, to make a square number." "It _must_ be done,"insisted the general. "All you have to do is to put the right number ofballs in your pyramids." "I've got it!" said a lieutenant, themathematical genius of the regiment. "Lay the balls out singly." "Bosh!"exclaimed the general. "You can't _pile_ one ball into a pyramid!" Is itreally possible to obey both orders?

139.--THE DUTCHMEN'S WIVES.

I wonder how many of my readers are acquainted with the puzzle of the"Dutchmen's Wives"--in which you have to determine the names of threemen's wives, or, rather, which wife belongs to each husband. Some thirtyyears ago it was "going the rounds," as something quite new, but Irecently discovered it in the _Ladies' Diary_ for 1739-40, so it wasclearly familiar to the fair sex over one hundred and seventy years ago.How many of our mothers, wives, sisters, daughters, and aunts couldsolve the puzzle to-day? A far greater proportion than then, let ushope.

Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives,Gurtruen, Katruen, and Anna, purchase hogs. Each buys as many as he (orshe) gives shillings for one. Each husband pays altogether three guineasmore than his wife. Hendrick buys twenty-three more hogs than Katruen,and Elas eleven more than Gurtruen. Now, what was the name of each man'swife?

[Illustration]

140.--FIND ADA'S SURNAME.

This puzzle closely resembles the last one, my remarks on the solutionof which the reader may like to apply in another case. It was recentlysubmitted to a Sydney evening newspaper that indulges in "intellectsharpeners," but was rejected with the remark that it is childish andthat they only published problems capable of solution! Five ladies,accompanied by their daughters, bought cloth at the same shop. Each ofthe ten paid as many farthings per foot as she bought feet, and eachmother spent 8s. 51/4d. more than her daughter. Mrs. Robinson spent 6s.more than Mrs. Evans, who spent about a quarter as much as Mrs. Jones.Mrs. Smith spent most of all. Mrs. Brown bought 21 yards more thanBessie--one of the girls. Annie bought 16 yards more than Mary and spentL3, 0s. 8d. more than Emily. The Christian name of the other girl wasAda. Now, what was her surname?

141.--SATURDAY MARKETING.

Here is an amusing little case of marketing which, although it dealswith a good many items of money, leads up to a question of a totallydifferent character. Four married couples went into their village on arecent Saturday night to do a little marketing. They had to be veryeconomical, for among them they only possessed forty shilling coins. Thefact is, Ann spent 1s., Mary spent 2s., Jane spent 3s., and Kate spent4s. The men were rather more extravagant than their wives, for Ned Smithspent as much as his wife, Tom Brown twice as much as his wife, BillJones three times as much as his wife, and Jack Robinson four times asmuch as his wife. On the way home somebody suggested that they shoulddivide what coin they had left equally among them. This was done, andthe puzzling question is simply this: What was the surname of eachwoman? Can you pair off the four couples?

GEOMETRICAL PROBLEMS.

"God geometrizes continually."

PLATO.

"There is no study," said Augustus de Morgan, "which presents so simplea beginning as that of geometry; there is none in which difficultiesgrow more rapidly as we proceed." This will be found when the readercomes to consider the following puzzles, though they are not arranged instrict order of difficulty. And the fact that they have interested andgiven pleasure to man for untold ages is no doubt due in some measure tothe appeal they make to the eye as well as to the brain. Sometimes analgebraical formula or theorem seems to give pleasure to themathematician's eye, but it is probably only an intellectual pleasure.But there can be no doubt that in the case of certain geometricalproblems, notably dissection or superposition puzzles, the aestheticfaculty in man contributes to the delight. For example, there areprobably few readers who will examine the various cuttings of the Greekcross in the following pages without being in some degree stirred by asense of beauty. Law and order in Nature are always pleasing tocontemplate, but when they come under the very eye they seem to make aspecially strong appeal. Even the person with no geometrical knowledgewhatever is induced after the inspection of such things to exclaim, "Howvery pretty!" In fact, I have known more than one person led on to astudy of geometry by the fascination of cutting-out puzzles. I have,therefore, thought it well to keep these dissection puzzles distinctfrom the geometrical problems on more general lines.

DISSECTION PUZZLES.

"Take him and cut him out in little stars."

_Romeo and Juliet_, iii. 2.

Puzzles have infinite variety, but perhaps there is no class moreancient than dissection, cutting-out, or superposition puzzles. Theywere certainly known to the Chinese several thousand years before theChristian era. And they are just as fascinating to-day as they can havebeen at any period of their history. It is supposed by those who haveinvestigated the matter that the ancient Chinese philosophers used thesepuzzles as a sort of kindergarten method of imparting the principles ofgeometry. Whether this was so or not, it is certain that all gooddissection puzzles (for the nursery type of jig-saw puzzle, which merelyconsists in cutting up a picture into pieces to be put together again,is not worthy of serious consideration) are really based on geometricallaws. This statement need not, however, frighten off the novice, for itmeans little more than this, that geometry will give us the "reasonwhy," if we are interested in knowing it, though the solutions may oftenbe discovered by any intelligent person after the exercise of patience,ingenuity, and common sagacity.

If we want to cut one plane figure into parts that by readjustment willform another figure, the first thing is to find a way of doing it atall, and then to discover how to do it in the fewest possible pieces.Often a dissection problem is quite easy apart from this limitation ofpieces. At the time of the publication in the _Weekly Dispatch_, in1902, of a method of cutting an equilateral triangle into four partsthat will form a square (see No. 26, "Canterbury Puzzles"), nogeometrician would have had any difficulty in doing what is required infive pieces: the whole point of the discovery lay in performing thelittle feat in four pieces only.

Mere approximations in the case of these problems are valueless; thesolution must be geometrically exact, or it is not a solution at all.Fallacies are cropping up now and again, and I shall have occasion torefer to one or two of these. They are interesting merely as fallacies.But I want to say something on two little points that are always arisingin cutting-out puzzles--the questions of "hanging by a thread" and"turning over." These points can best be illustrated by a puzzle that isfrequently to be found in the old books, but invariably with a falsesolution. The puzzle is to cut the figure shown in Fig. 1 into threepieces that will fit together and form a half-square triangle. Theanswer that is invariably given is that shown in Figs. 1 and 2. Now, itis claimed that the four pieces marked C are really only one piece,because they may be so cut that they are left "hanging together by amere thread." But no serious puzzle lover will ever admit this. If thecut is made so as to leave the four pieces joined in one, then it cannotresult in a perfectly exact solution. If, on the other hand, thesolution is to be exact, then there will be four pieces--or six piecesin all. It is, therefore, not a solution in three pieces.

[Illustration: Fig. 1]

[Illustration: Fig. 2]

If, however, the reader will look at the solution in Figs. 3 and 4, hewill see that no such fault can be found with it. There is no questionwhatever that there are three pieces, and the solution is in thisrespect quite satisfactory. But another question arises. It will befound on inspection that the piece marked F, in Fig. 3, is turned overin Fig. 4--that is to say, a different side has necessarily to bepresented. If the puzzle were merely to be cut out of cardboard or wood,there might be no objection to this reversal, but it is quite possiblethat the material would not admit of being reversed. There might be apattern, a polish, a difference of texture, that prevents it. But it isgenerally understood that in dissection puzzles you are allowed to turnpieces over unless it is distinctly stated that you may not do so. Andvery often a puzzle is greatly improved by the added condition, "nopiece may be turned over." I have often made puzzles, too, in which thediagram has a small repeated pattern, and the pieces have then so to becut that not only is there no turning over, but the pattern has to bematched, which cannot be done if the pieces are turned round, even withthe proper side uppermost.

[Illustration: Fig. 3]

[Illustration: Fig. 4]

Before presenting a varied series of cutting-out puzzles, some very easyand others difficult, I propose to consider one family alone--thoseproblems involving what is known as the Greek cross with the square.This will exhibit a great variety of curious transpositions, and, byhaving the solutions as we go along, the reader will be saved thetrouble of perpetually turning to another part of the book, and willhave everything under his eye. It is hoped that in this way the articlemay prove somewhat instructive to the novice and interesting to others.

GREEK CROSS PUZZLES.

"To fret thy soul with crosses."

SPENSER.

"But, for my part, it was Greek to me."

_Julius Caesar_, i. 2.

Many people are accustomed to consider the cross as a wholly Christiansymbol. This is erroneous: it is of very great antiquity. The ancientEgyptians employed it as a sacred symbol, and on Greek sculptures wefind representations of a cake (the supposed real origin of our hotcross buns) bearing a cross. Two such cakes were discovered atHerculaneum. Cecrops offered to Jupiter Olympus a sacred cake or _boun_of this kind. The cross and ball, so frequently found on Egyptianfigures, is a circle and the _tau_ cross. The circle signified theeternal preserver of the world, and the T, named from the Greek letter_tau_, is the monogram of Thoth, the Egyptian Mercury, meaning wisdom.This _tau_ cross is also called by Christians the cross of St. Anthony,and is borne on a badge in the bishop's palace at Exeter. As for theGreek or mundane cross, the cross with four equal arms, we are told bycompetent antiquaries that it was regarded by ancient occultists forthousands of years as a sign of the dual forces of Nature--the male andfemale spirit of everything that was everlasting.

[Illustration: Fig. 5.]

The Greek cross, as shown in Fig. 5, is formed by the assemblingtogether of five equal squares. We will start with what is known as theHindu problem, supposed to be upwards of three thousand years old. Itappears in the seal of Harvard College, and is often given in old worksas symbolical of mathematical science and exactitude. Cut the cross intofive pieces to form a square. Figs. 6 and 7 show how this is done. Itwas not until the middle of the nineteenth century that we found thatthe cross might be transformed into a square in only four pieces. Figs.8 and 9 will show how to do it, if we further require the four pieces tobe all of the same size and shape. This Fig. 9 is remarkable because,according to Dr. Le Plongeon and others, as expounded in a work byProfessor Wilson of the Smithsonian Institute, here we have the greatSwastika, or sign, of "good luck to you "--the most ancient symbol ofthe human race of which there is any record. Professor Wilson's workgives some four hundred illustrations of this curious sign as found inthe Aztec mounds of Mexico, the pyramids of Egypt, the ruins of Troy,and the ancient lore of India and China. One might almost say there is acurious affinity between the Greek cross and Swastika! If, however, werequire that the four pieces shall be produced by only two clips of thescissors (assuming the puzzle is in paper form), then we must cut as inFig. 10 to form Fig. 11, the first clip of the scissors being from ato b. Of course folding the paper, or holding the pieces togetherafter the first cut, would not in this case be allowed. But there is aninfinite number of different ways of making the cuts to solve the puzzlein four pieces. To this point I propose to return.

[Illustration: Fig. 6]

[Illustration: Fig. 7]

[Illustration: Fig. 8]

[Illustration: Fig. 9]

[Illustration: Fig. 10]

[Illustration: Fig. 11]

It will be seen that every one of these puzzles has its reversepuzzle--to cut a square into pieces to form a Greek cross. But as asquare has not so many angles as the cross, it is not always equallyeasy to discover the true directions of the cuts. Yet in the case of theexamples given, I will leave the reader to determine their direction forhimself, as they are rather obvious from the diagrams.

Cut a square into five pieces that will form two separate Greek crossesof _different sizes_. This is quite an easy puzzle. As will be seen inFig. 12, we have only to divide our square into 25 little squares andthen cut as shown. The cross A is cut out entire, and the pieces B, C,D, and E form the larger cross in Fig. 13. The reader may here like tocut the single piece, B, into four pieces all similar in shape toitself, and form a cross with them in the manner shown in Fig. 13. Ihardly need give the solution.

[Illustration: FIG. 12.]

[Illustration: FIG. 13.]

Cut a square into five pieces that will form two separate Greek crossesof exactly the _same size_. This is more difficult. We make the cuts asin Fig. 14, where the cross A comes out entire and the other four piecesform the cross in Fig. 15. The direction of the cuts is pretty obvious.It will be seen that the sides of the square in Fig. 14 are marked offinto six equal parts. The sides of the cross are found by ruling linesfrom certain of these points to others.

[Illustration: FIG. 14.]

[Illustration: FIG. 15.]

I will now explain, as I promised, why a Greek cross may be cut intofour pieces in an infinite number of different ways to make a square.Draw a cross, as in Fig. 16. Then draw on transparent paper the squareshown in Fig. 17, taking care that the distance c to d is exactlythe same as the distance a to b in the cross. Now place thetransparent paper over the cross and slide it about into differentpositions, only be very careful always to keep the square at the sameangle to the cross as shown, where a b is parallel to c d. Ifyou place the point c exactly over a the lines will indicate thesolution (Figs. 10 and 11). If you place c in the very centre of thedotted square, it will give the solution in Figs. 8 and 9. You will nowsee that by sliding the square about so that the point c is alwayswithin the dotted square you may get as many different solutions as youlike; because, since an infinite number of different points maytheoretically be placed within this square, there must be an infinitenumber of different solutions. But the point c need not necessarily beplaced within the dotted square. It may be placed, for example, at pointe to give a solution in four pieces. Here the joins at a and f maybe as slender as you like. Yet if you once get over the edge at a orf you no longer have a solution in four pieces. This proof will befound both entertaining and instructive. If you do not happen to haveany transparent paper at hand, any thin paper will of course do if youhold the two sheets against a pane of glass in the window.

[Illustration: FIG. 16.]

[Illustration: FIG. 17.]

It may have been noticed from the solutions of the puzzles that I havegiven that the side of the square formed from the cross is always equalto the distance a to b in Fig. 16. This must necessarily be so, andI will presently try to make the point quite clear.

We will now go one step further. I have already said that the idealsolution to a cutting-out puzzle is always that which requires thefewest possible pieces. We have just seen that two crosses of the samesize may be cut out of a square in five pieces. The reader whosucceeded in solving this perhaps asked himself: "Can it be done infewer pieces?" This is just the sort of question that the true puzzlelover is always asking, and it is the right attitude for him to adopt.The answer to the question is that the puzzle may be solved in fourpieces--the fewest possible. This, then, is a new puzzle. Cut a squareinto four pieces that will form two Greek crosses of the same size.

[Illustration: FIG. 18.]

[Illustration: FIG. 19.]

[Illustration: FIG. 20.]

The solution is very beautiful. If you divide by points the sides of thesquare into three equal parts, the directions of the lines in Fig. 18will be quite obvious. If you cut along these lines, the pieces A and Bwill form the cross in Fig. 19 and the pieces C and D the similar crossin Fig. 20. In this square we have another form of Swastika.

The reader will here appreciate the truth of my remark to the effectthat it is easier to find the directions of the cuts when transforming across to a square than when converting a square into a cross. Thus, inFigs. 6, 8, and 10 the directions of the cuts are more obvious than inFig. 14, where we had first to divide the sides of the square into sixequal parts, and in Fig. 18, where we divide them into three equalparts. Then, supposing you were required to cut two equal Greek crosses,each into two pieces, to form a square, a glance at Figs. 19 and 20 willshow how absurdly more easy this is than the reverse puzzle of cuttingthe square to make two crosses.

Referring to my remarks on "fallacies," I will now give a little exampleof these "solutions" that are not solutions. Some years ago a youngcorrespondent sent me what he evidently thought was a brilliant newdiscovery--the transforming of a square into a Greek cross in fourpieces by cuts all parallel to the sides of the square. I give hisattempt in Figs. 21 and 22, where it will be seen that the four piecesdo not form a symmetrical Greek cross, because the four arms are notreally squares but oblongs. To make it a true Greek cross we shouldrequire the additions that I have indicated with dotted lines. Of coursehis solution produces a cross, but it is not the symmetrical Greekvariety required by the conditions of the puzzle. My young friendthought his attempt was "near enough" to be correct; but if he bought apenny apple with a sixpence he probably would not have thought it "nearenough" if he had been given only fourpence change. As the readeradvances he will realize the importance of this question of exactitude.

[Illustration: FIG. 21.]

[Illustration: FIG. 22.]

In these cutting-out puzzles it is necessary not only to get thedirections of the cutting lines as correct as possible, but to rememberthat these lines have no width. If after cutting up one of the crossesin a manner indicated in these articles you find that the pieces do notexactly fit to form a square, you may be certain that the fault isentirely your own. Either your cross was not exactly drawn, or your cutswere not made quite in the right directions, or (if you used wood and afret-saw) your saw was not sufficiently fine. If you cut out the puzzlesin paper with scissors, or in cardboard with a penknife, no material islost; but with a saw, however fine, there is a certain loss. In the caseof most puzzles this slight loss is not sufficient to be appreciable,if the puzzle is cut out on a large scale, but there have beeninstances where I have found it desirable to draw and cut out each partseparately--not from one diagram--in order to produce a perfect result.

[Illustration: FIG. 23.]

[Illustration: FIG. 24.]

Now for another puzzle. If you have cut out the five pieces indicated inFig. 14, you will find that these can be put together so as to form thecurious cross shown in Fig. 23. So if I asked you to cut Fig. 24 intofive pieces to form either a square or two equal Greek crosses you wouldknow how to do it. You would make the cuts as in Fig. 23, and place themtogether as in Figs. 14 and 15. But I want something better than that,and it is this. Cut Fig. 24 into only four pieces that will fit togetherand form a square.

[Illustration: FIG. 25.]

[Illustration: FIG. 26.]

The solution to the puzzle is shown in Figs. 25 and 26. The direction ofthe cut dividing A and C in the first diagram is very obvious, and thesecond cut is made at right angles to it. That the four pieces shouldfit together and form a square will surprise the novice, who will dowell to study the puzzle with some care, as it is most instructive.

I will now explain the beautiful rule by which we determine the size ofa square that shall have the same area as a Greek cross, for it isapplicable, and necessary, to the solution of almost every dissectionpuzzle that we meet with. It was first discovered by the philosopherPythagoras, who died 500 B.C., and is the 47th proposition of Euclid.The young reader who knows nothing of the elements of geometry will getsome idea of the fascinating character of that science. The triangle ABCin Fig. 27 is what we call a right-angled triangle, because the side BCis at right angles to the side AB. Now if we build up a square on eachside of the triangle, the squares on AB and BC will together be exactlyequal to the square on the long side AC, which we call the hypotenuse.This is proved in the case I have given by subdividing the three squaresinto cells of equal dimensions.

[Illustration: FIG. 27.]

[Illustration: FIG. 28.]

It will be seen that 9 added to 16 equals 25, the number of cells in thelarge square. If you make triangles with the sides 5, 12 and 13, or with8, 15 and 17, you will get similar arithmetical proofs, for these areall "rational" right-angled triangles, but the law is equally true forall cases. Supposing we cut off the lower arm of a Greek cross and placeit to the left of the upper arm, as in Fig. 28, then the square on EFadded to the square on DE exactly equals a square on DF. Therefore weknow that the square of DF will contain the same area as the cross. Thisfact we have proved practically by the solutions of the earlier puzzlesof this series. But whatever length we give to DE and EF, we can nevergive the exact length of DF in numbers, because the triangle is not a"rational" one. But the law is none the less geometrically true.

[Illustration: FIG. 29.]

[Illustration: FIG. 30.]

Now look at Fig. 29, and you will see an elegant method for cutting apiece of wood of the shape of two squares (of any relative dimensions)into three pieces that will fit together and form a single square. Ifyou mark off the distance _ab_ equal to the side _cd_ the directions ofthe cuts are very evident. From what we have just been considering, youwill at once see why _bc_ must be the length of the side of the newsquare. Make the experiment as often as you like, taking differentrelative proportions for the two squares, and you will find the rulealways come true. If you make the two squares of exactly the same size,you will see that the diagonal of any square is always the side of asquare that is twice the size. All this, which is so simple that anybodycan understand it, is very essential to the solving of cutting-outpuzzles. It is in fact the key to most of them. And it is all sobeautiful that it seems a pity that it should not be familiar toeverybody.

We will now go one step further and deal with the half-square. Take asquare and cut it in half diagonally. Now try to discover how to cutthis triangle into four pieces that will form a Greek cross. Thesolution is shown in Figs. 31 and 32. In this case it will be seen thatwe divide two of the sides of the triangle into three equal parts andthe long side into four equal parts. Then the direction of the cuts willbe easily found. It is a pretty puzzle, and a little more difficult thansome of the others that I have given. It should be noted again that itwould have been much easier to locate the cuts in the reverse puzzle ofcutting the cross to form a half-square triangle.

[Illustration: FIG. 31.]

[Illustration: FIG. 32.]

[Illustration: FIG. 33.]

[Illustration: FIG. 34.]

Another ideal that the puzzle maker always keeps in mind is to contrivethat there shall, if possible, be only one correct solution. Thus, inthe case of the first puzzle, if we only require that a Greek crossshall be cut into four pieces to form a square, there is, as I haveshown, an infinite number of different solutions. It makes a betterpuzzle to add the condition that all the four pieces shall be of thesame size and shape, because it can then be solved in only one way, asin Figs. 8 and 9. In this way, too, a puzzle that is too easy to beinteresting may be improved by such an addition. Let us take an example.We have seen in Fig. 28 that Fig. 33 can be cut into two pieces to forma Greek cross. I suppose an intelligent child would do it in fiveminutes. But suppose we say that the puzzle has to be solved with apiece of wood that has a bad knot in the position shown in Fig. 33--aknot that we must not attempt to cut through--then a solution in twopieces is barred out, and it becomes a more interesting puzzle to solveit in three pieces. I have shown in Figs. 33 and 34 one way of doingthis, and it will be found entertaining to discover other ways of doingit. Of course I could bar out all these other ways by introducing moreknots, and so reduce the puzzle to a single solution, but it would thenbe overloaded with conditions.

And this brings us to another point in seeking the ideal. Do notoverload your conditions, or you will make your puzzle too complex to beinteresting. The simpler the conditions of a puzzle are, the better. Thesolution may be as complex and difficult as you like, or as happens, butthe conditions ought to be easily understood, or people will not attempta solution.

If the reader were now asked "to cut a half-square into as few pieces aspossible to form a Greek cross," he would probably produce our solution,Figs. 31-32, and confidently claim that he had solved the puzzlecorrectly. In this way he would be wrong, because it is not now statedthat the square is to be divided diagonally. Although we should alwaysobserve the exact conditions of a puzzle we must not read into itconditions that are not there. Many puzzles are based entirely on thetendency that people have to do this.

The very first essential in solving a puzzle is to be sure that youunderstand the exact conditions. Now, if you divided your square in halfso as to produce Fig. 35 it is possible to cut it into as few as threepieces to form a Greek cross. We thus save a piece.

I give another puzzle in Fig. 36. The dotted lines are added merely toshow the correct proportions of the figure--a square of 25 cells withthe four corner cells cut out. The puzzle is to cut this figure intofive pieces that will form a Greek cross (entire) and a square.

[Illustration: FIG. 35.]

[Illustration: FIG. 36.]

The solution to the first of the two puzzles last given--to cut arectangle of the shape of a half-square into three pieces that will forma Greek cross--is shown in Figs. 37 and 38. It will be seen that wedivide the long sides of the oblong into six equal parts and the shortsides into three equal parts, in order to get the points that willindicate the direction of the cuts. The reader should compare thissolution with some of the previous illustrations. He will see, forexample, that if we continue the cut that divides B and C in the cross,we get Fig. 15.

[Illustration: FIG. 37.]

[Illustration: FIG. 38.]

The other puzzle, like the one illustrated in Figs. 12 and 13, will showhow useful a little arithmetic may sometimes prove to be in the solutionof dissection puzzles. There are twenty-one of those little square cellsinto which our figure is subdivided, from which we have to form both asquare and a Greek cross. Now, as the cross is built up of five squares,and 5 from 21 leaves 16--a square number--we ought easily to be led tothe solution shown in Fig. 39. It will be seen that the cross is cut outentire, while the four remaining pieces form the square in Fig. 40.

[Illustration: FIG. 39]

[Illustration: FIG. 40]

Of course a half-square rectangle is the same as a double square, or twoequal squares joined together. Therefore, if you want to solve thepuzzle of cutting a Greek cross into four pieces to form two separatesquares of the same size, all you have to do is to continue the shortcut in Fig. 38 right across the cross, and you will have four pieces ofthe same size and shape. Now divide Fig. 37 into two equal squares by ahorizontal cut midway and you will see the four pieces forming the twosquares.

[Illustration: FIG. 41]

Cut a Greek cross into five pieces that will form two separate squares,one of which shall contain half the area of one of the arms of thecross. In further illustration of what I have already written, if thetwo squares of the same size A B C D and B C F E, in Fig. 41, are cut inthe manner indicated by the dotted lines, the four pieces will form thelarge square A G E C. We thus see that the diagonal A C is the side of asquare twice the size of A B C D. It is also clear that half thediagonal of any square is equal to the side of a square of half thearea. Therefore, if the large square in the diagram is one of the armsof your cross, the small square is the size of one of the squaresrequired in the puzzle.

The solution is shown in Figs. 42 and 43. It will be seen that the smallsquare is cut out whole and the large square composed of the four piecesB, C, D, and E. After what I have written, the reader will have nodifficulty in seeing that the square A is half the size of one of thearms of the cross, because the length of the diagonal of the former isclearly the same as the side of the latter. The thing is nowself-evident. I have thus tried to show that some of these puzzles thatmany people are apt to regard as quite wonderful and bewildering, arereally not difficult if only we use a little thought and judgment. Inconclusion of this particular subject I will give four Greek crosspuzzles, with detached solutions.

142.--THE SILK PATCHWORK.

The lady members of the Wilkinson family had made a simple patchworkquilt, as a small Christmas present, all composed of square pieces ofthe same size, as shown in the illustration. It only lacked the fourcorner pieces to make it complete. Somebody pointed out to them that ifyou unpicked the Greek cross in the middle and then cut the stitchesalong the dark joins, the four pieces all of the same size and shapewould fit together and form a square. This the reader knows, from thesolution in Fig. 39, is quite easily done. But George Wilkinson suddenlysuggested to them this poser. He said, "Instead of picking out the crossentire, and forming the square from four equal pieces, can you cut out asquare entire and four equal pieces that will form a perfect Greekcross?" The puzzle is, of course, now quite easy.

143.--TWO CROSSES FROM ONE.

Cut a Greek cross into five pieces that will form two such crosses, bothof the same size. The solution of this puzzle is very beautiful.

144.--THE CROSS AND THE TRIANGLE.

Cut a Greek cross into six pieces that will form an equilateraltriangle. This is another hard problem, and I will state here that asolution is practically impossible without a previous knowledge of mymethod of transforming an equilateral triangle into a square (see No.26, "Canterbury Puzzles").

145.--THE FOLDED CROSS.

Cut out of paper a Greek cross; then so fold it that with a singlestraight cut of the scissors the four pieces produced will form asquare.

VARIOUS DISSECTION PUZZLES.

We will now consider a small miscellaneous selection of cutting-outpuzzles, varying in degrees of difficulty.

146.--AN EASY DISSECTION PUZZLE.

First, cut out a piece of paper or cardboard of the shape shown in theillustration. It will be seen at once that the proportions are simplythose of a square attached to half of another similar square, divideddiagonally. The puzzle is to cut it into four pieces all of preciselythe same size and shape.

147.--AN EASY SQUARE PUZZLE.

If you take a rectangular piece of cardboard, twice as long as it isbroad, and cut it in half diagonally, you will get two of the piecesshown in the illustration. The puzzle is with five such pieces of equalsize to form a square. One of the pieces may be cut in two, but theothers must be used intact.

148.--THE BUN PUZZLE.

THE three circles represent three buns, and it is simply required toshow how these may be equally divided among four boys. The buns must beregarded as of equal thickness throughout and of equal thickness to eachother. Of course, they must be cut into as few pieces as possible. Tosimplify it I will state the rather surprising fact that only fivepieces are necessary, from which it will be seen that one boy gets hisshare in two pieces and the other three receive theirs in a singlepiece. I am aware that this statement "gives away" the puzzle, but itshould not destroy its interest to those who like to discover the"reason why."

149.--THE CHOCOLATE SQUARES.

Here is a slab of chocolate, indented at the dotted lines so that thetwenty squares can be easily separated. Make a copy of the slab in paperor cardboard and then try to cut it into nine pieces so that they willform four perfect squares all of exactly the same size.

150.--DISSECTING A MITRE.

The figure that is perplexing the carpenter in the illustrationrepresents a mitre. It will be seen that its proportions are those of asquare with one quarter removed. The puzzle is to cut it into fivepieces that will fit together and form a perfect square. I show anattempt, published in America, to perform the feat in four pieces, basedon what is known as the "step principle," but it is a fallacy.

[Illustration]

We are told first to cut oft the pieces 1 and 2 and pack them into thetriangular space marked off by the dotted line, and so form a rectangle.

So far, so good. Now, we are directed to apply the old step principle,as shown, and, by moving down the piece 4 one step, form the requiredsquare. But, unfortunately, it does _not_ produce a square: only anoblong. Call the three long sides of the mitre 84 in. each. Then, beforecutting the steps, our rectangle in three pieces will be 84 x 63. Thesteps must be 101/2 in. in height and 12 in. in breadth. Therefore, bymoving down a step we reduce by 12 in. the side 84 in. and increase by101/2 in. the side 63 in. Hence our final rectangle must be 72 in. x 731/2in., which certainly is not a square! The fact is, the step principlecan only be applied to rectangles with sides of particular relativelengths. For example, if the shorter side in this case were 61+5/7(instead of 63), then the step method would apply. For the steps wouldthen be 10+2/7 in. in height and 12 in. in breadth. Note that 61+5/7 x84 = the square of 72. At present no solution has been found in fourpieces, and I do not believe one possible.

151.--THE JOINER'S PROBLEM.

I have often had occasion to remark on the practical utility of puzzles,arising out of an application to the ordinary affairs of life of thelittle tricks and "wrinkles" that we learn while solving recreationproblems.

[Illustration]

The joiner, in the illustration, wants to cut the piece of wood into asfew pieces as possible to form a square table-top, without any waste ofmaterial. How should he go to work? How many pieces would you require?

152.--ANOTHER JOINER'S PROBLEM.

[Illustration]

A joiner had two pieces of wood of the shapes and relative proportionsshown in the diagram. He wished to cut them into as few pieces aspossible so that they could be fitted together, without waste, to form aperfectly square table-top. How should he have done it? There is nonecessity to give measurements, for if the smaller piece (which is halfa square) be made a little too large or a little too small it will notaffect the method of solution.

153--A CUTTING-OUT PUZZLE.

Here is a little cutting-out poser. I take a strip of paper, measuringfive inches by one inch, and, by cutting it into five pieces, the partsfit together and form a square, as shown in the illustration. Now, it isquite an interesting puzzle to discover how we can do this in only fourpieces.

[Illustration]

154.--MRS. HOBSON'S HEARTHRUG.

[Illustration]

Mrs. Hobson's boy had an accident when playing with the fire, and burnttwo of the corners of a pretty hearthrug. The damaged corners have beencut away, and it now has the appearance and proportions shown in mydiagram. How is Mrs. Hobson to cut the rug into the fewest possiblepieces that will fit together and form a perfectly square rug? It willbe seen that the rug is in the proportions 36 x 27 (it does not matterwhether we say inches or yards), and each piece cut away measured 12 and6 on the outside.

155.--THE PENTAGON AND SQUARE.

I wonder how many of my readers, amongst those who have not given anyclose attention to the elements of geometry, could draw a regularpentagon, or five-sided figure, if they suddenly required to do so. Aregular hexagon, or six-sided figure, is easy enough, for everybodyknows that all you have to do is to describe a circle and then, takingthe radius as the length of one of the sides, mark off the six pointsround the circumference. But a pentagon is quite another matter. So, asmy puzzle has to do with the cutting up of a regular pentagon, it willperhaps be well if I first show my less experienced readers how thisfigure is to be correctly drawn. Describe a circle and draw the twolines H B and D G, in the diagram, through the centre at right angles.Now find the point A, midway between C and B. Next place the point ofyour compasses at A and with the distance A D describe the arc cutting HB at E. Then place the point of your compasses at D and with thedistance D E describe the arc cutting the circumference at F. Now, D Fis one of the sides of your pentagon, and you have simply to mark offthe other sides round the circle. Quite simple when you know how, butotherwise somewhat of a poser.

[Illustration]

Having formed your pentagon, the puzzle is to cut it into the fewestpossible pieces that will fit together and form a perfect square.

[Illustration]

156.--THE DISSECTED TRIANGLE.

A good puzzle is that which the gentleman in the illustration is showingto his friends. He has simply cut out of paper an equilateraltriangle--that is, a triangle with all its three sides of the samelength. He proposes that it shall be cut into five pieces in such a waythat they will fit together and form either two or three smallerequilateral triangles, using all the material in each case. Can youdiscover how the cuts should be made?

Remember that when you have made your five pieces, you must be able, asdesired, to put them together to form either the single originaltriangle or to form two triangles or to form three triangles--allequilateral.

157.--THE TABLE-TOP AND STOOLS.

I have frequently had occasion to show that the published answers to agreat many of the oldest and most widely known puzzles are either quiteincorrect or capable of improvement. I propose to consider the old poserof the table-top and stools that most of my readers have probably seenin some form or another in books compiled for the recreation ofchildhood.

The story is told that an economical and ingenious schoolmaster oncewished to convert a circular table-top, for which he had no use, intoseats for two oval stools, each with a hand-hole in the centre. Heinstructed the carpenter to make the cuts as in the illustration andthen join the eight pieces together in the manner shown. So impressedwas he with the ingenuity of his performance that he set the puzzle tohis geometry class as a little study in dissection. But the remainder ofthe story has never been published, because, so it is said, it was acharacteristic of the principals of academies that they would neveradmit that they could err. I get my information from a descendant of theoriginal boy who had most reason to be interested in the matter.

The clever youth suggested modestly to the master that the hand-holeswere too big, and that a small boy might perhaps fall through them. Hetherefore proposed another way of making the cuts that would get overthis objection. For his impertinence he received such severechastisement that he became convinced that the larger the hand-hole inthe stools the more comfortable might they be.

[Illustration]

Now what was the method the boy proposed?

Can you show how the circular table-top may be cut into eight piecesthat will fit together and form two oval seats for stools (each ofexactly the same size and shape) and each having similar hand-holes ofsmaller dimensions than in the case shown above? Of course, all the woodmust be used.

158.--THE GREAT MONAD.

[Illustration]

Here is a symbol of tremendous antiquity which is worthy of notice. Itis borne on the Korean ensign and merchant flag, and has been adopted asa trade sign by the Northern Pacific Railroad Company, though probablyfew are aware that it is the Great Monad, as shown in the sketch below.This sign is to the Chinaman what the cross is to the Christian. It isthe sign of Deity and eternity, while the two parts into which thecircle is divided are called the Yin and the Yan--the male and femaleforces of nature. A writer on the subject more than three thousand yearsago is reported to have said in reference to it: "The illimitableproduces the great extreme. The great extreme produces the twoprinciples. The two principles produce the four quarters, and from thefour quarters we develop the quadrature of the eight diagrams ofFeuh-hi." I hope readers will not ask me to explain this, for I have notthe slightest idea what it means. Yet I am persuaded that for ages thesymbol has had occult and probably mathematical meanings for theesoteric student.

I will introduce the Monad in its elementary form. Here are three easyquestions respecting this great symbol:--

(I.) Which has the greater area, the inner circle containing the Yin andthe Yan, or the outer ring?

(II.) Divide the Yin and the Yan into four pieces of the same size andshape by one cut.

(III.) Divide the Yin and the Yan into four pieces of the same size, butdifferent shape, by one straight cut.

159.--THE SQUARE OF VENEER.

The following represents a piece of wood in my possession, 5 in. square.By markings on the surface it is divided into twenty-five square inches.I want to discover a way of cutting this piece of wood into the fewestpossible pieces that will fit together and form two perfect squares ofdifferent sizes and of known dimensions. But, unfortunately, at everyone of the sixteen intersections of the cross lines a small nail hasbeen driven in at some time or other, and my fret-saw will be injured ifit comes in contact with any of these. I have therefore to find a methodof doing the work that will not necessitate my cutting through any ofthose sixteen points. How is it to be done? Remember, the exactdimensions of the two squares must be given.

[Illustration]

160.--THE TWO HORSESHOES.

[Illustration]

Why horseshoes should be considered "lucky" is one of those thingswhich no man can understand. It is a very old superstition, and JohnAubrey (1626-1700) says, "Most houses at the West End of London have ahorseshoe on the threshold." In Monmouth Street there were seventeen in1813 and seven so late as 1855. Even Lord Nelson had one nailed to themast of the ship _Victory_. To-day we find it more conducive to "goodluck" to see that they are securely nailed on the feet of the horse weare about to drive.

Nevertheless, so far as the horseshoe, like the Swastika and otheremblems that I have had occasion at times to deal with, has served tosymbolize health, prosperity, and goodwill towards men, we may welltreat it with a certain amount of respectful interest. May there not,moreover, be some esoteric or lost mathematical mystery concealed in theform of a horseshoe? I have been looking into this matter, and I wish todraw my readers' attention to the very remarkable fact that the pair ofhorseshoes shown in my illustration are related in a striking andbeautiful manner to the circle, which is the symbol of eternity. Ipresent this fact in the form of a simple problem, so that it may beseen how subtly this relation has been concealed for ages and ages. Myreaders will, I know, be pleased when they find the key to the mystery.

Cut out the two horseshoes carefully round the outline and then cut theminto four pieces, all different in shape, that will fit together andform a perfect circle. Each shoe must be cut into two pieces and all thepart of the horse's hoof contained within the outline is to be used andregarded as part of the area.

161.--THE BETSY ROSS PUZZLE.

A correspondent asked me to supply him with the solution to an oldpuzzle that is attributed to a certain Betsy Ross, of Philadelphia, whoshowed it to George Washington. It consists in so folding a piece ofpaper that with one clip of the scissors a five-pointed star of Freedommay be produced. Whether the story of the puzzle's origin is a true oneor not I cannot say, but I have a print of the old house in Philadelphiawhere the lady is said to have lived, and I believe it still standsthere. But my readers will doubtless be interested in the little poser.

Take a circular piece of paper and so fold it that with one cut of thescissors you can produce a perfect five-pointed star.

162.--THE CARDBOARD CHAIN.

[Illustration]

Can you cut this chain out of a piece of cardboard without any joinwhatever? Every link is solid; without its having been split andafterwards joined at any place. It is an interesting old puzzle that Ilearnt as a child, but I have no knowledge as to its inventor.

163.--THE PAPER BOX.

It may be interesting to introduce here, though it is not strictly apuzzle, an ingenious method for making a paper box.

Take a square of stout paper and by successive foldings make all thecreases indicated by the dotted lines in the illustration. Then cut awaythe eight little triangular pieces that are shaded, and cut through thepaper along the dark lines. The second illustration shows the box halffolded up, and the reader will have no difficulty in effecting itscompletion. Before folding up, the reader might cut out the circularpiece indicated in the diagram, for a purpose I will now explain.

This box will be found to serve excellently for the production of vortexrings. These rings, which were discussed by Von Helmholtz in 1858, aremost interesting, and the box (with the hole cut out) will produce themto perfection. Fill the box with tobacco smoke by blowing it gentlythrough the hole. Now, if you hold it horizontally, and softly tap theside that is opposite to the hole, an immense number of perfect ringscan be produced from one mouthful of smoke. It is best that there shouldbe no currents of air in the room. People often do not realise thatthese rings are formed in the air when no smoke is used. The smoke onlymakes them visible. Now, one of these rings, if properly directed on itscourse, will travel across the room and put out the flame of a candle,and this feat is much more striking if you can manage to do it withoutthe smoke. Of course, with a little practice, the rings may be blownfrom the mouth, but the box produces them in much greater perfection,and no skill whatever is required. Lord Kelvin propounded the theorythat matter may consist of vortex rings in a fluid that fills all space,and by a development of the hypothesis he was able to explain chemicalcombination.

[Illustration:

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]

[Illustration]

164.--THE POTATO PUZZLE.

Take a circular slice of potato, place it on the table, and see into howlarge a number of pieces you can divide it with six cuts of a knife. Ofcourse you must not readjust the pieces or pile them after a cut. Whatis the greatest number of pieces you can make?

[Illustration:

-------- / \ 1/ \ / \ 2 \/ 3 / \ / \ /\ / \ / \ 4 \/ 5\/ 6 / \ | \ /\ /\ / | \ 7\/ 8\/ 9\/10 / \ /\ /\ /\ / \/11\/12\/13\/ \ /\ /\ / \/14\/15\/ \ /\ / \/16\/ -----

]

The illustration shows how to make sixteen pieces. This can, of course,be easily beaten.

165.--THE SEVEN PIGS.

[Illustration]

+------------------------------+ | | | P | | | | P | | P | | P | | P | | P | | P | | | +------------------------------+

Here is a little puzzle that was put to one of the sons of Erin theother day and perplexed him unduly, for it is really quite easy. It willbe seen from the illustration that he was shown a sketch of a square pencontaining seven pigs. He was asked how he would intersect the pen withthree straight fences so as to enclose every pig in a separate sty. Inother words, all you have to do is to take your pencil and, with threestraight strokes across the square, enclose each pig separately. Nothingcould be simpler.

[Illustration]

The Irishman complained that the pigs would not keep still while he wasputting up the fences. He said that they would all flock together, orone obstinate beast would go into a corner and flock all by himself. Itwas pointed out to him that for the purposes of the puzzle the pigs werestationary. He answered that Irish pigs are not stationery--they arepork. Being persuaded to make the attempt, he drew three lines, one ofwhich cut through a pig. When it was explained that this is not allowed,he protested that a pig was no use until you cut its throat. "Begorra,if it's bacon ye want without cutting your pig, it will be all gammon."We will not do the Irishman the injustice of suggesting that themiserable pun was intentional. However, he failed to solve the puzzle.Can you do it?

166.--THE LANDOWNER'S FENCES.

The landowner in the illustration is consulting with his bailiff over arather puzzling little question. He has a large plan of one of hisfields, in which there are eleven trees. Now, he wants to divide thefield into just eleven enclosures by means of straight fences, so thatevery enclosure shall contain one tree as a shelter for his cattle. Howis he to do it with as few fences as possible? Take your pencil and drawstraight lines across the field until you have marked off the elevenenclosures (and no more), and then see how many fences you require. Ofcourse the fences may cross one another.

167.--THE WIZARD'S CATS.

[Illustration]

A wizard placed ten cats inside a magic circle as shown in ourillustration, and hypnotized them so that they should remain stationaryduring his pleasure. He then proposed to draw three circles inside thelarge one, so that no cat could approach another cat without crossing amagic circle. Try to draw the three circles so that every cat has itsown enclosure and cannot reach another cat without crossing a line.

168.--THE CHRISTMAS PUDDING.

[Illustration]

"Speaking of Christmas puddings," said the host, as he glanced at theimposing delicacy at the other end of the table. "I am reminded of thefact that a friend gave me a new puzzle the other day respecting one.Here it is," he added, diving into his breast pocket.

"'Problem: To find the contents,' I suppose," said the Eton boy.

"No; the proof of that is in the eating. I will read you theconditions."

"'Cut the pudding into two parts, each of exactly the same size andshape, without touching any of the plums. The pudding is to be regardedas a flat disc, not as a sphere.'"

"Why should you regard a Christmas pudding as a disc? And why should anyreasonable person ever wish to make such an accurate division?" askedthe cynic.

"It is just a puzzle--a problem in dissection." All in turn had a lookat the puzzle, but nobody succeeded in solving it. It is a littledifficult unless you are acquainted with the principle involved in themaking of such puddings, but easy enough when you know how it is done.

169.--A TANGRAM PARADOX.

Many pastimes of great antiquity, such as chess, have so developed andchanged down the centuries that their original inventors would scarcelyrecognize them. This is not the case with Tangrams, a recreation thatappears to be at least four thousand years old, that has apparentlynever been dormant, and that has not been altered or "improved upon"since the legendary Chinaman Tan first cut out the seven pieces shown inDiagram I. If you mark the point B, midway between A and C, on one sideof a square of any size, and D, midway between C and E, on an adjoiningside, the direction of the cuts is too obvious to need furtherexplanation. Every design in this article is built up from the sevenpieces of blackened cardboard. It will at once be understood that thepossible combinations are infinite.

[Illustration]

The late Mr. Sam Loyd, of New York, who published a small book of veryingenious designs, possessed the manuscripts of the late Mr. Challenor,who made a long and close study of Tangrams. This gentleman, it is said,records that there were originally seven books of Tangrams, compiled inChina two thousand years before the Christian era. These books are sorare that, after forty years' residence in the country, he onlysucceeded in seeing perfect copies of the first and seventh volumes withfragments of the second. Portions of one of the books, printed in goldleaf upon parchment, were found in Peking by an English soldier and soldfor three hundred pounds.

A few years ago a little book came into my possession, from the libraryof the late Lewis Carroll, entitled _The Fashionable Chinese Puzzle_. Itcontains three hundred and twenty-three Tangram designs, mostlynondescript geometrical figures, to be constructed from the sevenpieces. It was "Published by J. and E. Wallis, 42 Skinner Street, and J.Wallis, Jun., Marine Library, Sidmouth" (South Devon). There is no date,but the following note fixes the time of publication pretty closely:"This ingenious contrivance has for some time past been the favouriteamusement of the ex-Emperor Napoleon, who, being now in a debilitatedstate and living very retired, passes many hours a day in thusexercising his patience and ingenuity." The reader will find, as did thegreat exile, that much amusement, not wholly uninstructive, may bederived from forming the designs of others. He will find many of theillustrations to this article quite easy to build up, and some ratherdifficult. Every picture may thus be regarded as a puzzle.

But it is another pastime altogether to create new and original designsof a pictorial character, and it is surprising what extraordinary scopethe Tangrams afford for producing pictures of real life--angular andoften grotesque, it is true, but full of character. I give an example ofa recumbent figure (2) that is particularly graceful, and only needssome slight reduction of its angularities to produce an entirelysatisfactory outline.

As I have referred to the author of _Alice in Wonderland_, I give alsomy designs of the March Hare (3) and the Hatter (4). I also give anattempt at Napoleon (5), and a very excellent Red Indian with his Squawby Mr. Loyd (6 and 7). A large number of other designs will be found inan article by me in _The Strand Magazine_ for November, 1908.

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

On the appearance of this magazine article, the late Sir James Murray,the eminent philologist, tried, with that amazing industry thatcharacterized all his work, to trace the word "tangram" to its source.At length he wrote as follows:--"One of my sons is a professor in theAnglo-Chinese college at Tientsin. Through him, his colleagues, and hisstudents, I was able to make inquiries as to the alleged Tan amongChinese scholars. Our Chinese professor here (Oxford) also took aninterest in the matter and obtained information from the secretary ofthe Chinese Legation in London, who is a very eminent representative ofthe Chinese literati."

[Illustration: 5]

"The result has been to show that the man Tan, the god Tan, and the'Book of Tan' are entirely unknown to Chinese literature, history, ortradition. By most of the learned men the name, or allegation of theexistence, of these had never been heard of. The puzzle is, of course,well known. It is called in Chinese _ch'i ch'iao t'u_; literally,'seven-ingenious-plan' or 'ingenious-puzzle figure of seven pieces.' Noname approaching 'tangram,' or even 'tan,' occurs in Chinese, and theonly suggestions for the latter were the Chinese _t'an_, 'to extend'; or_t'ang_, Cantonese dialect for 'Chinese.' It was suggested that probablysome American or Englishman who knew a little Chinese or Cantonese,wanting a name for the puzzle, might concoct one out of one of thesewords and the European ending 'gram.' I should say the name 'tangram'was probably invented by an American some little time before 1864 andafter 1847, but I cannot find it in print before the 1864 edition ofWebster. I have therefore had to deal very shortly with the word in thedictionary, telling what it is applied to and what conjectures orguesses have been made at the name, and giving a few quotations, onefrom your own article, which has enabled me to make more of the subjectthan I could otherwise have done."

[Illustration: 6]

[Illustration: 7]

Several correspondents have informed me that they possess, or hadpossessed, specimens of the old Chinese books. An American gentlemanwrites to me as follows:--"I have in my possession a book made of tissuepaper, printed in black (with a Chinese inscription on the front page),containing over three hundred designs, which belongs to the box of'tangrams,' which I also own. The blocks are seven in number, made ofmother-of-pearl, highly polished and finely engraved on either side.These are contained in a rosewood box 2+1/8 in. square. My great uncle,----, was one of the first missionaries to visit China. This box andbook, along with quite a collection of other relics, were sent to mygrandfather and descended to myself."

My correspondent kindly supplied me with rubbings of the Tangrams, fromwhich it is clear that they are cut in the exact proportions that I haveindicated. I reproduce the Chinese inscription (8) for this reason. Theowner of the book informs me that he has submitted it to a number ofChinamen in the United States and offered as much as a dollar for atranslation. But they all steadfastly refused to read the words,offering the lame excuse that the inscription is Japanese. Natives ofJapan, however, insist that it is Chinese. Is there something occult andesoteric about Tangrams, that it is so difficult to lift the veil?Perhaps this page will come under the eye of some reader acquainted withthe Chinese language, who will supply the required translation, whichmay, or may not, throw a little light on this curious question.

[Illustration: 8]

By using several sets of Tangrams at the same time we may construct moreambitious pictures. I was advised by a friend not to send my picture, "AGame of Billiards" (9), to the Academy. He assured me that it would notbe accepted because the "judges are so hide-bound by convention."Perhaps he was right, and it will be more appreciated byPost-impressionists and Cubists. The players are considering a verydelicate stroke at the top of the table. Of course, the two men, thetable, and the clock are formed from four sets of Tangrams. My secondpicture is named "The Orchestra" (10), and it was designed for thedecoration of a large hall of music. Here we have the conductor, thepianist, the fat little cornet-player, the left-handed player of thedouble-bass, whose attitude is life-like, though he does stand at anunusual distance from his instrument, and the drummer-boy, with hisimposing music-stand. The dog at the back of the pianoforte is nothowling: he is an appreciative listener.

[Illustration: 9]

[Illustration: 10]

One remarkable thing about these Tangram pictures is that they suggestto the imagination such a lot that is not really there. Who, forexample, can look for a few minutes at Lady Belinda (11) and the Dutchgirl (12) without soon feeling the haughty expression in the one caseand the arch look in the other? Then look again at the stork (13), andsee how it is suggested to the mind that the leg is actually much moreslender than any one of the pieces employed. It is really an opticalillusion. Again, notice in the case of the yacht (14) how, by leavingthat little angular point at the top, a complete mast is suggested. Ifyou place your Tangrams together on white paper so that they do notquite touch one another, in some cases the effect is improved by thewhite lines; in other cases it is almost destroyed.

[Illustration: 11]

[Illustration: 12]

Finally, I give an example from the many curious paradoxes that onehappens upon in manipulating Tangrams. I show designs of two dignifiedindividuals (15 and 16) who appear to be exactly alike, except for thefact that one has a foot and the other has not. Now, both of thesefigures are made from the same seven Tangrams. Where does the second manget his foot from?

[Illustration: 13]

[Illustration: 14]

[Illustration: 15]

[Illustration: 16]

PATCHWORK PUZZLES.

"Of shreds and patches."--_Hamlet_, iii. 4.

170.--THE CUSHION COVERS.

[Illustration]

The above represents a square of brocade. A lady wishes to cut it infour pieces so that two pieces will form one perfectly square cushiontop, and the remaining two pieces another square cushion top. How is sheto do it? Of course, she can only cut along the lines that divide thetwenty-five squares, and the pattern must "match" properly without anyirregularity whatever in the design of the material. There is only oneway of doing it. Can you find it?

171.--THE BANNER PUZZLE.

[Illustration]

A Lady had a square piece of bunting with two lions on it, of which theillustration is an exactly reproduced reduction. She wished to cut thestuff into pieces that would fit together and form two square bannerswith a lion on each banner. She discovered that this could be done in asfew as four pieces. How did she manage it? Of course, to cut the BritishLion would be an unpardonable offence, so you must be careful that nocut passes through any portion of either of them. Ladies are informedthat no allowance whatever has to be made for "turnings," and no part ofthe material may be wasted. It is quite a simple little dissectionpuzzle if rightly attacked. Remember that the banners have to be perfectsquares, though they need not be both of the same size.

172.--MRS. SMILEY'S CHRISTMAS PRESENT.

Mrs. Smiley's expression of pleasure was sincere when her sixgranddaughters sent to her, as a Christmas present, a very prettypatchwork quilt, which they had made with their own hands. It wasconstructed of square pieces of silk material, all of one size, and asthey made a large quilt with fourteen of these little squares on eachside, it is obvious that just 196 pieces had been stitched into it. Now,the six granddaughters each contributed a part of the work in the formof a perfect square (all six portions being different in size), but inorder to join them up to form the square quilt it was necessary that thework of one girl should be unpicked into three separate pieces. Can youshow how the joins might have been made? Of course, no portion can beturned over.

[Illustration]

173.--MRS. PERKINS'S QUILT.

[Illustration]

It will be seen that in this case the square patchwork quilt is built upof 169 pieces. The puzzle is to find the smallest possible number ofsquare portions of which the quilt could be composed and show how theymight be joined together. Or, to put it the reverse way, divide thequilt into as few square portions as possible by merely cutting thestitches.

174.--THE SQUARES OF BROCADE.

[Illustration]

I happened to be paying a call at the house of a lady, when I took upfrom a table two lovely squares of brocade. They were beautifulspecimens of Eastern workmanship--both of the same design, a delicatechequered pattern.

"Are they not exquisite?" said my friend. "They were brought to me by acousin who has just returned from India. Now, I want you to give me alittle assistance. You see, I have decided to join them together so asto make one large square cushion-cover. How should I do this so as tomutilate the material as little as possible? Of course I propose to makemy cuts only along the lines that divide the little chequers."

[Illustration]

I cut the two squares in the manner desired into four pieces that wouldfit together and form another larger square, taking care that thepattern should match properly, and when I had finished I noticed thattwo of the pieces were of exactly the same area; that is, each of thetwo contained the same number of chequers. Can you show how the cutswere made in accordance with these conditions?

175--ANOTHER PATCHWORK PUZZLE.

[Illustration]

A lady was presented, by two of her girl friends, with the pretty piecesof silk patchwork shown in our illustration. It will be seen that bothpieces are made up of squares all of the same size--one 12 x 12 and theother 5 x 5. She proposes to join them together and make one squarepatchwork quilt, 13 x 13, but, of course, she will not cut any of thematerial--merely cut the stitches where necessary and join togetheragain. What perplexes her is this. A friend assures her that there needbe no more than four pieces in all to join up for the new quilt. Couldyou show her how this little needlework puzzle is to be solved in so fewpieces?

176.--LINOLEUM CUTTING.

[Illustration]

The diagram herewith represents two separate pieces of linoleum. Thechequered pattern is not repeated at the back, so that the pieces cannotbe turned over. The puzzle is to cut the two squares into four pieces sothat they shall fit together and form one perfect square 10 x 10, sothat the pattern shall properly match, and so that the larger pieceshall have as small a portion as possible cut from it.

177.--ANOTHER LINOLEUM PUZZLE.

[Illustration]

Can you cut this piece of linoleum into four pieces that will fittogether and form a perfect square? Of course the cuts may only be madealong the lines.

VARIOUS GEOMETRICAL PUZZLES.

"So various are the tastes of men." MARK AKENSIDE.

178.--THE CARDBOARD BOX.

This puzzle is not difficult, but it will be found entertaining todiscover the simple rule for its solution. I have a rectangularcardboard box. The top has an area of 120 square inches, the side 96square inches, and the end 80 square inches. What are the exactdimensions of the box?

179.--STEALING THE BELL-ROPES.

Two men broke into a church tower one night to steal the bell-ropes. Thetwo ropes passed through holes in the wooden ceiling high above them,and they lost no time in climbing to the top. Then one man drew hisknife and cut the rope above his head, in consequence of which he fellto the floor and was badly injured. His fellow-thief called out that itserved him right for being such a fool. He said that he should have doneas he was doing, upon which he cut the rope below the place at which heheld on. Then, to his dismay, he found that he was in no better plight,for, after hanging on as long as his strength lasted, he was compelledto let go and fall beside his comrade. Here they were both found thenext morning with their limbs broken. How far did they fall? One of theropes when they found it was just touching the floor, and when youpulled the end to the wall, keeping the rope taut, it touched a pointjust three inches above the floor, and the wall was four feet from therope when it hung at rest. How long was the rope from floor to ceiling?

180.--THE FOUR SONS.

Readers will recognize the diagram as a familiar friend of their youth.A man possessed a square-shaped estate. He bequeathed to his widow thequarter of it that is shaded off. The remainder was to be dividedequitably amongst his four sons, so that each should receive land ofexactly the same area and exactly similar in shape. We are shown howthis was done. But the remainder of the story is not so generally known.In the centre of the estate was a well, indicated by the dark spot, andBenjamin, Charles, and David complained that the division was not"equitable," since Alfred had access to this well, while they could notreach it without trespassing on somebody else's land. The puzzle is toshow how the estate is to be apportioned so that each son shall haveland of the same shape and area, and each have access to the wellwithout going off his own land.

[Illustration]

181.--THE THREE RAILWAY STATIONS.

As I sat in a railway carriage I noticed at the other end of thecompartment a worthy squire, whom I knew by sight, engaged inconversation with another passenger, who was evidently a friend of his.

"How far have you to drive to your place from the railway station?"asked the stranger.

"Well," replied the squire, "if I get out at Appleford, it is just thesame distance as if I go to Bridgefield, another fifteen miles fartheron; and if I changed at Appleford and went thirteen miles from there toCarterton, it would still be the same distance. You see, I amequidistant from the three stations, so I get a good choice of trains."

Now I happened to know that Bridgefield is just fourteen miles fromCarterton, so I amused myself in working out the exact distance that thesquire had to drive home whichever station he got out at. What was thedistance?

182.--THE GARDEN PUZZLE.

Professor Rackbrain tells me that he was recently smoking a friendlypipe under a tree in the garden of a country acquaintance. The gardenwas enclosed by four straight walls, and his friend informed him that hehad measured these and found the lengths to be 80, 45, 100, and 63 yardsrespectively. "Then," said the professor, "we can calculate the exactarea of the garden." "Impossible," his host replied, "because you canget an infinite number of different shapes with those four sides." "Butyou forget," Rackbrane said, with a twinkle in his eye, "that you toldme once you had planted this tree equidistant from all the four cornersof the garden." Can you work out the garden's area?

183.--DRAWING A SPIRAL.

If you hold the page horizontally and give it a quick rotary motionwhile looking at the centre of the spiral, it will appear to revolve.Perhaps a good many readers are acquainted with this little opticalillusion. But the puzzle is to show how I was able to draw this spiralwith so much exactitude without using anything but a pair of compassesand the sheet of paper on which the diagram was made. How would youproceed in such circumstances?

[Illustration]

184.--HOW TO DRAW AN OVAL.

Can you draw a perfect oval on a sheet of paper with one sweep of thecompasses? It is one of the easiest things in the world when you knowhow.

185.--ST. GEORGE'S BANNER.

At a celebration of the national festival of St. George's Day I wascontemplating the familiar banner of the patron saint of our country. Weall know the red cross on a white ground, shown in our illustration.This is the banner of St. George. The banner of St. Andrew (Scotland) isa white "St. Andrew's Cross" on a blue ground. That of St. Patrick(Ireland) is a similar cross in red on a white ground. These three areunited in one to form our Union Jack.

Now on looking at St. George's banner it occurred to me that thefollowing question would make a simple but pretty little puzzle.Supposing the flag measures four feet by three feet, how wide must thearm of the cross be if it is required that there shall be used just thesame quantity of red and of white bunting?

[Illustration]

186.--THE CLOTHES LINE PUZZLE.

A boy tied a clothes line from the top of each of two poles to the baseof the other. He then proposed to his father the following question. Asone pole was exactly seven feet above the ground and the other exactlyfive feet, what was the height from the ground where the two cordscrossed one another?

187.--THE MILKMAID PUZZLE.

[Illustration]

Here is a little pastoral puzzle that the reader may, at first sight, beled into supposing is very profound, involving deep calculations. He mayeven say that it is quite impossible to give any answer unless we aretold something definite as to the distances. And yet it is really quite"childlike and bland."

In the corner of a field is seen a milkmaid milking a cow, and on theother side of the field is the dairy where the extract has to bedeposited. But it has been noticed that the young woman always goes downto the river with her pail before returning to the dairy. Here thesuspicious reader will perhaps ask why she pays these visits to theriver. I can only reply that it is no business of ours. The alleged milkis entirely for local consumption.

"Where are you going to, my pretty maid?" "Down to the river, sir," she said. "I'll _not_ choose your dairy, my pretty maid." "Nobody axed you, sir," she said.

If one had any curiosity in the matter, such an independent spirit wouldentirely disarm one. So we will pass from the point of commercialmorality to the subject of the puzzle.

Draw a line from the milking-stool down to the river and thence to thedoor of the dairy, which shall indicate the shortest possible route forthe milkmaid. That is all. It is quite easy to indicate the exact spoton the bank of the river to which she should direct her steps if shewants as short a walk as possible. Can you find that spot?

188.--THE BALL PROBLEM.

[Illustration]

A stonemason was engaged the other day in cutting out a round ball forthe purpose of some architectural decoration, when a smart schoolboycame upon the scene.

"Look here," said the mason, "you seem to be a sharp youngster, can youtell me this? If I placed this ball on the level ground, how many otherballs of the same size could I lay around it (also on the ground) sothat every ball should touch this one?"

The boy at once gave the correct answer, and then put this littlequestion to the mason:--

"If the surface of that ball contained just as many square feet as itsvolume contained cubic feet, what would be the length of its diameter?"

The stonemason could not give an answer. Could you have repliedcorrectly to the mason's and the boy's questions?

189.--THE YORKSHIRE ESTATES.

[Illustration]

I was on a visit to one of the large towns of Yorkshire. While walkingto the railway station on the day of my departure a man thrust ahand-bill upon me, and I took this into the railway carriage and read itat my leisure. It informed me that three Yorkshire neighbouring estateswere to be offered for sale. Each estate was square in shape, and theyjoined one another at their corners, just as shown in the diagram.Estate A contains exactly 370 acres, B contains 116 acres, and C 74acres.

Now, the little triangular bit of land enclosed by the three squareestates was not offered for sale, and, for no reason in particular, Ibecame curious as to the area of that piece. How many acres did itcontain?

190.--FARMER WURZEL'S ESTATE.

[Illustration]

I will now present another land problem. The demonstration of the answerthat I shall give will, I think, be found both interesting and easy ofcomprehension.

Farmer Wurzel owned the three square fields shown in the annexed plan,containing respectively 18, 20, and 26 acres. In order to get aring-fence round his property he bought the four intervening triangularfields. The puzzle is to discover what was then the whole area of hisestate.

191.--THE CRESCENT PUZZLE.

[Illustration]

Here is an easy geometrical puzzle. The crescent is formed by twocircles, and C is the centre of the larger circle. The width of thecrescent between B and D is 9 inches, and between E and F 5 inches. Whatare the diameters of the two circles?

192.--THE PUZZLE WALL.

[Illustration]

There was a small lake, around which four poor men built their cottages.Four rich men afterwards built their mansions, as shown in theillustration, and they wished to have the lake to themselves, so theyinstructed a builder to put up the shortest possible wall that wouldexclude the cottagers, but give themselves free access to the lake. Howwas the wall to be built?

193.--THE SHEEPFOLD.

It is a curious fact that the answers always given to some of thebest-known puzzles that appear in every little book of firesiderecreations that has been published for the last fifty or a hundredyears are either quite unsatisfactory or clearly wrong. Yet nobody everseems to detect their faults. Here is an example:--A farmer had a penmade of fifty hurdles, capable of holding a hundred sheep only.Supposing he wanted to make it sufficiently large to hold double thatnumber, how many additional hurdles must he have?

194.--THE GARDEN WALLS.

[Illustration]

A speculative country builder has a circular field, on which he haserected four cottages, as shown in the illustration. The field issurrounded by a brick wall, and the owner undertook to put up threeother brick walls, so that the neighbours should not be overlooked byeach other, but the four tenants insist that there shall be nofavouritism, and that each shall have exactly the same length of wallspace for his wall fruit trees. The puzzle is to show how the threewalls may be built so that each tenant shall have the same area ofground, and precisely the same length of wall.

Of course, each garden must be entirely enclosed by its walls, and itmust be possible to prove that each garden has exactly the same lengthof wall. If the puzzle is properly solved no figures are necessary.

195.--LADY BELINDA'S GARDEN.

Lady Belinda is an enthusiastic gardener. In the illustration she isdepicted in the act of worrying out a pleasant little problem which Iwill relate. One of her gardens is oblong in shape, enclosed by a highholly hedge, and she is turning it into a rosary for the cultivation ofsome of her choicest roses. She wants to devote exactly half of the areaof the garden to the flowers, in one large bed, and the other half to bea path going all round it of equal breadth throughout. Such a garden isshown in the diagram at the foot of the picture. How is she to mark outthe garden under these simple conditions? She has only a tape, thelength of the garden, to do it with, and, as the holly hedge is so thickand dense, she must make all her measurements inside. Lady Belinda didnot know the exact dimensions of the garden, and, as it was notnecessary for her to know, I also give no dimensions. It is quite asimple task no matter what the size or proportions of the garden may be.Yet how many lady gardeners would know just how to proceed? The tape maybe quite plain--that is, it need not be a graduated measure.

[Illustration]

196.--THE TETHERED GOAT.

[Illustration]

Here is a little problem that everybody should know how to solve. Thegoat is placed in a half-acre meadow, that is in shape an equilateraltriangle. It is tethered to a post at one corner of the field. Whatshould be the length of the tether (to the nearest inch) in order thatthe goat shall be able to eat just half the grass in the field? It isassumed that the goat can feed to the end of the tether.

197.--THE COMPASSES PUZZLE.

It is curious how an added condition or restriction will sometimesconvert an absurdly easy puzzle into an interesting and perhapsdifficult one. I remember buying in the street many years ago a littlemechanical puzzle that had a tremendous sale at the time. It consistedof a medal with holes in it, and the puzzle was to work a ring with agap in it from hole to hole until it was finally detached. As I waswalking along the street I very soon acquired the trick of taking offthe ring with one hand while holding the puzzle in my pocket. A friendto whom I showed the little feat set about accomplishing it himself, andwhen I met him some days afterwards he exhibited his proficiency in theart. But he was a little taken aback when I then took the puzzle fromhim and, while simply holding the medal between the finger and thumb ofone hand, by a series of little shakes and jerks caused the ring,without my even touching it, to fall off upon the floor. The followinglittle poser will probably prove a rather tough nut for a great manyreaders, simply on account of the restricted conditions:--

Show how to find exactly the middle of any straight line by means of thecompasses only. You are not allowed to use any ruler, pencil, or otherarticle--only the compasses; and no trick or dodge, such as folding thepaper, will be permitted. You must simply use the compasses in theordinary legitimate way.

198.--THE EIGHT STICKS.

I have eight sticks, four of them being exactly half the length of theothers. I lay every one of these on the table, so that they enclosethree squares, all of the same size. How do I do it? There must be noloose ends hanging over.

199.--PAPA'S PUZZLE.

Here is a puzzle by Pappus, who lived at Alexandria about the end of thethird century. It is the fifth proposition in the eighth book of his_Mathematical Collections_. I give it in the form that I presented itsome years ago under the title "Papa's Puzzle," just to see how manyreaders would discover that it was by Pappus himself. "The little maid'spapa has taken two different-sized rectangular pieces of cardboard, andhas clipped off a triangular piece from one of them, so that when it issuspended by a thread from the point A it hangs with the long sideperfectly horizontal, as shown in the illustration. He has perplexed thechild by asking her to find the point A on the other card, so as toproduce a similar result when cut and suspended by a thread." Of course,the point must not be found by trial clippings. A curious and prettypoint is involved in this setting of the puzzle. Can the reader discoverit?

[Illustration]

200.--A KITE-FLYING PUZZLE.

While accompanying my friend Professor Highflite during a scientifickite-flying competition on the South Downs of Sussex I was led into alittle calculation that ought to interest my readers. The Professor waspaying out the wire to which his kite was attached from a winch on whichit had been rolled into a perfectly spherical form. This ball of wirewas just two feet in diameter, and the wire had a diameter ofone-hundredth of an inch. What was the length of the wire?

Now, a simple little question like this that everybody can perfectlyunderstand will puzzle many people to answer in any way. Let us seewhether, without going into any profound mathematical calculations, wecan get the answer roughly--say, within a mile of what is correct! Wewill assume that when the wire is all wound up the ball is perfectlysolid throughout, and that no allowance has to be made for the axle thatpasses through it. With that simplification, I wonder how many readerscan state within even a mile of the correct answer the length of thatwire.

201.--HOW TO MAKE CISTERNS.

[Illustration]

Our friend in the illustration has a large sheet of zinc, measuring(before cutting) eight feet by three feet, and he has cut out squarepieces (all of the same size) from the four corners and now proposes tofold up the sides, solder the edges, and make a cistern. But the pointthat puzzles him is this: Has he cut out those square pieces of thecorrect size in order that the cistern may hold the greatest possiblequantity of water? You see, if you cut them very small you get a veryshallow cistern; if you cut them large you get a tall and slender one.It is all a question of finding a way of cutting put these four squarepieces exactly the right size. How are we to avoid making them too smallor too large?

202.--THE CONE PUZZLE.

[Illustration]

I have a wooden cone, as shown in Fig. 1. How am I to cut out of it thegreatest possible cylinder? It will be seen that I can cut out one thatis long and slender, like Fig. 2, or short and thick, like Fig. 3. Butneither is the largest possible. A child could tell you where to cut, ifhe knew the rule. Can you find this simple rule?

203.--CONCERNING WHEELS.

[Illustration]

There are some curious facts concerning the movements of wheels that areapt to perplex the novice. For example: when a railway train istravelling from London to Crewe certain parts of the train at any givenmoment are actually moving from Crewe towards London. Can you indicatethose parts? It seems absurd that parts of the same train can at anytime travel in opposite directions, but such is the case.

In the accompanying illustration we have two wheels. The lower one issupposed to be fixed and the upper one running round it in the directionof the arrows. Now, how many times does the upper wheel turn on its ownaxis in making a complete revolution of the other wheel? Do not be in ahurry with your answer, or you are almost certain to be wrong.Experiment with two pennies on the table and the correct answer willsurprise you, when you succeed in seeing it.

204.--A NEW MATCH PUZZLE.

[Illustration]

In the illustration eighteen matches are shown arranged so that theyenclose two spaces, one just twice as large as the other. Can yourearrange them (1) so as to enclose two four-sided spaces, one exactlythree times as large as the other, and (2) so as to enclose twofive-sided spaces, one exactly three times as large as the other? Allthe eighteen matches must be fairly used in each case; the two spacesmust be quite detached, and there must be no loose ends or duplicatedmatches.

205.--THE SIX SHEEP-PENS.

[Illustration]

Here is a new little puzzle with matches. It will be seen in theillustration that thirteen matches, representing a farmer's hurdles,have been so placed that they enclose six sheep-pens all of the samesize. Now, one of these hurdles was stolen, and the farmer wanted stillto enclose six pens of equal size with the remaining twelve. How was heto do it? All the twelve matches must be fairly used, and there must beno duplicated matches or loose ends.

POINTS AND LINES PROBLEMS.

"Line upon line, line upon line; here a little and there alittle."--_Isa_. xxviii. 10.

What are known as "Points and Lines" puzzles are found very interestingby many people. The most familiar example, here given, to plant ninetrees so that they shall form ten straight rows with three trees inevery row, is attributed to Sir Isaac Newton, but the earliestcollection of such puzzles is, I believe, in a rare little book that Ipossess--published in 1821--_Rational Amusement for Winter Evenings_, byJohn Jackson. The author gives ten examples of "Trees planted in Rows."

These tree-planting puzzles have always been a matter of greatperplexity. They are real "puzzles," in the truest sense of the word,because nobody has yet succeeded in finding a direct and certain way ofsolving them. They demand the exercise of sagacity, ingenuity, andpatience, and what we call "luck" is also sometimes of service. Perhapssome day a genius will discover the key to the whole mystery. Rememberthat the trees must be regarded as mere points, for if we were allowedto make our trees big enough we might easily "fudge" our diagrams andget in a few extra straight rows that were more apparent than real.

[Illustration]

206.--THE KING AND THE CASTLES.

There was once, in ancient times, a powerful king, who had eccentricideas on the subject of military architecture. He held that there wasgreat strength and economy in symmetrical forms, and always cited theexample of the bees, who construct their combs in perfect hexagonalcells, to prove that he had nature to support him. He resolved to buildten new castles in his country all to be connected by fortified walls,which should form five lines with four castles in every line. The royalarchitect presented his preliminary plan in the form I have shown. Butthe monarch pointed out that every castle could be approached from theoutside, and commanded that the plan should be so modified that as manycastles as possible should be free from attack from the outside, andcould only be reached by crossing the fortified walls. The architectreplied that he thought it impossible so to arrange them that even onecastle, which the king proposed to use as a royal residence, could be soprotected, but his majesty soon enlightened him by pointing out how itmight be done. How would you have built the ten castles andfortifications so as best to fulfil the king's requirements? Rememberthat they must form five straight lines with four castles in every line.

[Illustration]

207.--CHERRIES AND PLUMS.

[Illustration]

The illustration is a plan of a cottage as it stands surrounded by anorchard of fifty-five trees. Ten of these trees are cherries, ten areplums, and the remainder apples. The cherries are so planted as to formfive straight lines, with four cherry trees in every line. The plumtrees are also planted so as to form five straight lines with four plumtrees in every line. The puzzle is to show which are the ten cherrytrees and which are the ten plums. In order that the cherries and plumsshould have the most favourable aspect, as few as possible (under theconditions) are planted on the north and east sides of the orchard. Ofcourse in picking out a group of ten trees (cherry or plum, as the casemay be) you ignore all intervening trees. That is to say, four trees maybe in a straight line irrespective of other trees (or the house) beingin between. After the last puzzle this will be quite easy.

208.--A PLANTATION PUZZLE.

[Illustration]

A man had a square plantation of forty-nine trees, but, as will be seenby the omissions in the illustration, four trees were blown down andremoved. He now wants to cut down all the remainder except ten trees,which are to be so left that they shall form five straight rows withfour trees in every row. Which are the ten trees that he must leave?

209.--THE TWENTY-ONE TREES.

A gentleman wished to plant twenty-one trees in his park so that theyshould form twelve straight rows with five trees in every row. Could youhave supplied him with a pretty symmetrical arrangement that wouldsatisfy these conditions?

210.--THE TEN COINS.

Place ten pennies on a large sheet of paper or cardboard, as shown inthe diagram, five on each edge. Now remove four of the coins, withoutdisturbing the others, and replace them on the paper so that the tenshall form five straight lines with four coins in every line. This initself is not difficult, but you should try to discover in how manydifferent ways the puzzle may be solved, assuming that in every case thetwo rows at starting are exactly the same.

[Illustration]

211.--THE TWELVE MINCE-PIES.

It will be seen in our illustration how twelve mince-pies may be placedon the table so as to form six straight rows with four pies in everyrow. The puzzle is to remove only four of them to new positions so thatthere shall be _seven_ straight rows with four in every row. Which fourwould you remove, and where would you replace them?

[Illustration]

212.--THE BURMESE PLANTATION.

[Illustration]

A short time ago I received an interesting communication from theBritish chaplain at Meiktila, Upper Burma, in which my correspondentinformed me that he had found some amusement on board ship on his wayout in trying to solve this little poser.

If he has a plantation of forty-nine trees, planted in the form of asquare as shown in the accompanying illustration, he wishes to know howhe may cut down twenty-seven of the trees so that the twenty-two leftstanding shall form as many rows as possible with four trees in everyrow.

Of course there may not be more than four trees in any row.

213.--TURKS AND RUSSIANS.

This puzzle is on the lines of the Afridi problem published by me in_Tit-Bits_ some years ago.

On an open level tract of country a party of Russian infantry, no two ofwhom were stationed at the same spot, were suddenly surprised bythirty-two Turks, who opened fire on the Russians from all directions.Each of the Turks simultaneously fired a bullet, and each bullet passedimmediately over the heads of three Russian soldiers. As each of thesebullets when fired killed a different man, the puzzle is to discoverwhat is the smallest possible number of soldiers of which the Russianparty could have consisted and what were the casualties on each side.

MOVING COUNTER PROBLEMS.

"I cannot do't without counters."

_Winter's Tale_, iv. 3.

Puzzles of this class, except so far as they occur in connection withactual games, such as chess, seem to be a comparatively modernintroduction. Mathematicians in recent times, notably Vandermonde andReiss, have devoted some attention to them, but they do not appear tohave been considered by the old writers. So far as games with countersare concerned, perhaps the most ancient and widely known in old times is"Nine Men's Morris" (known also, as I shall show, under a great manyother names), unless the simpler game, distinctly mentioned in the worksof Ovid (No. 110, "Ovid's Game," in _The Canterbury Puzzles_), fromwhich "Noughts and Crosses" seems to be derived, is still more ancient.

In France the game is called Marelle, in Poland Siegen Wulf Myll(She-goat Wolf Mill, or Fight), in Germany and Austria it is calledMuhle (the Mill), in Iceland it goes by the name of Mylla, while theBogas (or native bargees) of South America are said to play it, and onthe Amazon it is called Trique, and held to be of Indian origin. In ourown country it has different names in different districts, such as MegMerrylegs, Peg Meryll, Nine Peg o'Merryal, Nine-Pin Miracle, Merry Peg,and Merry Hole. Shakespeare refers to it in "Midsummer Night's Dream"(Act ii., scene 1):--

"The nine-men's morris is filled up with mud; And the quaint mazes in the wanton green, For lack of tread, are undistinguishable."

It was played by the shepherds with stones in holes cut in the turf.John Clare, the peasant poet of Northamptonshire, in "The Shepherd Boy"(1835) says:--"Oft we track his haunts .... By nine-peg-morris nickedupon the green." It is also mentioned by Drayton in his "Polyolbion."

It was found on an old Roman tile discovered during the excavations atSilchester, and cut upon the steps of the Acropolis at Athens. Whenvisiting the Christiania Museum a few years ago I was shown the greatViking ship that was discovered at Gokstad in 1880. On the oak planksforming the deck of the vessel were found boles and lines marking outthe game, the holes being made to receive pegs. While inspecting theancient oak furniture in the Rijks Museum at Amsterdam I becameinterested in an old catechumen's settle, and was surprised to find thegame diagram cut in the centre of the seat--quite conveniently forsurreptitious play. It has been discovered cut in the choir stalls ofseveral of our English cathedrals. In the early eighties it was foundscratched upon a stone built into a wall (probably about the date 1200),during the restoration of Hargrave church in Northamptonshire. Thisstone is now in the Northampton Museum. A similar stone has since beenfound at Sempringham, Lincolnshire. It is to be seen on an ancienttombstone in the Isle of Man, and painted on old Dutch tiles. And in1901 a stone was dug out of a gravel pit near Oswestry bearing anundoubted diagram of the game.

The game has been played with different rules at different periods andplaces. I give a copy of the board. Sometimes the diagonal lines areomitted, but this evidently was not intended to affect the play: itsimply meant that the angles alone were thought sufficient to indicatethe points. This is how Strutt, in _Sports and Pastimes_, describes thegame, and it agrees with the way I played it as a boy:--"Two persons,having each of them nine pieces, or men, lay them down alternately, oneby one, upon the spots; and the business of either party is to preventhis antagonist from placing three of his pieces so as to form a row ofthree, without the intervention of an opponent piece. If a row beformed, he that made it is at liberty to take up one of his competitor'spieces from any part he thinks most to his advantage; excepting he hasmade a row, which must not be touched if he have another piece upon theboard that is not a component part of that row. When all the pieces arelaid down, they are played backwards and forwards, in any direction thatthe lines run, but only can move from one spot to another (next to it)at one time. He that takes off all his antagonist's pieces is theconqueror."

[Illustration]

214.--THE SIX FROGS.

[Illustration]

The six educated frogs in the illustration are trained to reverse theirorder, so that their numbers shall read 6, 5, 4, 3, 2, 1, with the blanksquare in its present position. They can jump to the next square (ifvacant) or leap over one frog to the next square beyond (if vacant),just as we move in the game of draughts, and can go backwards orforwards at pleasure. Can you show how they perform their feat in thefewest possible moves? It is quite easy, so when you have done it add aseventh frog to the right and try again. Then add more frogs until youare able to give the shortest solution for any number. For it can alwaysbe done, with that single vacant square, no matter how many frogs thereare.

215.--THE GRASSHOPPER PUZZLE.

It has been suggested that this puzzle was a great favourite among theyoung apprentices of the City of London in the sixteenth and seventeenthcenturies. Readers will have noticed the curious brass grasshopper onthe Royal Exchange. This long-lived creature escaped the fires of 1666and 1838. The grasshopper, after his kind, was the crest of Sir ThomasGresham, merchant grocer, who died in 1579, and from this cause it hasbeen used as a sign by grocers in general. Unfortunately for the legendas to its origin, the puzzle was only produced by myself so late as theyear 1900. On twelve of the thirteen black discs are placed numberedcounters or grasshoppers. The puzzle is to reverse their order, so thatthey shall read, 1, 2, 3, 4, etc., in the opposite direction, with thevacant disc left in the same position as at present. Move one at a timein any order, either to the adjoining vacant disc or by jumping over onegrasshopper, like the moves in draughts. The moves or leaps may be madein either direction that is at any time possible. What are the fewestpossible moves in which it can be done?

[Illustration]

216.--THE EDUCATED FROGS.

[Illustration]

Our six educated frogs have learnt a new and pretty feat. When placed onglass tumblers, as shown in the illustration, they change sides so thatthe three black ones are to the left and the white frogs to the right,with the unoccupied tumbler at the opposite end--No. 7. They can jump tothe next tumbler (if unoccupied), or over one, or two, frogs to anunoccupied tumbler. The jumps can be made in either direction, and afrog may jump over his own or the opposite colour, or both colours. Foursuccessive specimen jumps will make everything quite plain: 4 to 1, 5 to4, 3 to 5, 6 to 3. Can you show how they do it in ten jumps?

217.--THE TWICKENHAM PUZZLE.

[Illustration:

( I ) ((N))

( M ) ((A))

( H ) ((T))

( E ) ((W))

( C ) ((K)) ( )

]

In the illustration we have eleven discs in a circle. On five of thediscs we place white counters with black letters--as shown--and on fiveother discs the black counters with white letters. The bottom disc isleft vacant. Starting thus, it is required to get the counters intoorder so that they spell the word "Twickenham" in a clockwise direction,leaving the vacant disc in the original position. The black countersmove in the direction that a clock-hand revolves, and the white countersgo the opposite way. A counter may jump over one of the opposite colourif the vacant disc is next beyond. Thus, if your first move is with K,then C can jump over K. If then K moves towards E, you may next jump Wover C, and so on. The puzzle may be solved in twenty-six moves.Remember a counter cannot jump over one of its own colour.

218.--THE VICTORIA CROSS PUZZLE.

[Illustration:

+---------------------+ | \... A .../ | | (I) |.......| (V) | |\_____|_______|_____/| |......| |------| |.. R .| |. I ..| |......| |......| | _____|_______|_____ | |/ |.......| \| | (O) |.. T ..| (C) | | /.........\ | +---------------------+

]

The puzzle-maker is peculiarly a "snapper-up of unconsidered trifles,"and his productions are often built up with the slenderest materials.Trivialities that might entirely escape the observation of others, or,if they were observed, would be regarded as of no possible moment, oftensupply the man who is in quest of posers with a pretty theme or an ideathat he thinks possesses some "basal value."

When seated opposite to a lady in a railway carriage at the time ofQueen Victoria's Diamond Jubilee, my attention was attracted to a broochthat she was wearing. It was in the form of a Maltese or Victoria Cross,and bore the letters of the word VICTORIA. The number and arrangement ofthe letters immediately gave me the suggestion for the puzzle which Inow present.

The diagram, it will be seen, is composed of nine divisions. The puzzleis to place eight counters, bearing the letters of the word VICTORIA,exactly in the manner shown, and then slide one letter at a time fromblack to white and white to black alternately, until the word readsround in the same direction, only with the initial letter V on one ofthe black arms of the cross. At no time may two letters be in the samedivision. It is required to find the shortest method.

Leaping moves are, of course, not permitted. The first move mustobviously be made with A, I, T, or R. Supposing you move T to thecentre, the next counter played will be O or C, since I or R cannot bemoved. There is something a little remarkable in the solution of thispuzzle which I will explain.

219.--THE LETTER BLOCK PUZZLE.

[Illustration:

+-----+-----+-----+\ | | | | | | G | E | F | | | | | | | +-----+-----+-----+\| | | | | | | H | C | B | | | | | | | +-----+-----+-----+\| | |\____| | | | D || | A | | | || | | | +-----+-----+-----+ | \_________________\|

]

Here is a little reminiscence of our old friend the Fifteen BlockPuzzle. Eight wooden blocks are lettered, and are placed in a box, asshown in the illustration. It will be seen that you can only move oneblock at a time to the place vacant for the time being, as no block maybe lifted out of the box. The puzzle is to shift them about until youget them in the order--

A B C D E F G H

This you will find by no means difficult if you are allowed as manymoves as you like. But the puzzle is to do it in the fewest possiblemoves. I will not say what this smallest number of moves is, because thereader may like to discover it for himself. In writing down your movesyou will find it necessary to record no more than the letters in theorder that they are shifted. Thus, your first five moves might be C, H,G, E, F; and this notation can have no possible ambiguity. In practiceyou only need eight counters and a simple diagram on a sheet of paper.

220.--A LODGING-HOUSE DIFFICULTY.

[Illustration]

The Dobsons secured apartments at Slocomb-on-Sea. There were six roomson the same floor, all communicating, as shown in the diagram. The roomsthey took were numbers 4, 5, and 6, all facing the sea. But a littledifficulty arose. Mr. Dobson insisted that the piano and the bookcaseshould change rooms. This was wily, for the Dobsons were not musical,but they wanted to prevent any one else playing the instrument. Now, therooms were very small and the pieces of furniture indicated were verybig, so that no two of these articles could be got into any room at thesame time. How was the exchange to be made with the least possiblelabour? Suppose, for example, you first move the wardrobe into No. 2;then you can move the bookcase to No. 5 and the piano to No. 6, and soon. It is a fascinating puzzle, but the landlady had reasons for notappreciating it. Try to solve her difficulty in the fewest possibleremovals with counters on a sheet of paper.

221.--THE EIGHT ENGINES.

The diagram represents the engine-yard of a railway company undereccentric management. The engines are allowed to be stationary only atthe nine points indicated, one of which is at present vacant. It isrequired to move the engines, one at a time, from point to point, inseventeen moves, so that their numbers shall be in numerical order roundthe circle, with the central point left vacant. But one of the engineshas had its fire drawn, and therefore cannot move. How is the thing tobe done? And which engine remains stationary throughout?

[Illustration]

222.--A RAILWAY PUZZLE.

[Illustration]

Make a diagram, on a large sheet of paper, like the illustration, andhave three counters marked A, three marked B, and three marked C. Itwill be seen that at the intersection of lines there are ninestopping-places, and a tenth stopping-place is attached to the outercircle like the tail of a Q. Place the three counters or engines markedA, the three marked B, and the three marked C at the places indicated.The puzzle is to move the engines, one at a time, along the lines, fromstopping-place to stopping-place, until you succeed in getting an A, aB, and a C on each circle, and also A, B, and C on each straight line.You are required to do this in as few moves as possible. How many movesdo you need?

223.--A RAILWAY MUDDLE.

The plan represents a portion of the line of the London, Clodville, andMudford Railway Company. It is a single line with a loop. There is onlyroom for eight wagons, or seven wagons and an engine, between B and C oneither the left line or the right line of the loop. It happened that twogoods trains (each consisting of an engine and sixteen wagons) got intothe position shown in the illustration. It looked like a hopelessdeadlock, and each engine-driver wanted the other to go back to the nextstation and take off nine wagons. But an ingenious stoker undertook topass the trains and send them on their respective journeys with theirengines properly in front. He also contrived to reverse the engines thefewest times possible. Could you have performed the feat? And how manytimes would you require to reverse the engines? A "reversal" means achange of direction, backward or forward. No rope-shunting,fly-shunting, or other trick is allowed. All the work must be donelegitimately by the two engines. It is a simple but interesting puzzleif attempted with counters.

[Illustration]

224.--THE MOTOR-GARAGE PUZZLE.

[Illustration]

The difficulties of the proprietor of a motor garage are converted intoa little pastime of a kind that has a peculiar fascination. All you needis to make a simple plan or diagram on a sheet of paper or cardboard andnumber eight counters, 1 to 8. Then a whole family can enter into anamusing competition to find the best possible solution of thedifficulty.

The illustration represents the plan of a motor garage, withaccommodation for twelve cars. But the premises are so inconvenientlyrestricted that the proprietor is often caused considerable perplexity.Suppose, for example, that the eight cars numbered 1 to 8 are in thepositions shown, how are they to be shifted in the quickest possible wayso that 1, 2, 3, and 4 shall change places with 5, 6, 7, and 8--that is,with the numbers still running from left to right, as at present, butthe top row exchanged with the bottom row? What are the fewest possiblemoves?

One car moves at a time, and any distance counts as one move. To preventmisunderstanding, the stopping-places are marked in squares, and onlyone car can be in a square at the same time.

225.--THE TEN PRISONERS.

If prisons had no other use, they might still be preserved for thespecial benefit of puzzle-makers. They appear to be an inexhaustiblemine of perplexing ideas. Here is a little poser that will perhapsinterest the reader for a short period. We have in the illustration aprison of sixteen cells. The locations of the ten prisoners will beseen. The jailer has queer superstitions about odd and even numbers, andhe wants to rearrange the ten prisoners so that there shall be as manyeven rows of men, vertically, horizontally, and diagonally, aspossible. At present it will be seen, as indicated by the arrows, thatthere are only twelve such rows of 2 and 4. I will state at once thatthe greatest number of such rows that is possible is sixteen. But thejailer only allows four men to be removed to other cells, and informs methat, as the man who is seated in the bottom right-hand corner isinfirm, he must not be moved. Now, how are we to get those sixteen rowsof even numbers under such conditions?

[Illustration]

226.--ROUND THE COAST.

[Illustration]

Here is a puzzle that will, I think, be found as amusing as instructive.We are given a ring of eight circles. Leaving circle 8 blank, we arerequired to write in the name of a seven-lettered port in the UnitedKingdom in this manner. Touch a blank circle with your pencil, then jumpover two circles in either direction round the ring, and write down thefirst letter. Then touch another vacant circle, jump over two circles,and write down your second letter. Proceed similarly with the otherletters in their proper order until you have completed the word. Thus,suppose we select "Glasgow," and proceed as follows: 6--1, 7--2, 8--3,7--4, 8--5, which means that we touch 6, jump over 7 and and write down"G" on 1; then touch 7, jump over 8 and 1, and write down "l" on 2; andso on. It will be found that after we have written down the first fiveletters--"Glasg"--as above, we cannot go any further. Either there issomething wrong with "Glasgow," or we have not managed our jumpsproperly. Can you get to the bottom of the mystery?

227.--CENTRAL SOLITAIRE.

[Illustration]

This ancient puzzle was a great favourite with our grandmothers, andmost of us, I imagine, have on occasions come across a "Solitaire"board--a round polished board with holes cut in it in a geometricalpattern, and a glass marble in every hole. Sometimes I have noticed oneon a side table in a suburban front parlour, or found one on a shelf ina country cottage, or had one brought under my notice at a wayside inn.Sometimes they are of the form shown above, but it is equally common forthe board to have four more holes, at the points indicated by dots. Iselect the simpler form.

Though "Solitaire" boards are still sold at the toy shops, it will besufficient if the reader will make an enlarged copy of the above on asheet of cardboard or paper, number the "holes," and provide himselfwith 33 counters, buttons, or beans. Now place a counter in every holeexcept the central one, No. 17, and the puzzle is to take off all thecounters in a series of jumps, except the last counter, which must beleft in that central hole. You are allowed to jump one counter over thenext one to a vacant hole beyond, just as in the game of draughts, andthe counter jumped over is immediately taken off the board. Onlyremember every move must be a jump; consequently you will take off acounter at each move, and thirty-one single jumps will of course removeall the thirty-one counters. But compound moves are allowed (as indraughts, again), for so long as one counter continues to jump, thejumps all count as one move.

Here is the beginning of an imaginary solution which will serve to makethe manner of moving perfectly plain, and show how the solver shouldwrite out his attempts: 5-17, 12-10, 26-12, 24-26 (13-11, 11-25), 9-11(26-24, 24-10, 10-12), etc., etc. The jumps contained within bracketscount as one move, because they are made with the same counter. Find thefewest possible moves. Of course, no diagonal jumps are permitted; youcan only jump in the direction of the lines.

228.--THE TEN APPLES.

[Illustration]

The family represented in the illustration are amusing themselves withthis little puzzle, which is not very difficult but quite interesting.They have, it will be seen, placed sixteen plates on the table in theform of a square, and put an apple in each of ten plates. They want tofind a way of removing all the apples except one by jumping over one ata time to the next vacant square, as in draughts; or, better, as insolitaire, for you are not allowed to make any diagonal moves--onlymoves parallel to the sides of the square. It is obvious that as theapples stand no move can be made, but you are permitted to transfer anysingle apple you like to a vacant plate before starting. Then the movesmust be all leaps, taking off the apples leaped over.

229.--THE NINE ALMONDS.

"Here is a little puzzle," said a Parson, "that I have found peculiarlyfascinating. It is so simple, and yet it keeps you interestedindefinitely."

The reverend gentleman took a sheet of paper and divided it off intotwenty-five squares, like a square portion of a chessboard. Then heplaced nine almonds on the central squares, as shown in theillustration, where we have represented numbered counters forconvenience in giving the solution.

"Now, the puzzle is," continued the Parson, "to remove eight of thealmonds and leave the ninth in the central square. You make the removalsby jumping one almond over another to the vacant square beyond andtaking off the one jumped over--just as in draughts, only here you canjump in any direction, and not diagonally only. The point is to do thething in the fewest possible moves."

The following specimen attempt will make everything clear. Jump 4 over1, 5 over 9, 3 over 6, 5 over 3, 7 over 5 and 2, 4 over 7, 8 over 4. But8 is not left in the central square, as it should be. Remember to removethose you jump over. Any number of jumps in succession with the samealmond count as one move.

[Illustration]

230.--THE TWELVE PENNIES.

Here is a pretty little puzzle that only requires twelve pennies orcounters. Arrange them in a circle, as shown in the illustration. Nowtake up one penny at a time and, passing it over two pennies, place iton the third penny. Then take up another single penny and do the samething, and so on, until, in six such moves, you have the coins in sixpairs in the positions 1, 2, 3, 4, 5, 6. You can move in eitherdirection round the circle at every play, and it does not matterwhether the two jumped over are separate or a pair. This is quite easyif you use just a little thought.

[Illustration]

231.--PLATES AND COINS.

Place twelve plates, as shown, on a round table, with a penny or orangein every plate. Start from any plate you like and, always going in onedirection round the table, take up one penny, pass it over two otherpennies, and place it in the next plate. Go on again; take up anotherpenny and, having passed it over two pennies, place it in a plate; andso continue your journey. Six coins only are to be removed, and whenthese have been placed there should be two coins in each of six platesand six plates empty. An important point of the puzzle is to go roundthe table as few times as possible. It does not matter whether the twocoins passed over are in one or two plates, nor how many empty platesyou pass a coin over. But you must always go in one direction round thetable and end at the point from which you set out. Your hand, that is tosay, goes steadily forward in one direction, without ever movingbackwards.

[Illustration]

232.--CATCHING THE MICE.

[Illustration]

"Play fair!" said the mice. "You know the rules of the game."

"Yes, I know the rules," said the cat. "I've got to go round and roundthe circle, in the direction that you are looking, and eat everythirteenth mouse, but I must keep the white mouse for a tit-bit at thefinish. Thirteen is an unlucky number, but I will do my best to obligeyou."

"Hurry up, then!" shouted the mice.

"Give a fellow time to think," said the cat. "I don't know which of youto start at. I must figure it out."

While the cat was working out the puzzle he fell asleep, and, the spellbeing thus broken, the mice returned home in safety. At which mouseshould the cat have started the count in order that the white mouseshould be the last eaten?

When the reader has solved that little puzzle, here is a second one forhim. What is the smallest number that the cat can count round and roundthe circle, if he must start at the white mouse (calling that "one" inthe count) and still eat the white mouse last of all?

And as a third puzzle try to discover what is the smallest number thatthe cat can count round and round if she must start at the white mouse(calling that "one") and make the white mouse the third eaten.

233.--THE ECCENTRIC CHEESEMONGER.

[Illustration]

The cheesemonger depicted in the illustration is an inveterate puzzlelover. One of his favourite puzzles is the piling of cheeses in hiswarehouse, an amusement that he finds good exercise for the body as wellas for the mind. He places sixteen cheeses on the floor in a straightrow and then makes them into four piles, with four cheeses in everypile, by always passing a cheese over four others. If you use sixteencounters and number them in order from 1 to 16, then you may place 1 on6, 11 on 1, 7 on 4, and so on, until there are four in every pile. Itwill be seen that it does not matter whether the four passed over arestanding alone or piled; they count just the same, and you can alwayscarry a cheese in either direction. There are a great many differentways of doing it in twelve moves, so it makes a good game of "patience"to try to solve it so that the four piles shall be left in differentstipulated places. For example, try to leave the piles at the extremeends of the row, on Nos. 1, 2, 15 and 16; this is quite easy. Then tryto leave three piles together, on Nos. 13, 14, and 15. Then again playso that they shall be left on Nos. 3, 5, 12, and 14.

234.--THE EXCHANGE PUZZLE.

Here is a rather entertaining little puzzle with moving counters. Youonly need twelve counters--six of one colour, marked A, C, E, G, I, andK, and the other six marked B, D, F, H, J, and L. You first place themon the diagram, as shown in the illustration, and the puzzle is to getthem into regular alphabetical order, as follows:--

A B C D E F G H I J K L

The moves are made by exchanges of opposite colours standing on the sameline. Thus, G and J may exchange places, or F and A, but you cannotexchange G and C, or F and D, because in one case they are both whiteand in the other case both black. Can you bring about the requiredarrangement in seventeen exchanges?

[Illustration]

It cannot be done in fewer moves. The puzzle is really much easier thanit looks, if properly attacked.

235.--TORPEDO PRACTICE.

[Illustration]

If a fleet of sixteen men-of-war were lying at anchor and surrounded bythe enemy, how many ships might be sunk if every torpedo, projected in astraight line, passed under three vessels and sank the fourth? In thediagram we have arranged the fleet in square formation, where it will beseen that as many as seven ships may be sunk (those in the top row andfirst column) by firing the torpedoes indicated by arrows. Anchoring thefleet as we like, to what extent can we increase this number? Rememberthat each successive ship is sunk before another torpedo is launched,and that every torpedo proceeds in a different direction; otherwise, byplacing the ships in a straight line, we might sink as many as thirteen!It is an interesting little study in naval warfare, and eminentlypractical--provided the enemy will allow you to arrange his fleet foryour convenience and promise to lie still and do nothing!

236.--THE HAT PUZZLE.

Ten hats were hung on pegs as shown in the illustration--five silk hatsand five felt "bowlers," alternately silk and felt. The two pegs at theend of the row were empty.

[Illustration]

The puzzle is to remove two contiguous hats to the vacant pegs, then twoother adjoining hats to the pegs now unoccupied, and so on until fivepairs have been moved and the hats again hang in an unbroken row, butwith all the silk ones together and all the felt hats together.

Remember, the two hats removed must always be contiguous ones, and youmust take one in each hand and place them on their new pegs withoutreversing their relative position. You are not allowed to cross yourhands, nor to hang up one at a time.

Can you solve this old puzzle, which I give as introductory to the next?Try it with counters of two colours or with coins, and remember that thetwo empty pegs must be left at one end of the row.

237.--BOYS AND GIRLS.

If you mark off ten divisions on a sheet of paper to represent thechairs, and use eight numbered counters for the children, you will havea fascinating pastime. Let the odd numbers represent boys and evennumbers girls, or you can use counters of two colours, or coins.

The puzzle is to remove two children who are occupying adjoining chairsand place them in two empty chairs, _making them first change sides_;then remove a second pair of children from adjoining chairs and placethem in the two now vacant, making them change sides; and so on, untilall the boys are together and all the girls together, with the twovacant chairs at one end as at present. To solve the puzzle you must dothis in five moves. The two children must always be taken from chairsthat are next to one another; and remember the important point of makingthe two children change sides, as this latter is the distinctive featureof the puzzle. By "change sides" I simply mean that if, for example, youfirst move 1 and 2 to the vacant chairs, then the first (the outside)chair will be occupied by 2 and the second one by 1.

[Illustration]

238.--ARRANGING THE JAMPOTS.

I happened to see a little girl sorting out some jam in a cupboard forher mother. She was putting each different kind of preserve apart on theshelves. I noticed that she took a pot of damson in one hand and a potof gooseberry in the other and made them change places; then she changeda strawberry with a raspberry, and so on. It was interesting to observewhat a lot of unnecessary trouble she gave herself by making moreinterchanges than there was any need for, and I thought it would workinto a good puzzle.

It will be seen in the illustration that little Dorothy has tomanipulate twenty-four large jampots in as many pigeon-holes. She wantsto get them in correct numerical order--that is, 1, 2, 3, 4, 5, 6 on thetop shelf, 7, 8, 9, 10, 11, 12 on the next shelf, and so on. Now, if shealways takes one pot in the right hand and another in the left and makesthem change places, how many of these interchanges will be necessary toget all the jampots in proper order? She would naturally first changethe 1 and the 3, then the 2 and the 3, when she would have the firstthree pots in their places. How would you advise her to go on then?Place some numbered counters on a sheet of paper divided into squaresfor the pigeon-holes, and you will find it an amusing puzzle.

[Illustration]

UNICURSAL AND ROUTE PROBLEMS.

"I see them on their winding way." REGINALD HEBER.

It is reasonable to suppose that from the earliest ages one man hasasked another such questions as these: "Which is the nearest way home?""Which is the easiest or pleasantest way?" "How can we find a way thatwill enable us to dodge the mastodon and the plesiosaurus?" "How can weget there without ever crossing the track of the enemy?" All these areelementary route problems, and they can be turned into good puzzles bythe introduction of some conditions that complicate matters. A varietyof such complications will be found in the following examples. I havealso included some enumerations of more or less difficulty. These affordexcellent practice for the reasoning faculties, and enable one togeneralize in the case of symmetrical forms in a manner that is mostinstructive.

239.--A JUVENILE PUZZLE.

For years I have been perpetually consulted by my juvenile friends aboutthis little puzzle. Most children seem to know it, and yet, curiouslyenough, they are invariably unacquainted with the answer. The questionthey always ask is, "Do, please, tell me whether it is really possible."I believe Houdin the conjurer used to be very fond of giving it to hischild friends, but I cannot say whether he invented the little puzzle ornot. No doubt a large number of my readers will be glad to have themystery of the solution cleared up, so I make no apology for introducingthis old "teaser."

The puzzle is to draw with three strokes of the pencil the diagram thatthe little girl is exhibiting in the illustration. Of course, you mustnot remove your pencil from the paper during a stroke or go over thesame line a second time. You will find that you can get in a good dealof the figure with one continuous stroke, but it will always appear asif four strokes are necessary.

[Illustration]

Another form of the puzzle is to draw the diagram on a slate and thenrub it out in three rubs.

240.--THE UNION JACK.

[Illustration]

The illustration is a rough sketch somewhat resembling the British flag,the Union Jack. It is not possible to draw the whole of it withoutlifting the pencil from the paper or going over the same line twice. Thepuzzle is to find out just _how much_ of the drawing it is possible tomake without lifting your pencil or going twice over the same line. Takeyour pencil and see what is the best you can do.

241.--THE DISSECTED CIRCLE.

How many continuous strokes, without lifting your pencil from the paper,do you require to draw the design shown in our illustration? Directlyyou change the direction of your pencil it begins a new stroke. You maygo over the same line more than once if you like. It requires just alittle care, or you may find yourself beaten by one stroke.

[Illustration]

242.--THE TUBE INSPECTOR'S PUZZLE.

The man in our illustration is in a little dilemma. He has just beenappointed inspector of a certain system of tube railways, and it is hisduty to inspect regularly, within a stated period, all the company'sseventeen lines connecting twelve stations, as shown on the big posterplan that he is contemplating. Now he wants to arrange his route so thatit shall take him over all the lines with as little travelling aspossible. He may begin where he likes and end where he likes. What ishis shortest route?

Could anything be simpler? But the reader will soon find that, howeverhe decides to proceed, the inspector must go over some of the lines morethan once. In other words, if we say that the stations are a mile apart,he will have to travel more than seventeen miles to inspect every line.There is the little difficulty. How far is he compelled to travel, andwhich route do you recommend?

[Illustration]

243.--VISITING THE TOWNS.

[Illustration]

A traveller, starting from town No. 1, wishes to visit every one of thetowns once, and once only, going only by roads indicated by straightlines. How many different routes are there from which he can select? Ofcourse, he must end his journey at No. 1, from which he started, andmust take no notice of cross roads, but go straight from town to town.This is an absurdly easy puzzle, if you go the right way to work.

244.--THE FIFTEEN TURNINGS.

Here is another queer travelling puzzle, the solution of which calls foringenuity. In this case the traveller starts from the black town andwishes to go as far as possible while making only fifteen turnings andnever going along the same road twice. The towns are supposed to be amile apart. Supposing, for example, that he went straight to A, thenstraight to B, then to C, D, E, and F, you will then find that he hastravelled thirty-seven miles in five turnings. Now, how far can he go infifteen turnings?

[Illustration]

245.--THE FLY ON THE OCTAHEDRON.

"Look here," said the professor to his colleague, "I have been watchingthat fly on the octahedron, and it confines its walks entirely to theedges. What can be its reason for avoiding the sides?"

"Perhaps it is trying to solve some route problem," suggested the other."Supposing it to start from the top point, how many different routes arethere by which it may walk over all the edges, without ever going twicealong the same edge in any route?"

[Illustration]

The problem was a harder one than they expected, and after working at itduring leisure moments for several days their results did not agree--infact, they were both wrong. If the reader is surprised at their failure,let him attempt the little puzzle himself. I will just explain that theoctahedron is one of the five regular, or Platonic, bodies, and iscontained under eight equal and equilateral triangles. If you cut outthe two pieces of cardboard of the shape shown in the margin of theillustration, cut half through along the dotted lines and then bend themand put them together, you will have a perfect octahedron. In any routeover all the edges it will be found that the fly must end at the pointof departure at the top.

246.--THE ICOSAHEDRON PUZZLE.

The icosahedron is another of the five regular, or Platonic, bodieshaving all their sides, angles, and planes similar and equal. It isbounded by twenty similar equilateral triangles. If you cut out a pieceof cardboard of the form shown in the smaller diagram, and cut halfthrough along the dotted lines, it will fold up and form a perfecticosahedron.

Now, a Platonic body does not mean a heavenly body; but it will suit thepurpose of our puzzle if we suppose there to be a habitable planet ofthis shape. We will also suppose that, owing to a superfluity of water,the only dry land is along the edges, and that the inhabitants have noknowledge of navigation. If every one of those edges is 10,000 mileslong and a solitary traveller is placed at the North Pole (the highestpoint shown), how far will he have to travel before he will have visitedevery habitable part of the planet--that is, have traversed every one ofthe edges?

[Illustration]

247.--INSPECTING A MINE.

The diagram is supposed to represent the passages or galleries in amine. We will assume that every passage, A to B, B to C, C to H, H to I,and so on, is one furlong in length. It will be seen that there arethirty-one of these passages. Now, an official has to inspect all ofthem, and he descends by the shaft to the point A. How far must hetravel, and what route do you recommend? The reader may at first say,"As there are thirty-one passages, each a furlong in length, he willhave to travel just thirty-one furlongs." But this is assuming that heneed never go along a passage more than once, which is not the case.Take your pencil and try to find the shortest route. You will soondiscover that there is room for considerable judgment. In fact, it is aperplexing puzzle.

[Illustration]

248.--THE CYCLISTS' TOUR.

Two cyclists were consulting a road map in preparation for a little tourtogether. The circles represent towns, and all the good roads arerepresented by lines. They are starting from the town with a star, andmust complete their tour at E. But before arriving there they want tovisit every other town once, and only once. That is the difficulty. Mr.Spicer said, "I am certain we can find a way of doing it;" but Mr. Maggsreplied, "No way, I'm sure." Now, which of them was correct? Take yourpencil and see if you can find any way of doing it. Of course you mustkeep to the roads indicated.

[Illustration]

249.--THE SAILOR'S PUZZLE.

The sailor depicted in the illustration stated that he had since hisboyhood been engaged in trading with a small vessel among some twentylittle islands in the Pacific. He supplied the rough chart of which Ihave given a copy, and explained that the lines from island to islandrepresented the only routes that he ever adopted. He always started fromisland A at the beginning of the season, and then visited every islandonce, and once only, finishing up his tour at the starting-point A. Buthe always put off his visit to C as long as possible, for trade reasonsthat I need not enter into. The puzzle is to discover his exact route,and this can be done with certainty. Take your pencil and, starting atA, try to trace it out. If you write down the islands in the order inwhich you visit them--thus, for example, A, I, O, L, G, etc.--you can atonce see if you have visited an island twice or omitted any. Of course,the crossings of the lines must be ignored--that is, you must continueyour route direct, and you are not allowed to switch off at a crossingand proceed in another direction. There is no trick of this kind in thepuzzle. The sailor knew the best route. Can you find it?

[Illustration]

250.--THE GRAND TOUR.

One of the everyday puzzles of life is the working out of routes. If youare taking a holiday on your bicycle, or a motor tour, there alwaysarises the question of how you are to make the best of your time andother resources. You have determined to get as far as some particularplace, to include visits to such-and-such a town, to try to seesomething of special interest elsewhere, and perhaps to try to look upan old friend at a spot that will not take you much out of your way.Then you have to plan your route so as to avoid bad roads, uninterestingcountry, and, if possible, the necessity of a return by the same waythat you went. With a map before you, the interesting puzzle is attackedand solved. I will present a little poser based on these lines.

I give a rough map of a country--it is not necessary to say whatparticular country--the circles representing towns and the dotted linesthe railways connecting them. Now there lived in the town marked A a manwho was born there, and during the whole of his life had never once lefthis native place. From his youth upwards he had been very industrious,sticking incessantly to his trade, and had no desire whatever to roamabroad. However, on attaining his fiftieth birthday he decided to seesomething of his country, and especially to pay a visit to a very oldfriend living at the town marked Z. What he proposed was this: that hewould start from his home, enter every town once and only once, andfinish his journey at Z. As he made up his mind to perform this grandtour by rail only, he found it rather a puzzle to work out his route,but he at length succeeded in doing so. How did he manage it? Do notforget that every town has to be visited once, and not more than once.

[Illustration]

251.--WATER, GAS, AND ELECTRICITY.

There are some half-dozen puzzles, as old as the hills, that areperpetually cropping up, and there is hardly a month in the year thatdoes not bring inquiries as to their solution. Occasionally one ofthese, that one had thought was an extinct volcano, bursts into eruptionin a surprising manner. I have received an extraordinary number ofletters respecting the ancient puzzle that I have called "Water, Gas,and Electricity." It is much older than electric lighting, or even gas,but the new dress brings it up to date. The puzzle is to lay on water,gas, and electricity, from W, G, and E, to each of the three houses, A,B, and C, without any pipe crossing another. Take your pencil and drawlines showing how this should be done. You will soon find yourselflanded in difficulties.

[Illustration]

252.--A PUZZLE FOR MOTORISTS.

[Illustration]

Eight motorists drove to church one morning. Their respective housesand churches, together with the only roads available (the dotted lines),are shown. One went from his house A to his church A, another from hishouse B to his church B, another from C to C, and so on, but it wasafterwards found that no driver ever crossed the track of another car.Take your pencil and try to trace out their various routes.

253.--A BANK HOLIDAY PUZZLE.

Two friends were spending their bank holiday on a cycling trip. Stoppingfor a rest at a village inn, they consulted a route map, which isrepresented in our illustration in an exceedingly simplified form, forthe puzzle is interesting enough without all the original complexities.They started from the town in the top left-hand corner marked A. It willbe seen that there are one hundred and twenty such towns, all connectedby straight roads. Now they discovered that there are exactly 1,365different routes by which they may reach their destination, alwaystravelling either due south or due east. The puzzle is to discover whichtown is their destination.

[Illustration]

Of course, if you find that there are more than 1,365 different routesto a town it cannot be the right one.

254.--THE MOTOR-CAR TOUR.

[Illustration]

In the above diagram the circles represent towns and the lines goodroads. In just how many different ways can a motorist, starting fromLondon (marked with an L), make a tour of all these towns, visitingevery town once, and only once, on a tour, and always coming back toLondon on the last ride? The exact reverse of any route is not countedas different.

255.--THE LEVEL PUZZLE.

[Illustration]

This is a simple counting puzzle. In how many different ways can youspell out the word LEVEL by placing the point of your pencil on an L andthen passing along the lines from letter to letter. You may go in anydirection, backwards or forwards. Of course you are not allowed to missletters--that is to say, if you come to a letter you must use it.

256.--THE DIAMOND PUZZLE.

IN how many different ways may the word DIAMOND be read in thearrangement shown? You may start wherever you like at a D and go up ordown, backwards or forwards, in and out, in any direction you like, solong as you always pass from one letter to another that adjoins it. Howmany ways are there?

[Illustration]

257.--THE DEIFIED PUZZLE.

In how many different ways may the word DEIFIED be read in thisarrangement under the same conditions as in the last puzzle, with theaddition that you can use any letters twice in the same reading?

[Illustration]

258.--THE VOTERS' PUZZLE.

[Illustration]

Here we have, perhaps, the most interesting form of the puzzle. In howmany different ways can you read the political injunction, "RISE TOVOTE, SIR," under the same conditions as before? In this case everyreading of the palindrome requires the use of the central V as themiddle letter.

259.--HANNAH'S PUZZLE.

A man was in love with a young lady whose Christian name was Hannah.When he asked her to be his wife she wrote down the letters of her namein this manner:--

H H H H H H H A A A A H H A N N A H H A N N A H H A A A A H H H H H H H

and promised that she would be his if he could tell her correctly in howmany different ways it was possible to spell out her name, alwayspassing from one letter to another that was adjacent. Diagonal steps arehere allowed. Whether she did this merely to tease him or to test hiscleverness is not recorded, but it is satisfactory to know that hesucceeded. Would you have been equally successful? Take your pencil andtry. You may start from any of the H's and go backwards or forwards andin any direction, so long as all the letters in a spelling are adjoiningone another. How many ways are there, no two exactly alike?

260.--THE HONEYCOMB PUZZLE.

[Illustration]

Here is a little puzzle with the simplest possible conditions. Place thepoint of your pencil on a letter in one of the cells of the honeycomb,and trace out a very familiar proverb by passing always from a cell toone that is contiguous to it. If you take the right route you will havevisited every cell once, and only once. The puzzle is much easier thanit looks.

261.--THE MONK AND THE BRIDGES.

In this case I give a rough plan of a river with an island and fivebridges. On one side of the river is a monastery, and on the other sideis seen a monk in the foreground. Now, the monk has decided that he willcross every bridge once, and only once, on his return to the monastery.This is, of course, quite easy to do, but on the way he thought tohimself, "I wonder how many different routes there are from which Imight have selected." Could you have told him? That is the puzzle. Takeyour pencil and trace out a route that will take you once over all thefive bridges. Then trace out a second route, then a third, and see ifyou can count all the variations. You will find that the difficulty istwofold: you have to avoid dropping routes on the one hand and countingthe same routes more than once on the other.

[Illustration]

COMBINATION AND GROUP PROBLEMS.

"A combination and a form indeed." _Hamlet_, iii. 4.

Various puzzles in this class might be termed problems in the "geometryof situation," but their solution really depends on the theory ofcombinations which, in its turn, is derived directly from the theory ofpermutations. It has seemed convenient to include here certain grouppuzzles and enumerations that might, perhaps, with equal reason havebeen placed elsewhere; but readers are again asked not to be toocritical about the classification, which is very difficult andarbitrary. As I have included my problem of "The Round Table" (No. 273),perhaps a few remarks on another well-known problem of the same class,known by the French as La Probleme des Menages, may be interesting. Ifn married ladies are seated at a round table in any determined order,in how many different ways may their n husbands be placed so thatevery man is between two ladies but never next to his own wife?

This difficult problem was first solved by Laisant, and the method shownin the following table is due to Moreau:--

4 0 2 5 3 13 6 13 80 7 83 579 8 592 4738 9 4821 43387 10 43979 439792

The first column shows the number of married couples. The numbers in thesecond column are obtained in this way: 5 x 3 + 0 - 2 = 13; 6 x 13 + 3 +2 = 83; 7 x 83 + 13 - 2 = 592; 8 x 592 + 83 + 2 = 4821; and so on. Findall the numbers, except 2, in the table, and the method will be evident.It will be noted that the 2 is subtracted when the first number (thenumber of couples) is odd, and added when that number is even. Thenumbers in the third column are obtained thus: 13 - 0 = 13; 83 - 3 = 80;592 - 13 = 579; 4821 - 83 = 4738; and so on. The numbers in this lastcolumn give the required solutions. Thus, four husbands may be seated intwo ways, five husbands may be placed in thirteen ways, and six husbandsin eighty ways.

The following method, by Lucas, will show the remarkable way in whichchessboard analysis may be applied to the solution of a circular problemof this kind. Divide a square into thirty-six cells, six by six, andstrike out all the cells in the long diagonal from the bottom left-handcorner to the top right-hand corner, also the five cells in the diagonalnext above it and the cell in the bottom right-hand corner. The answerfor six couples will be the same as the number of ways in which you canplace six rooks (not using the cancelled cells) so that no rook shallever attack another rook. It will be found that the six rooks may beplaced in eighty different ways, which agrees with the above table.

262.--THOSE FIFTEEN SHEEP.

A certain cyclopaedia has the following curious problem, I am told:"Place fifteen sheep in four pens so that there shall be the same numberof sheep in each pen." No answer whatever is vouchsafed, so I thought Iwould investigate the matter. I saw that in dealing with apples orbricks the thing would appear to be quite impossible, since four timesany number must be an even number, while fifteen is an odd number. Ithought, therefore, that there must be some quality peculiar to thesheep that was not generally known. So I decided to interview somefarmers on the subject. The first one pointed out that if we put one peninside another, like the rings of a target, and placed all sheep in thesmallest pen, it would be all right. But I objected to this, because youadmittedly place all the sheep in one pen, not in four pens. The secondman said that if I placed four sheep in each of three pens and threesheep in the last pen (that is fifteen sheep in all), and one of theewes in the last pen had a lamb during the night, there would be thesame number in each pen in the morning. This also failed to satisfy me.

[Illustration]

The third farmer said, "I've got four hurdle pens down in one of myfields, and a small flock of wethers, so if you will just step down withme I will show you how it is done." The illustration depicts my friendas he is about to demonstrate the matter to me. His lucid explanationwas evidently that which was in the mind of the writer of the article inthe cyclopaedia. What was it? Can you place those fifteen sheep?

263.--KING ARTHUR'S KNIGHTS.

King Arthur sat at the Round Table on three successive evenings with hisknights--Beleobus, Caradoc, Driam, Eric, Floll, and Galahad--but on nooccasion did any person have as his neighbour one who had before satnext to him. On the first evening they sat in alphabetical order roundthe table. But afterwards King Arthur arranged the two next sittings sothat he might have Beleobus as near to him as possible and Galahad asfar away from him as could be managed. How did he seat the knights tothe best advantage, remembering that rule that no knight may have thesame neighbour twice?

264.--THE CITY LUNCHEONS.

Twelve men connected with a large firm in the City of London sit down toluncheon together every day in the same room. The tables are small onesthat only accommodate two persons at the same time. Can you show howthese twelve men may lunch together on eleven days in pairs, so that notwo of them shall ever sit twice together? We will represent the men bythe first twelve letters of the alphabet, and suppose the first day'spairing to be as follows--

(A B) (C D) (E F) (G H) (I J) (K L).

Then give any pairing you like for the next day, say--

(A C) (B D) (E G) (F H) (I K) (J L),

and so on, until you have completed your eleven lines, with no pair everoccurring twice. There are a good many different arrangements possible.Try to find one of them.

265.--A PUZZLE FOR CARD-PLAYERS.

Twelve members of a club arranged to play bridge together on elevenevenings, but no player was ever to have the same partner more thanonce, or the same opponent more than twice. Can you draw up a schemeshowing how they may all sit down at three tables every evening? Callthe twelve players by the first twelve letters of the alphabet and tryto group them.

266.--A TENNIS TOURNAMENT.

Four married couples played a "mixed double" tennis tournament, a manand a lady always playing against a man and a lady. But no person everplayed with or against any other person more than once. Can you show howthey all could have played together in the two courts on threesuccessive days? This is a little puzzle of a quite practical kind, andit is just perplexing enough to be interesting.

267.--THE WRONG HATS.

"One of the most perplexing things I have come across lately," said Mr.Wilson, "is this. Eight men had been dining not wisely but too well at acertain London restaurant. They were the last to leave, but not one manwas in a condition to identify his own hat. Now, considering that theytook their hats at random, what are the chances that every man took ahat that did not belong to him?"

"The first thing," said Mr. Waterson, "is to see in how many differentways the eight hats could be taken."

"That is quite easy," Mr. Stubbs explained. "Multiply together thenumbers, 1, 2, 3, 4, 5, 6, 7, and 8. Let me see--half a minute--yes;there are 40,320 different ways."

"Now all you've got to do is to see in how many of these cases no manhas his own hat," said Mr. Waterson.

"Thank you, I'm not taking any," said Mr. Packhurst. "I don't envy theman who attempts the task of writing out all those forty-thousand-oddcases and then picking out the ones he wants."

They all agreed that life is not long enough for that sort of amusement;and as nobody saw any other way of getting at the answer, the matter waspostponed indefinitely. Can you solve the puzzle?

268.--THE PEAL OF BELLS.

A correspondent, who is apparently much interested in campanology, asksme how he is to construct what he calls a "true and correct" peal forfour bells. He says that every possible permutation of the four bellsmust be rung once, and once only. He adds that no bell must move morethan one place at a time, that no bell must make more than twosuccessive strokes in either the first or the last place, and that thelast change must be able to pass into the first. These fantasticconditions will be found to be observed in the little peal for threebells, as follows:--

1 2 3 2 1 3 2 3 1 3 2 1 3 1 2 1 3 2

How are we to give him a correct solution for his four bells?

269.--THREE MEN IN A BOAT.

A certain generous London manufacturer gives his workmen every year aweek's holiday at the seaside at his own expense. One year fifteen ofhis men paid a visit to Herne Bay. On the morning of their departurefrom London they were addressed by their employer, who expressed thehope that they would have a very pleasant time.

"I have been given to understand," he added, "that some of you fellowsare very fond of rowing, so I propose on this occasion to provide youwith this recreation, and at the same time give you an amusing littlepuzzle to solve. During the seven days that you are at Herne Bay everyone of you will go out every day at the same time for a row, but theremust always be three men in a boat and no more. No two men may ever goout in a boat together more than once, and no man is allowed to go outtwice in the same boat. If you can manage to do this, and use as fewdifferent boats as possible, you may charge the firm with the expense."

One of the men tells me that the experience he has gained in suchmatters soon enabled him to work out the answer to the entiresatisfaction of themselves and their employer. But the amusing part ofthe thing is that they never really solved the little mystery. I findtheir method to have been quite incorrect, and I think it will amuse myreaders to discover how the men should have been placed in the boats. Astheir names happen to have been Andrews, Baker, Carter, Danby, Edwards,Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, andOnslow, we can call them by their initials and write out the five groupsfor each of the seven days in the following simple way:

1 2 3 4 5 First Day: (ABC) (DEF) (GHI) (JKL) (MNO).

The men within each pair of brackets are here seen to be in the sameboat, and therefore A can never go out with B or with C again, and C cannever go out again with B. The same applies to the other four boats. Thefigures show the number on the boat, so that A, B, or C, for example,can never go out in boat No. 1 again.

270.--THE GLASS BALLS.

A number of clever marksmen were staying at a country house, and thehost, to provide a little amusement, suspended strings of glass balls,as shown in the illustration, to be fired at. After they had all puttheir skill to a sufficient test, somebody asked the following question:"What is the total number of different ways in which these sixteen ballsmay be broken, if we must always break the lowest ball that remains onany string?" Thus, one way would be to break all the four balls on eachstring in succession, taking the strings from left to right. Anotherwould be to break all the fourth balls on the four strings first, thenbreak the three remaining on the first string, then take the balls onthe three other strings alternately from right to left, and so on. Thereis such a vast number of different ways (since every little variation oforder makes a different way) that one is apt to be at first impressed bythe great difficulty of the problem. Yet it is really quite simple whenonce you have hit on the proper method of attacking it. How manydifferent ways are there?

[Illustration]

271.--FIFTEEN LETTER PUZZLE.

ALE FOE HOD BGN CAB HEN JOG KFM HAG GEM MOB BFH FAN KIN JEK DFL JAM HIM GCL LJH AID JIB FCJ NJD OAK FIG HCK MLN BED OIL MCD BLK ICE CON DGK

The above is the solution of a puzzle I gave in _Tit-bits_ in the summerof 1896. It was required to take the letters, A, B, C, D, E, F, G, H, I,J, K, L, M, N, and O, and with them form thirty-five groups of threeletters so that the combinations should include the greatest numberpossible of common English words. No two letters may appear together ina group more than once. Thus, A and L having been together in ALE, mustnever be found together again; nor may A appear again in a group with E,nor L with E. These conditions will be found complied with in the abovesolution, and the number of words formed is twenty-one. Many personshave since tried hard to beat this number, but so far have notsucceeded.

More than thirty-five combinations of the fifteen letters cannot beformed within the conditions. Theoretically, there cannot possibly bemore than twenty-three words formed, because only this number ofcombinations is possible with a vowel or vowels in each. And as noEnglish word can be formed from three of the given vowels (A, E, I, andO), we must reduce the number of possible words to twenty-two. This iscorrect theoretically, but practically that twenty-second word cannot begot in. If JEK, shown above, were a word it would be all right; but itis not, and no amount of juggling with the other letters has resulted ina better answer than the one shown. I should, say that proper nouns andabbreviations, such as Joe, Jim, Alf, Hal, Flo, Ike, etc., aredisallowed.

Now, the present puzzle is a variation of the above. It is simply this:Instead of using the fifteen letters given, the reader is allowed toselect any fifteen different letters of the alphabet that he may prefer.Then construct thirty-five groups in accordance with the conditions, andshow as many good English words as possible.

272.--THE NINE SCHOOLBOYS.

This is a new and interesting companion puzzle to the "FifteenSchoolgirls" (see solution of No. 269), and even in the simplestpossible form in which I present it there are unquestionabledifficulties. Nine schoolboys walk out in triplets on the six week daysso that no boy ever walks _side by side_ with any other boy more thanonce. How would you arrange them?

If we represent them by the first nine letters of the alphabet, theymight be grouped on the first day as follows:--

A B C D E F G H I

Then A can never walk again side by side with B, or B with C, or D withE, and so on. But A can, of course, walk side by side with C. It is herenot a question of being together in the same triplet, but of walkingside by side in a triplet. Under these conditions they can walk out onsix days; under the "Schoolgirls" conditions they can only walk on fourdays.

273.--THE ROUND TABLE.

Seat the same n persons at a round table on

(n - 1)(n - 2) -------------- 2

occasions so that no person shall ever have the same two neighbourstwice. This is, of course, equivalent to saying that every person mustsit once, and once only, between every possible pair.

274.--THE MOUSE-TRAP PUZZLE.

[Illustration

6 20 2 19 13 21 7 5 3 18 17 8 15 11 14 16 1 9 10 4 12

]

This is a modern version, with a difference, of an old puzzle of thesame name. Number twenty-one cards, 1, 2, 3, etc., up to 21, and placethem in a circle in the particular order shown in the illustration.These cards represent mice. You start from any card, calling that card"one," and count, "one, two, three," etc., in a clockwise direction, andwhen your count agrees with the number on the card, you have made a"catch," and you remove the card. Then start at the next card, callingthat "one," and try again to make another "catch." And so on. Supposingyou start at 18, calling that card "one," your first "catch" will be 19.Remove 19 and your next "catch" is 10. Remove 10 and your next "catch"is 1. Remove the 1, and if you count up to 21 (you must never gobeyond), you cannot make another "catch." Now, the ideal is to "catch"all the twenty-one mice, but this is not here possible, and if it wereit would merely require twenty-one different trials, at the most, tosucceed. But the reader may make any two cards change places before hebegins. Thus, you can change the 6 with the 2, or the 7 with the 11, orany other pair. This can be done in several ways so as to enable you to"catch" all the twenty-one mice, if you then start at the right place.You may never pass over a "catch"; you must always remove the card andstart afresh.

275.--THE SIXTEEN SHEEP.

[Illustration:

+========================+ || | | | || || 0 | 0 | 0 | 0 || +-----+-----+-----+------+ || | | | || || 0 | 0 | 0 | 0 || +========================+ || || | || || || 0 || 0 | 0 || 0 || +-----+=====+=====+------+ || | || | || || 0 | 0 || 0 | 0 || +========================+

]

Here is a new puzzle with matches and counters or coins. In theillustration the matches represent hurdles and the counters sheep. Thesixteen hurdles on the outside, and the sheep, must be regarded asimmovable; the puzzle has to do entirely with the nine hurdles on theinside. It will be seen that at present these nine hurdles enclose fourgroups of 8, 3, 3, and 2 sheep. The farmer requires to readjust some ofthe hurdles so as to enclose 6, 6, and 4 sheep. Can you do it by onlyreplacing two hurdles? When you have succeeded, then try to do it byreplacing three hurdles; then four, five, six, and seven in succession.Of course, the hurdles must be legitimately laid on the dotted lines,and no such tricks are allowed as leaving unconnected ends of hurdles,or two hurdles placed side by side, or merely making hurdles changeplaces. In fact, the conditions are so simple that any farm labourerwill understand it directly.

276.--THE EIGHT VILLAS.

In one of the outlying suburbs of London a man had a square plot ofground on which he decided to build eight villas, as shown in theillustration, with a common recreation ground in the middle. After thehouses were completed, and all or some of them let, he discovered thatthe number of occupants in the three houses forming a side of the squarewas in every case nine. He did not state how the occupants weredistributed, but I have shown by the numbers on the sides of the housesone way in which it might have happened. The puzzle is to discover thetotal number of ways in which all or any of the houses might beoccupied, so that there should be nine persons on each side. In orderthat there may be no misunderstanding, I will explain that although B iswhat we call a reflection of A, these would count as two differentarrangements, while C, if it is turned round, will give fourarrangements; and if turned round in front of a mirror, four otherarrangements. All eight must be counted.

[Illustration:

/\ /\ /\ |2 | |5 | |2 |

/\ /\ |5 | |5 |

/\ /\ /\ |2 | |5 | |2 |

+---+---+---+ +---+---+---+ +---+---+---+ | 1 | 6 | 2 | | 2 | 6 | 1 | | 1 | 6 | 2 | +---+---+---+ +---+---+---+ +---+---+---+ | 6 | | 6 | | 6 | | 6 | | 4 | | 4 | +---+---+---+ +---+---+---+ +---+---+---+ | 2 | 6 | 1 | | 1 | 6 | 2 | | 4 | 2 | 3 | +---+---+---+ +---+---+---+ +---+---+---+ A B C

]

277.--COUNTER CROSSES.

All that we need for this puzzle is nine counters, numbered 1, 2, 3, 4,5, 6, 7, 8, and 9. It will be seen that in the illustration A these arearranged so as to form a Greek cross, while in the case of B they form aLatin cross. In both cases the reader will find that the sum of thenumbers in the upright of the cross is the same as the sum of thenumbers in the horizontal arm. It is quite easy to hit on such anarrangement by trial, but the problem is to discover in exactly how manydifferent ways it may be done in each case. Remember that reversals andreflections do not count as different. That is to say, if you turn thispage round you get four arrangements of the Greek cross, and if you turnit round again in front of a mirror you will get four more. But theseeight are all regarded as one and the same. Now, how many different waysare there in each case?

[Illustration:

(1) (2)

(2) (4) (5) (1) (6) (7)

(3) (4) (9) (5) (6) (3)

(7) (8)

A (8) B (9)

]

278.--A DORMITORY PUZZLE.

In a certain convent there were eight large dormitories on one floor,approached by a spiral staircase in the centre, as shown in our plan. Onan inspection one Monday by the abbess it was found that the southaspect was so much preferred that six times as many nuns slept on thesouth side as on each of the other three sides. She objected to thisovercrowding, and ordered that it should be reduced. On Tuesday shefound that five times as many slept on the south side as on each of theother sides. Again she complained. On Wednesday she found four times asmany on the south side, on Thursday three times as many, and on Fridaytwice as many. Urging the nuns to further efforts, she was pleased tofind on Saturday that an equal number slept on each of the four sides ofthe house. What is the smallest number of nuns there could have been,and how might they have arranged themselves on each of the six nights?No room may ever be unoccupied.

[Illustration

+---+---+---+ | | | | | | | | | | | | +---+---+---+ | |\|/| | | |-*-| | | |/|\| | +---+---+---+ | | | | | | | | | | | | +---+---+---+

]

279.--THE BARRELS OF BALSAM.

A merchant of Bagdad had ten barrels of precious balsam for sale. Theywere numbered, and were arranged in two rows, one on top of the other,as shown in the picture. The smaller the number on the barrel, thegreater was its value. So that the best quality was numbered "1" and theworst numbered "10," and all the other numbers of graduating values.Now, the rule of Ahmed Assan, the merchant, was that he never put abarrel either beneath or to the right of one of less value. Thearrangement shown is, of course, the simplest way of complying with thiscondition. But there are many other ways--such, for example, as this:--

1 2 5 7 8 3 4 6 9 10

Here, again, no barrel has a smaller number than itself on its right orbeneath it. The puzzle is to discover in how many different ways themerchant of Bagdad might have arranged his barrels in the two rowswithout breaking his rule. Can you count the number of ways?

280.--BUILDING THE TETRAHEDRON.

I possess a tetrahedron, or triangular pyramid, formed of six sticksglued together, as shown in the illustration. Can you count correctlythe number of different ways in which these six sticks might have beenstuck together so as to form the pyramid?

Some friends worked at it together one evening, each person providinghimself with six lucifer matches to aid his thoughts; but it was foundthat no two results were the same. You see, if we remove one of thesticks and turn it round the other way, that will be a differentpyramid. If we make two of the sticks change places the result willagain be different. But remember that every pyramid may be made to standon either of its four sides without being a different one. How many waysare there altogether?

[Illustration]

281.--PAINTING A PYRAMID.

This puzzle concerns the painting of the four sides of a tetrahedron, ortriangular pyramid. If you cut out a piece of cardboard of thetriangular shape shown in Fig. 1, and then cut half through along thedotted lines, it will fold up and form a perfect triangular pyramid. AndI would first remind my readers that the primary colours of the solarspectrum are seven--violet, indigo, blue, green, yellow, orange, andred. When I was a child I was taught to remember these by the ungainlyword formed by the initials of the colours, "Vibgyor."

[Illustration]

In how many different ways may the triangular pyramid be coloured, usingin every case one, two, three, or four colours of the solar spectrum? Ofcourse a side can only receive a single colour, and no side can be leftuncoloured. But there is one point that I must make quite clear. Thefour sides are not to be regarded as individually distinct. That is tosay, if you paint your pyramid as shown in Fig. 2 (where the bottom sideis green and the other side that is out of view is yellow), and thenpaint another in the order shown in Fig. 3, these are really both thesame and count as one way. For if you tilt over No. 2 to the right itwill so fall as to represent No. 3. The avoidance of repetitions of thiskind is the real puzzle of the thing. If a coloured pyramid cannot beplaced so that it exactly resembles in its colours and their relativeorder another pyramid, then they are different. Remember that one waywould be to colour all the four sides red, another to colour two sidesgreen, and the remaining sides yellow and blue; and so on.

282.--THE ANTIQUARY'S CHAIN.

An antiquary possessed a number of curious old links, which he took to ablacksmith, and told him to join together to form one straight piece ofchain, with the sole condition that the two circular links were not tobe together. The following illustration shows the appearance of thechain and the form of each link. Now, supposing the owner shouldseparate the links again, and then take them to another smith and repeathis former instructions exactly, what are the chances against the linksbeing put together exactly as they were by the first man? Remember thatevery successive link can be joined on to another in one of two ways,just as you can put a ring on your finger in two ways, or link yourforefingers and thumbs in two ways.

[Illustration]

283.--THE FIFTEEN DOMINOES.

In this case we do not use the complete set of twenty-eight dominoes tobe found in the ordinary box. We dispense with all those dominoes thathave a five or a six on them and limit ourselves to the fifteen thatremain, where the double-four is the highest.

In how many different ways may the fifteen dominoes be arranged in astraight line in accordance with the simple rule of the game that anumber must always be placed against a similar number--that is, a fouragainst a four, a blank against a blank, and so on? Left to right andright to left of the same arrangement are to be counted as two differentways.

384.--THE CROSS TARGET.

+-+-+ |*|*| +-+-+ |*|*| +-+-+-+-+-+-+ | | | |*| | | +-+-+-+-+-+-+ | | |*| |*| | +-+-+-+-+-+-+ | |*| +-+-+ | | | +-+-+

In the illustration we have a somewhat curious target designed by aneccentric sharpshooter. His idea was that in order to score you must hitfour circles in as many shots so that those four shots shall form asquare. It will be seen by the results recorded on the target that twoattempts have been successful. The first man hit the four circles at thetop of the cross, and thus formed his square. The second man intended tohit the four in the bottom arm, but his second shot, on the left, wenttoo high. This compelled him to complete his four in a different waythan he intended. It will thus be seen that though it is immaterialwhich circle you hit at the first shot, the second shot may commit youto a definite procedure if you are to get your square. Now, the puzzleis to say in just how many different ways it is possible to form asquare on the target with four shots.

285.--THE FOUR POSTAGE STAMPS.

+---+----+----+----+ | 1 | 2 | 3 | 4 | +---+----+----+----+ | 5 | 6 | 7 | 8 | +---+----+----+----+ | 9 | 10 | 11 | 12 | +---+----+----+----+

"It is as easy as counting," is an expression one sometimes hears. Butmere counting may be puzzling at times. Take the following simpleexample. Suppose you have just bought twelve postage stamps, in thisform--three by four--and a friend asks you to oblige him with fourstamps, all joined together--no stamp hanging on by a mere corner. Inhow many different ways is it possible for you to tear off those fourstamps? You see, you can give him 1, 2, 3, 4, or 2, 3, 6, 7, or 1, 2, 3,6, or 1, 2, 3, 7, or 2, 3, 4, 8, and so on. Can you count the number ofdifferent ways in which those four stamps might be delivered? There arenot many more than fifty ways, so it is not a big count. Can you get theexact number?

286.--PAINTING THE DIE.

In how many different ways may the numbers on a single die be marked,with the only condition that the 1 and 6, the 2 and 5, and the 3 and 4must be on opposite sides? It is a simple enough question, and yet itwill puzzle a good many people.

287.--AN ACROSTIC PUZZLE.

In the making or solving of double acrostics, has it ever occurred toyou to consider the variety and limitation of the pair of initial andfinal letters available for cross words? You may have to find a wordbeginning with A and ending with B, or A and C, or A and D, and so on.Some combinations are obviously impossible--such, for example, as thosewith Q at the end. But let us assume that a good English word can befound for every case. Then how many possible pairs of letters areavailable?

CHESSBOARD PROBLEMS.

"You and I will goe to the chesse."

GREENE'S _Groatsworth of Wit._

During a heavy gale a chimney-pot was hurled through the air, andcrashed upon the pavement just in front of a pedestrian. He quite calmlysaid, "I have no use for it: I do not smoke." Some readers, when theyhappen to see a puzzle represented on a chessboard with chess pieces,are apt to make the equally inconsequent remark, "I have no use for it:I do not play chess." This is largely a result of the common, buterroneous, notion that the ordinary chess puzzle with which we arefamiliar in the press (dignified, for some reason, with the name"problem") has a vital connection with the game of chess itself. Butthere is no condition in the game that you shall checkmate your opponentin two moves, in three moves, or in four moves, while the majority ofthe positions given in these puzzles are such that one player would haveso great a superiority in pieces that the other would have resignedbefore the situations were reached. And the solving of them helps youbut little, and that quite indirectly, in playing the game, it beingwell known that, as a rule, the best "chess problemists" are indifferentplayers, and _vice versa_. Occasionally a man will be found strong onboth subjects, but he is the exception to the rule.

Yet the simple chequered board and the characteristic moves of thepieces lend themselves in a very remarkable manner to the devising ofthe most entertaining puzzles. There is room for such infinite varietythat the true puzzle lover cannot afford to neglect them. It was with aview to securing the interest of readers who are frightened off by themere presentation of a chessboard that so many puzzles of this classwere originally published by me in various fanciful dresses. Some ofthese posers I still retain in their disguised form; others I havetranslated into terms of the chessboard. In the majority of cases thereader will not need any knowledge whatever of chess, but I have thoughtit best to assume throughout that he is acquainted with the terminology,the moves, and the notation of the game.

I first deal with a few questions affecting the chessboard itself; thenwith certain statical puzzles relating to the Rook, the Bishop, theQueen, and the Knight in turn; then dynamical puzzles with the pieces inthe same order; and, finally, with some miscellaneous puzzles on thechessboard. It is hoped that the formulae and tables given at the end ofthe statical puzzles will be of interest, as they are, for the mostpart, published for the first time.

THE CHESSBOARD.

"Good company's a chessboard." BYRON'S _Don Juan_, xiii. 89.

A chessboard is essentially a square plane divided into sixty-foursmaller squares by straight lines at right angles. Originally it was notchequered (that is, made with its rows and columns alternately black andwhite, or of any other two colours), and this improvement was introducedmerely to help the eye in actual play. The utility of the chequers isunquestionable. For example, it facilitates the operation of thebishops, enabling us to see at the merest glance that our king or pawnson black squares are not open to attack from an opponent's bishoprunning on the white diagonals. Yet the chequering of the board is notessential to the game of chess. Also, when we are propounding puzzles onthe chessboard, it is often well to remember that additional interestmay result from "generalizing" for boards containing any number ofsquares, or from limiting ourselves to some particular chequeredarrangement, not necessarily a square. We will give a few puzzlesdealing with chequered boards in this general way.

288.--CHEQUERED BOARD DIVISIONS.

I recently asked myself the question: In how many different ways may achessboard be divided into two parts of the same size and shape by cutsalong the lines dividing the squares? The problem soon proved to be bothfascinating and bristling with difficulties. I present it in asimplified form, taking a board of smaller dimensions.

[Illustration:

+---+---*---+---+ +---+---+---*---+ +---+---+---*---+ | | H | | | | | H | | | | H | +---+---*---+---+ +---+---*===*---+ +---*===*---*---+ | | H | | | | H | | | H H H | +---+---*---+---+ +---+---*---+---+ +---*---*---*---+ | | H | | | | H | | | H H H | +---+---*---+---+ +---*===*---+---+ +---*---*===*---+ | | H | | | H | | | | H | | | +---+---*---+---+ +---*---+---+---+ +---*---+---+---+

+---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+

+---+---*---+---+ +---+---+---*---+ +---+---+---*---+ | | H | | | | | H | | | | H | +---*===*---+---+ +---*===*===*---+ +---+---*===*---+ | H | | | | H | | | | | H | | +---*===*===*---+ +---*===*===*---+ +---+---*---+---+ | | | H | | | | H | | | H | | +---+---*===*---+ +---*===*===*---+ +---*===*---+---+ | | H | | | H | | | | H | | | +---+---*---+---+ +---*---+---+---+ +---*---+---+---+

]

It is obvious that a board of four squares can only be so divided in oneway--by a straight cut down the centre--because we shall not countreversals and reflections as different. In the case of a board ofsixteen squares--four by four--there are just six different ways. I havegiven all these in the diagram, and the reader will not find any others.Now, take the larger board of thirty-six squares, and try to discover inhow many ways it may be cut into two parts of the same size and shape.

289.--LIONS AND CROWNS.

The young lady in the illustration is confronted with a littlecutting-out difficulty in which the reader may be glad to assist her.She wishes, for some reason that she has not communicated to me, to cutthat square piece of valuable material into four parts, all of exactlythe same size and shape, but it is important that every piece shallcontain a lion and a crown. As she insists that the cuts can only bemade along the lines dividing the squares, she is considerably perplexedto find out how it is to be done. Can you show her the way? There isonly one possible method of cutting the stuff.

[Illustration:

+-+-+-+-+-+-+ | | | | | | | +-+-+-+-+-+-+ | |L|L|L| | | +-+-+-+-+-+-+ | | |C|C| | | +-+-+-+-+-+-+ | | |C|C| | | +-+-+-+-+-+-+ |L| | | | | | +-+-+-+-+-+-+ | | | | | | | +-+-+-+-+-+-+

]

290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

We will here consider the question of those boards that contain an oddnumber of squares. We will suppose that the central square is first cutout, so as to leave an even number of squares for division. Now, it isobvious that a square three by three can only be divided in one way, asshown in Fig. 1. It will be seen that the pieces A and B are of the samesize and shape, and that any other way of cutting would only produce thesame shaped pieces, so remember that these variations are not counted asdifferent ways. The puzzle I propose is to cut the board five by five(Fig. 2) into two pieces of the same size and shape in as many differentways as possible. I have shown in the illustration one way of doing it.How many different ways are there altogether? A piece which when turnedover resembles another piece is not considered to be of a differentshape.

[Illustration:

+---*---+---+ | H | | +---*===*---+ | HHHHH | +---*===*---+ | | H | +---+---*---+

Fig 1]

[Illustration:

+---+---+---+---+---+ | | | | | | *===*===*===*---+---+ | | | H | | +---+---*===*---+---+ | | HHHHH | | +---+---*===*---+---+ | | H | | | +---+---*===*===*===* | H | | | | +---*---+---+---+---+

Fig 2]

291.--THE GRAND LAMA'S PROBLEM.

Once upon a time there was a Grand Lama who had a chessboard made ofpure gold, magnificently engraved, and, of course, of great value. Everyyear a tournament was held at Lhassa among the priests, and whenever anyone beat the Grand Lama it was considered a great honour, and his namewas inscribed on the back of the board, and a costly jewel set in theparticular square on which the checkmate had been given. After thissovereign pontiff had been defeated on four occasions he died--possiblyof chagrin.

[Illustration:

+---+---+---+---+---+---+---+---+ | * | | | | | | | | +---+---+---+---+---+---+---+---+ | | * | | | | | | | +---+---+---+---+---+---+---+---+ | | | * | | | | | | +---+---+---+---+---+---+---+---+ | | | | * | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+

]

Now the new Grand Lama was an inferior chess-player, and preferred otherforms of innocent amusement, such as cutting off people's heads. So hediscouraged chess as a degrading game, that did not improve either themind or the morals, and abolished the tournament summarily. Then he sentfor the four priests who had had the effrontery to play better than aGrand Lama, and addressed them as follows: "Miserable and heathenishmen, calling yourselves priests! Know ye not that to lay claim to acapacity to do anything better than my predecessor is a capital offence?Take that chessboard and, before day dawns upon the torture chamber, cutit into four equal parts of the same shape, each containing sixteenperfect squares, with one of the gems in each part! If in this you fail,then shall other sports be devised for your special delectation. Go!"The four priests succeeded in their apparently hopeless task. Can youshow how the board may be divided into four equal parts, each ofexactly the same shape, by cuts along the lines dividing the squares,each part to contain one of the gems?

292.--THE ABBOT'S WINDOW.

[Illustration]

Once upon a time the Lord Abbot of St. Edmondsbury, in consequence of"devotions too strong for his head," fell sick and was unable to leavehis bed. As he lay awake, tossing his head restlessly from side to side,the attentive monks noticed that something was disturbing his mind; butnobody dared ask what it might be, for the abbot was of a sterndisposition, and never would brook inquisitiveness. Suddenly he calledfor Father John, and that venerable monk was soon at the bedside.

"Father John," said the Abbot, "dost thou know that I came into thiswicked world on a Christmas Even?"

The monk nodded assent.

"And have I not often told thee that, having been born on ChristmasEven, I have no love for the things that are odd? Look there!"

The Abbot pointed to the large dormitory window, of which I give asketch. The monk looked, and was perplexed.

"Dost thou not see that the sixty-four lights add up an even numbervertically and horizontally, but that all the _diagonal_ lines, exceptfourteen are of a number that is odd? Why is this?"

"Of a truth, my Lord Abbot, it is of the very nature of things, andcannot be changed."

"Nay, but it _shall_ be changed. I command thee that certain of thelights be closed this day, so that every line shall have an even numberof lights. See thou that this be done without delay, lest the cellars belocked up for a month and other grievous troubles befall thee."

Father John was at his wits' end, but after consultation with one whowas learned in strange mysteries, a way was found to satisfy the whim ofthe Lord Abbot. Which lights were blocked up, so that those whichremained added up an even number in every line horizontally, vertically,and diagonally, while the least possible obstruction of light wascaused?

293.--THE CHINESE CHESSBOARD.

Into how large a number of different pieces may the chessboard be cut(by cuts along the lines only), no two pieces being exactly alike?Remember that the arrangement of black and white constitutes adifference. Thus, a single black square will be different from a singlewhite square, a row of three containing two white squares will differfrom a row of three containing two black, and so on. If two piecescannot be placed on the table so as to be exactly alike, they count asdifferent. And as the back of the board is plain, the pieces cannot beturned over.

294.--THE CHESSBOARD SENTENCE.

[Illustration]

I once set myself the amusing task of so dissecting an ordinarychessboard into letters of the alphabet that they would form a completesentence. It will be seen from the illustration that the piecesassembled give the sentence, "CUT THY LIFE," with the stops between. Theideal sentence would, of course, have only one full stop, but that I didnot succeed in obtaining.

The sentence is an appeal to the transgressor to cut himself adrift fromthe evil life he is living. Can you fit these pieces together to form aperfect chessboard?

STATICAL CHESS PUZZLES.

"They also serve who only stand and wait." MILTON.

295.--THE EIGHT ROOKS.

[Illustration:

+---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | R | R | R | R | R | R | R | R | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+

FIG. 1.]

[Illustration:

+---+---+---+---+---+---+---+---+ | R | | | | | | | | +---+---+---+---+---+---+---+---+ | | R | | | | | | | +---+---+---+---+---+---+---+---+ | | | R | | | | | | +---+---+---+---+---+---+---+---+ | | | | R | | | | | +---+---+---+---+---+---+---+---+ | | | | | R | | | | +---+---+---+---+---+---+---+---+ | | | | | | R | | | +---+---+---+---+---+---+---+---+ | | | | | | | R | | +---+---+---+---+---+---+---+---+ | | | | | | | | R | +---+---+---+---+---+---+---+---+

FIG. 2.]

It will be seen in the first diagram that every square on the board iseither occupied or attacked by a rook, and that every rook is "guarded"(if they were alternately black and white rooks we should say"attacked") by another rook. Placing the eight rooks on any row or fileobviously will have the same effect. In diagram 2 every square is againeither occupied or attacked, but in this case every rook is unguarded.Now, in how many different ways can you so place the eight rooks on theboard that every square shall be occupied or attacked and no rook everguarded by another? I do not wish to go into the question of reversalsand reflections on this occasion, so that placing the rooks on the otherdiagonal will count as different, and similarly with other repetitionsobtained by turning the board round.

296.--THE FOUR LIONS.

The puzzle is to find in how many different ways the four lions may beplaced so that there shall never be more than one lion in any row orcolumn. Mere reversals and reflections will not count as different.Thus, regarding the example given, if we place the lions in the otherdiagonal, it will be considered the same arrangement. For if you holdthe second arrangement in front of a mirror or give it a quarter turn,you merely get the first arrangement. It is a simple little puzzle, butrequires a certain amount of careful consideration.

[Illustration

+---+---+---+---+ | L | | | | +---+---+---+---+ | | L | | | +---+---+---+---+ | | | L | | +---+---+---+---+ | | | | L | +---+---+---+---+

]

297.--BISHOPS--UNGUARDED.

Place as few bishops as possible on an ordinary chessboard so that everysquare of the board shall be either occupied or attacked. It will beseen that the rook has more scope than the bishop: for wherever youplace the former, it will always attack fourteen other squares; whereasthe latter will attack seven, nine, eleven, or thirteen squares,according to the position of the diagonal on which it is placed. And itis well here to state that when we speak of "diagonals" in connectionwith the chessboard, we do not limit ourselves to the two long diagonalsfrom corner to corner, but include all the shorter lines that areparallel to these. To prevent misunderstanding on future occasions, itwill be well for the reader to note carefully this fact.

298.--BISHOPS--GUARDED.

Now, how many bishops are necessary in order that every square shall beeither occupied or attacked, and every bishop guarded by another bishop?And how may they be placed?

299.--BISHOPS IN CONVOCATION.

[Illustration:

+---+---+---+---+---+---+---+---+ | B | B | B | B | B | B | B | B | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | B | B | B | B | B | B | | +---+---+---+---+---+---+---+---+

]

The greatest number of bishops that can be placed at the same time onthe chessboard, without any bishop attacking another, is fourteen. Ishow, in diagram, the simplest way of doing this. In fact, on a squarechequered board of any number of squares the greatest number of bishopsthat can be placed without attack is always two less than twice thenumber of squares on the side. It is an interesting puzzle to discoverin just how many different ways the fourteen bishops may be so placedwithout mutual attack. I shall give an exceedingly simple rule fordetermining the number of ways for a square chequered board of anynumber of squares.

300.--THE EIGHT QUEENS.

[Illustration:

+---+---+---+---+---+---+---+---+ | | | | ..Q | | | | +---+---+---+...+---+---+---+---+ | | ..Q.. | | | | | +---+...+---+---+---+---+---+---+ | Q.. | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | Q | | +---+---+---+---+---+---+---+---+ | | Q | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | ..Q | +---+---+---+---+---+---+...+---+ | | | | | ..Q.. | | +---+---+---+---+...+---+---+---+ | | | | Q.. | | | | +---+---+---+---+---+---+---+---+

]

The queen is by far the strongest piece on the chessboard. If you placeher on one of the four squares in the centre of the board, she attacksno fewer than twenty-seven other squares; and if you try to hide her ina corner, she still attacks twenty-one squares. Eight queens may beplaced on the board so that no queen attacks another, and it is an oldpuzzle (first proposed by Nauck in 1850, and it has quite a littleliterature of its own) to discover in just how many different ways thismay be done. I show one way in the diagram, and there are in all twelveof these fundamentally different ways. These twelve produce ninety-twoways if we regard reversals and reflections as different. The diagram isin a way a symmetrical arrangement. If you turn the page upside down, itwill reproduce itself exactly; but if you look at it with one of theother sides at the bottom, you get another way that is not identical.Then if you reflect these two ways in a mirror you get two more ways.Now, all the other eleven solutions are non-symmetrical, and thereforeeach of them may be presented in eight ways by these reversals andreflections. It will thus be seen why the twelve fundamentally differentsolutions produce only ninety-two arrangements, as I have said, and notninety-six, as would happen if all twelve were non-symmetrical. It iswell to have a clear understanding on the matter of reversals andreflections when dealing with puzzles on the chessboard.

Can the reader place the eight queens on the board so that no queenshall attack another and so that no three queens shall be in a straightline in any oblique direction? Another glance at the diagram will showthat this arrangement will not answer the conditions, for in the twodirections indicated by the dotted lines there are three queens in astraight line. There is only one of the twelve fundamental ways thatwill solve the puzzle. Can you find it?

301.--THE EIGHT STARS.

[Illustration:

+---+---+---+---+---+---+---+---+ |///| | | | | | |///| +---+---+---+---+---+---+---+---+ | |///| | | | |///| * | +---+---+---+---+---+---+---+---+ | | |///| | |///| | | +---+---+---+---+---+---+---+---+ | | | |///|///| | | | +---+---+---+---+---+---+---+---+ | | | |///|///| | | | +---+---+---+---+---+---+---+---+ | | |///| | |///| | | +---+---+---+---+---+---+---+---+ | |///| | | | |///| | +---+---+---+---+---+---+---+---+ |///| | | | | | |///| +---+---+---+---+---+---+---+---+

]

The puzzle in this case is to place eight stars in the diagram so thatno star shall be in line with another star horizontally, vertically, ordiagonally. One star is already placed, and that must not be moved, sothere are only seven for the reader now to place. But you must not placea star on any one of the shaded squares. There is only one way ofsolving this little puzzle.

302.--A PROBLEM IN MOSAICS.

The art of producing pictures or designs by means of joining togetherpieces of hard substances, either naturally or artificially coloured, isof very great antiquity. It was certainly known in the time of thePharaohs, and we find a reference in the Book of Esther to "a pavementof red, and blue, and white, and black marble." Some of this ancientwork that has come down to us, especially some of the Roman mosaics,would seem to show clearly, even where design is not at first evident,that much thought was bestowed upon apparently disorderly arrangements.Where, for example, the work has been produced with a very limitednumber of colours, there are evidences of great ingenuity in preventingthe same tints coming in close proximity. Lady readers who are familiarwith the construction of patchwork quilts will know how desirable it issometimes, when they are limited in the choice of material, to preventpieces of the same stuff coming too near together. Now, this puzzle willapply equally to patchwork quilts or tesselated pavements.

It will be seen from the diagram how a square piece of flooring may bepaved with sixty-two square tiles of the eight colours violet, red,yellow, green, orange, purple, white, and blue (indicated by the initialletters), so that no tile is in line with a similarly coloured tile,vertically, horizontally, or diagonally. Sixty-four such tiles could notpossibly be placed under these conditions, but the two shaded squareshappen to be occupied by iron ventilators.

[Illustration:

+---+---+---+---+---+---+---+---+ | V | R | Y | G | O | P | W | B | +---+---+---+---+---+---+---+---+ | W | B | O | P | Y | G | V | R | +---+---*===*---+---*===*---+---+ | G | P H W H V | B H R H Y | O | +---+---*===*---+---*===*---+---+ | R | Y | B | O | G | V | P | W | +---+---+---+---+---+---+---+---+ | B | G | R | Y | P | W | O | V | +---+---+---+---+---+---+---+---+ | O | V | P | W | R | Y | B | G | +---+---+---+---+---+---+---+---+ | P | W | G | B | V | O | R | Y | +---+---+---+---+---+---+---+---+ |///| O | V | R | W | B | G |///| +---+---+---+---+---+---+---+---+

]

The puzzle is this. These two ventilators have to be removed to thepositions indicated by the darkly bordered tiles, and two tiles placedin those bottom corner squares. Can you readjust the thirty-two tiles sothat no two of the same colour shall still be in line?

303.--UNDER THE VEIL.

[Illustration:

+---+---+---+---+---+---+---+---+ | | | V | E | I | L | | | +---+---+---+---+---+---+---+---+ | | | I | L | V | E | | | +---+---+---+---+---+---+---+---+ | I | V | | | | | L | E | +---+---+---+---+---+---+---+---+ | L | E | | | | | I | V | +---+---+---+---+---+---+---+---+ | V | I | | | | | E | L | +---+---+---+---+---+---+---+---+ | E | L | | | | | V | I | +---+---+---+---+---+---+---+---+ | | | E | V | L | I | | | +---+---+---+---+---+---+---+---+ | | | L | I | E | V | | | +---+---+---+---+---+---+---+---+

]

If the reader will examine the above diagram, he will see that I have soplaced eight V's, eight E's, eight I's, and eight L's in the diagramthat no letter is in line with a similar one horizontally, vertically,or diagonally. Thus, no V is in line with another V, no E with anotherE, and so on. There are a great many different ways of arranging theletters under this condition. The puzzle is to find an arrangement thatproduces the greatest possible number of four-letter words, readingupwards and downwards, backwards and forwards, or diagonally. Allrepetitions count as different words, and the five variations that maybe used are: VEIL, VILE, LEVI, LIVE, and EVIL.

This will be made perfectly clear when I say that the above arrangementscores eight, because the top and bottom row both give VEIL; the secondand seventh columns both give VEIL; and the two diagonals, starting fromthe L in the 5th row and E in the 8th row, both give LIVE and EVIL.There are therefore eight different readings of the words in all.

This difficult word puzzle is given as an example of the use ofchessboard analysis in solving such things. Only a person who isfamiliar with the "Eight Queens" problem could hope to solve it.

304.--BACHET'S SQUARE.

One of the oldest card puzzles is by Claude Caspar Bachet de Meziriac,first published, I believe, in the 1624 edition of his work. Rearrangethe sixteen court cards (including the aces) in a square so that in norow of four cards, horizontal, vertical, or diagonal, shall be found twocards of the same suit or the same value. This in itself is easy enough,but a point of the puzzle is to find in how many different ways this maybe done. The eminent French mathematician A. Labosne, in his modernedition of Bachet, gives the answer incorrectly. And yet the puzzle isreally quite easy. Any arrangement produces seven more by turning thesquare round and reflecting it in a mirror. These are counted asdifferent by Bachet.

Note "row of four cards," so that the only diagonals we have here toconsider are the two long ones.

305.--THE THIRTY-SIX LETTER-BLOCKS.

[Illustration]

The illustration represents a box containing thirty-six letter-blocks.The puzzle is to rearrange these blocks so that no A shall be in a linevertically, horizontally, or diagonally with another A, no B withanother B, no C with another C, and so on. You will find it impossibleto get all the letters into the box under these conditions, but thepoint is to place as many as possible. Of course no letters other thanthose shown may be used.

306.--THE CROWDED CHESSBOARD.

[Illustration]

The puzzle is to rearrange the fifty-one pieces on the chessboard sothat no queen shall attack another queen, no rook attack another rook,no bishop attack another bishop, and no knight attack another knight. Nonotice is to be taken of the intervention of pieces of another type fromthat under consideration--that is, two queens will be considered toattack one another although there may be, say, a rook, a bishop, and aknight between them. And so with the rooks and bishops. It is notdifficult to dispose of each type of piece separately; the difficultycomes in when you have to find room for all the arrangements on theboard simultaneously.

307.--THE COLOURED COUNTERS.

[Illustration]

The diagram represents twenty-five coloured counters, Red, Blue, Yellow,Orange, and Green (indicated by their initials), and there are five ofeach colour, numbered 1, 2, 3, 4, and 5. The problem is so to place themin a square that neither colour nor number shall be found repeated inany one of the five rows, five columns, and two diagonals. Can you sorearrange them?

308.--THE GENTLE ART OF STAMP-LICKING.

The Insurance Act is a most prolific source of entertaining puzzles,particularly entertaining if you happen to be among the exempt. One'sinitiation into the gentle art of stamp-licking suggests the followinglittle poser: If you have a card divided into sixteen spaces (4 x 4),and are provided with plenty of stamps of the values 1d., 2d., 3d., 4d.,and 5d., what is the greatest value that you can stick on the card ifthe Chancellor of the Exchequer forbids you to place any stamp in astraight line (that is, horizontally, vertically, or diagonally) withanother stamp of similar value? Of course, only one stamp can be affixedin a space. The reader will probably find, when he sees the solution,that, like the stamps themselves, he is licked He will most likely betwopence short of the maximum. A friend asked the Post Office how it wasto be done; but they sent him to the Customs and Excise officer, whosent him to the Insurance Commissioners, who sent him to an approvedsociety, who profanely sent him--but no matter.

309.--THE FORTY-NINE COUNTERS.

[Illustration]

Can you rearrange the above forty-nine counters in a square so that noletter, and also no number, shall be in line with a similar one,vertically, horizontally, or diagonally? Here I, of course, mean in thelines parallel with the diagonals, in the chessboard sense.

310.--THE THREE SHEEP.

[Illustration]

A farmer had three sheep and an arrangement of sixteen pens, divided offby hurdles in the manner indicated in the illustration. In how manydifferent ways could he place those sheep, each in a separate pen, sothat every pen should be either occupied or in line (horizontally,vertically, or diagonally) with at least one sheep? I have given onearrangement that fulfils the conditions. How many others can you find?Mere reversals and reflections must not be counted as different. Thereader may regard the sheep as queens. The problem is then to place thethree queens so that every square shall be either occupied or attackedby at least one queen--in the maximum number of different ways.

311.--THE FIVE DOGS PUZZLE.

In 1863, C.F. de Jaenisch first discussed the "Five Queens Puzzle"--toplace five queens on the chessboard so that every square shall beattacked or occupied--which was propounded by his friend, a "Mr. de R."Jaenisch showed that if no queen may attack another there are ninety-onedifferent ways of placing the five queens, reversals and reflections notcounting as different. If the queens may attack one another, I haverecorded hundreds of ways, but it is not practicable to enumerate themexactly.

[Illustration]

The illustration is supposed to represent an arrangement of sixty-fourkennels. It will be seen that five kennels each contain a dog, and onfurther examination it will be seen that every one of the sixty-fourkennels is in a straight line with at least one dog--eitherhorizontally, vertically, or diagonally. Take any kennel you like, andyou will find that you can draw a straight line to a dog in one or otherof the three ways mentioned. The puzzle is to replace the five dogs anddiscover in just how many different ways they may be placed in fivekennels _in a straight row_, so that every kennel shall always be inline with at least one dog. Reversals and reflections are here countedas different.

312.--THE FIVE CRESCENTS OF BYZANTIUM.

When Philip of Macedon, the father of Alexander the Great, found himselfconfronted with great difficulties in the siege of Byzantium, he set hismen to undermine the walls. His desires, however, miscarried, for nosooner had the operations been begun than a crescent moon suddenlyappeared in the heavens and discovered his plans to his adversaries. TheByzantines were naturally elated, and in order to show their gratitudethey erected a statue to Diana, and the crescent became thenceforward asymbol of the state. In the temple that contained the statue was asquare pavement composed of sixty-four large and costly tiles. Thesewere all plain, with the exception of five, which bore the symbol of thecrescent. These five were for occult reasons so placed that every tileshould be watched over by (that is, in a straight line, vertically,horizontally, or diagonally with) at least one of the crescents. Thearrangement adopted by the Byzantine architect was as follows:--

[Illustration]

Now, to cover up one of these five crescents was a capital offence, thedeath being something very painful and lingering. But on a certainoccasion of festivity it was necessary to lay down on this pavement asquare carpet of the largest dimensions possible, and I have shown inthe illustration by dark shading the largest dimensions that would beavailable.

The puzzle is to show how the architect, if he had foreseen thisquestion of the carpet, might have so arranged his five crescent tilesin accordance with the required conditions, and yet have allowed for thelargest possible square carpet to be laid down without any one of thefive crescent tiles being covered, or any portion of them.

313.--QUEENS AND BISHOP PUZZLE.

It will be seen that every square of the board is either occupied orattacked. The puzzle is to substitute a bishop for the rook on the samesquare, and then place the four queens on other squares so that everysquare shall again be either occupied or attacked.

[Illustration]

314.--THE SOUTHERN CROSS.

[Illustration]

In the above illustration we have five Planets and eighty-one FixedStars, five of the latter being hidden by the Planets. It will be foundthat every Star, with the exception of the ten that have a black spot intheir centres, is in a straight line, vertically, horizontally, ordiagonally, with at least one of the Planets. The puzzle is so torearrange the Planets that all the Stars shall be in line with one ormore of them.

In rearranging the Planets, each of the five may be moved once in astraight line, in either of the three directions mentioned. They will,of course, obscure five other Stars in place of those at presentcovered.

315.--THE HAT-PEG PUZZLE.

Here is a five-queen puzzle that I gave in a fanciful dress in 1897. Asthe queens were there represented as hats on sixty-four pegs, I willkeep to the title, "The Hat-Peg Puzzle." It will be seen that everysquare is occupied or attacked. The puzzle is to remove one queen to adifferent square so that still every square is occupied or attacked,then move a second queen under a similar condition, then a third queen,and finally a fourth queen. After the fourth move every square must beattacked or occupied, but no queen must then attack another. Of course,the moves need not be "queen moves;" you can move a queen to any part ofthe board.

[Illustration]

316.--THE AMAZONS.

[Illustration]

This puzzle is based on one by Captain Turton. Remove three of thequeens to other squares so that there shall be eleven squares on theboard that are not attacked. The removal of the three queens need not beby "queen moves." You may take them up and place them anywhere. There isonly one solution.

317.--A PUZZLE WITH PAWNS.

Place two pawns in the middle of the chessboard, one at Q 4 and theother at K 5. Now, place the remaining fourteen pawns (sixteen in all)so that no three shall be in a straight line in any possible direction.

Note that I purposely do not say queens, because by the words "anypossible direction" I go beyond attacks on diagonals. The pawns must beregarded as mere points in space--at the centres of the squares. Seedotted lines in the case of No. 300, "The Eight Queens."

318.--LION-HUNTING.

[Illustration]

My friend Captain Potham Hall, the renowned hunter of big game, saysthere is nothing more exhilarating than a brush with a herd--a pack--ateam--a flock--a swarm (it has taken me a full quarter of an hour torecall the right word, but I have it at last)--a _pride_ of lions. Why anumber of lions are called a "pride," a number of whales a "school," anda number of foxes a "skulk" are mysteries of philology into which I willnot enter.

Well, the captain says that if a spirited lion crosses your path in thedesert it becomes lively, for the lion has generally been looking forthe man just as much as the man has sought the king of the forest. Andyet when they meet they always quarrel and fight it out. A littlecontemplation of this unfortunate and long-standing feud between twoestimable families has led me to figure out a few calculations as to theprobability of the man and the lion crossing one another's path in thejungle. In all these cases one has to start on certain more or lessarbitrary assumptions. That is why in the above illustration I havethought it necessary to represent the paths in the desert with suchrigid regularity. Though the captain assures me that the tracks of thelions usually run much in this way, I have doubts.

The puzzle is simply to find out in how many different ways the man andthe lion may be placed on two different spots that are not on the samepath. By "paths" it must be understood that I only refer to the ruledlines. Thus, with the exception of the four corner spots, each combatantis always on two paths and no more. It will be seen that there is a lotof scope for evading one another in the desert, which is just what onehas always understood.

319.--THE KNIGHT-GUARDS.

[Illustration]

The knight is the irresponsible low comedian of the chessboard. "He is avery uncertain, sneaking, and demoralizing rascal," says an Americanwriter. "He can only move two squares, but makes up in the quality ofhis locomotion for its quantity, for he can spring one square sidewaysand one forward simultaneously, like a cat; can stand on one leg in themiddle of the board and jump to any one of eight squares he chooses; canget on one side of a fence and blackguard three or four men on theother; has an objectionable way of inserting himself in safe placeswhere he can scare the king and compel him to move, and then gobble aqueen. For pure cussedness the knight has no equal, and when you chasehim out of one hole he skips into another." Attempts have been made overand over again to obtain a short, simple, and exact definition of themove of the knight--without success. It really consists in moving onesquare like a rook, and then another square like a bishop--the twooperations being done in one leap, so that it does not matter whetherthe first square passed over is occupied by another piece or not. It is,in fact, the only leaping move in chess. But difficult as it is todefine, a child can learn it by inspection in a few minutes.

I have shown in the diagram how twelve knights (the fewest possible thatwill perform the feat) may be placed on the chessboard so that everysquare is either occupied or attacked by a knight. Examine every squarein turn, and you will find that this is so. Now, the puzzle in this caseis to discover what is the smallest possible number of knights that isrequired in order that every square shall be either occupied orattacked, and every knight protected by another knight. And how wouldyou arrange them? It will be found that of the twelve shown in thediagram only four are thus protected by being a knight's move fromanother knight.

THE GUARDED CHESSBOARD.

On an ordinary chessboard, 8 by 8, every square can be guarded--that is,either occupied or attacked--by 5 queens, the fewest possible. There areexactly 91 fundamentally different arrangements in which no queenattacks another queen. If every queen must attack (or be protected by)another queen, there are at fewest 41 arrangements, and I have recordedsome 150 ways in which some of the queens are attacked and some not, butthis last case is very difficult to enumerate exactly.

On an ordinary chessboard every square can be guarded by 8 rooks (thefewest possible) in 40,320 ways, if no rook may attack another rook, butit is not known how many of these are fundamentally different. (Seesolution to No. 295, "The Eight Rooks.") I have not enumerated the waysin which every rook shall be protected by another rook.

On an ordinary chessboard every square can be guarded by 8 bishops (thefewest possible), if no bishop may attack another bishop. Ten bishopsare necessary if every bishop is to be protected. (See Nos. 297 and 298,"Bishops unguarded" and "Bishops guarded.")

On an ordinary chessboard every square can be guarded by 12 knights ifall but 4 are unprotected. But if every knight must be protected, 14 arenecessary. (See No. 319, "The Knight-Guards.")

Dealing with the queen on n squared boards generally, where n is lessthan 8, the following results will be of interest:--

1 queen guards 2 squared board in 1 fundamental way.

1 queen guards 3 squared board in 1 fundamental way.

2 queens guard 4 squared board in 3 fundamental ways (protected).

3 queens guard 4 squared board in 2 fundamental ways (not protected).

3 queens guard 5 squared board in 37 fundamental ways (protected).

3 queens guard 5 squared board in 2 fundamental ways (not protected).

3 queens guard 6 squared board in 1 fundamental way (protected).

4 queens guard 6 squared board in 17 fundamental ways (not protected).

4 queens guard 7 squared board in 5 fundamental ways (protected).

4 queens guard 7 squared board in 1 fundamental way (not protected).

NON-ATTACKING CHESSBOARD ARRANGEMENTS.

We know that n queens may always be placed on a square board of n squaredsquares (if n be greater than 3) without any queen attacking anotherqueen. But no general formula for enumerating the number of differentways in which it may be done has yet been discovered; probably it isundiscoverable. The known results are as follows:--

Where n = 4 there is 1 fundamental solution and 2 in all.

Where n = 5 there are 2 fundamental solutions and 10 in all.

Where n = 6 there is 1 fundamental solution and 4 in all.

Where n = 7 there are 6 fundamental solutions and 40 in all.

Where n = 8 there are 12 fundamental solutions and 92 in all.

Where n = 9 there are 46 fundamental solutions.

Where n = 10 there are 92 fundamental solutions.

Where n = 11 there are 341 fundamental solutions.

Obviously n rooks may be placed without attack on an n squared board in n!ways, but how many of these are fundamentally different I have onlyworked out in the four cases where n equals 2, 3, 4, and 5. The answershere are respectively 1, 2, 7, and 23. (See No. 296, "The Four Lions.")

We can place 2n-2 bishops on an n squared board in 2^{n} ways. (See No. 299,"Bishops in Convocation.") For boards containing 2, 3, 4, 5, 6, 7, 8squares, on a side there are respectively 1, 2, 3, 6, 10, 20, 36fundamentally different arrangements. Where n is odd there are2^{1/2(n-1)} such arrangements, each giving 4 by reversals andreflections, and 2^{n-3} - 2^{1/2(n-3)} giving 8. Where n is even thereare 2^{1/2(n-2)}, each giving 4 by reversals and reflections, and 2^{n-3}- 2^{1/2(n-4)}, each giving 8.

We can place 1/2(n squared+1) knights on an n squared board without attack, when nis odd, in 1 fundamental way; and 1/2n squared knights on an n squared board, whenn is even, in 1 fundamental way. In the first case we place all theknights on the same colour as the central square; in the second case weplace them all on black, or all on white, squares.

THE TWO PIECES PROBLEM.

On a board of n squared squares, two queens, two rooks, two bishops, or twoknights can always be placed, irrespective of attack or not, in 1/2(n^{4}- n squared) ways. The following formulae will show in how many of these waysthe two pieces may be placed with attack and without:--

With Attack. Without Attack.

2 Queens 5n cubed - 6n squared + n 3n^{4} - 10n cubed + 9n squared - 2n ------------------- ------------------------------ 3 6

2 Rooks n cubed - n squared n^{4} - 2n cubed + n squared ---------------------- 2

2 Bishops 4n cubed - 6n squared + 2n 3n^{4} - 4n cubed + 3n squared - 2n -------------------- ----------------------------- 6 6

2 Knights 4n squared - 12n + 8 n^{4} - 9n squared + 24n -------------------- 2

(See No. 318, " Lion Hunting.")

DYNAMICAL CHESS PUZZLES.

"Push on--keep moving." THOS. MORTON: _Cure for the Heartache_.

320.--THE ROOK'S TOUR.

[Illustration:

+---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | R | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+

]

The puzzle is to move the single rook over the whole board, so that itshall visit every square of the board once, and only once, and end itstour on the square from which it starts. You have to do this in as fewmoves as possible, and unless you are very careful you will take justone move too many. Of course, a square is regarded equally as "visited"whether you merely pass over it or make it a stopping-place, and we willnot quibble over the point whether the original square is actuallyvisited twice. We will assume that it is not.

321.--THE ROOK'S JOURNEY.

This puzzle I call "The Rook's Journey," because the word "tour"(derived from a turner's wheel) implies that we return to the point fromwhich we set out, and we do not do this in the present case. We shouldnot be satisfied with a personally conducted holiday tour that ended byleaving us, say, in the middle of the Sahara. The rook here makestwenty-one moves, in the course of which journey it visits every squareof the board once and only once, stopping at the square marked 10 at theend of its tenth move, and ending at the square marked 21. Twoconsecutive moves cannot be made in the same direction--that is to say,you must make a turn after every move.

[Illustration:

+---+---+---+---+---+---+---+---+ | | | | | | | | R | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | 21| | 10| | | | | +---+---+---+---+---+---+---+---+

]

322.--THE LANGUISHING MAIDEN.

[Illustration:

--+-----+-----+-----+-----+-----+-----+-----+ | | | | | | | | | Kt | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | M | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

A wicked baron in the good old days imprisoned an innocent maiden in oneof the deepest dungeons beneath the castle moat. It will be seen fromour illustration that there were sixty-three cells in the dungeon, allconnected by open doors, and the maiden was chained in the cell in whichshe is shown. Now, a valiant knight, who loved the damsel, succeeded inrescuing her from the enemy. Having gained an entrance to the dungeon atthe point where he is seen, he succeeded in reaching the maiden afterentering every cell once and only once. Take your pencil and try totrace out such a route. When you have succeeded, then try to discover aroute in twenty-two straight paths through the cells. It can be done inthis number without entering any cell a second time.

323.--A DUNGEON PUZZLE.

[Illustration:

+-----+-----+-----+-----+-----+-----+-----+-----+ | | | | | | | | | | ............. ....... ............. | | . | | . | . | . | . | | . | +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ | . | | . | . | . | . | | . | | ....... ....... ....... ....... | | | . | | | | | . | | +-- --+--.--+-- --+-- --+-- --+-- --+--.--+-- --+ | | . | | | | | . | | | ....... ....... ....... ....... | | . | | . | . | . | . | | . | +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ | . | | . | . | . | . | | . | | ............. ....... . ....... | | | | | | | . | . | | +-- --+-- --+-- --+-- --+-- --+--.--+--.--+-- --+ | | | | | | . | . | | | ............. ....... . ....... | | . | | . | . | . | . | | . | +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ | . | | . | . | . | . | | . | | ....... ....... ....... ....... | | | . | | | | | . | | +-- --+--.--+-- --+-- --+-- --+-- --+--.--+-- --+ | | . | | | | | . | | | ....... ....... ....... ....... | | . | | . | . | . | . | | . | +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ | . | | . | . | . | . | | . | | ............. . P ............. | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

A French prisoner, for his sins (or other people's), was confined in anunderground dungeon containing sixty-four cells, all communicating withopen doorways, as shown in our illustration. In order to reduce thetedium of his restricted life, he set himself various puzzles, and thisis one of them. Starting from the cell in which he is shown, how couldhe visit every cell once, and only once, and make as many turnings aspossible? His first attempt is shown by the dotted track. It will befound that there are as many as fifty-five straight lines in his path,but after many attempts he improved upon this. Can you get more thanfifty-five? You may end your path in any cell you like. Try the puzzlewith a pencil on chessboard diagrams, or you may regard them as rooks'moves on a board.

324.--THE LION AND THE MAN.

In a public place in Rome there once stood a prison divided intosixty-four cells, all open to the sky and all communicating with oneanother, as shown in the illustration. The sports that here took placewere watched from a high tower. The favourite game was to place aChristian in one corner cell and a lion in the diagonally oppositecorner and then leave them with all the inner doors open. The consequenteffect was sometimes most laughable. On one occasion the man was given asword. He was no coward, and was as anxious to find the lion as the lionundoubtedly was to find him.

[Illustration:

+-----+-----+-----+-----+-----+-----+-----+-----+ | | | | | | | | | | L | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | C | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

The man visited every cell once and only once in the fewest possiblestraight lines until he reached the lion's cell. The lion, curiouslyenough, also visited every cell once and only once in the fewestpossible straight lines until he finally reached the man's cell. Theystarted together and went at the same speed; yet, although theyoccasionally got glimpses of one another, they never once met. Thepuzzle is to show the route that each happened to take.

325.--AN EPISCOPAL VISITATION.

The white squares on the chessboard represent the parishes of a diocese.Place the bishop on any square you like, and so contrive that (using theordinary bishop's move of chess) he shall visit every one of hisparishes in the fewest possible moves. Of course, all the parishespassed through on any move are regarded as "visited." You can visit anysquares more than once, but you are not allowed to move twice betweenthe same two adjoining squares. What are the fewest possible moves? Thebishop need not end his visitation at the parish from which he first setout.

326.--A NEW COUNTER PUZZLE.

Here is a new puzzle with moving counters, or coins, that at firstglance looks as if it must be absurdly simple. But it will be foundquite a little perplexity. I give it in this place for a reason that Iwill explain when we come to the next puzzle. Copy the simple diagram,enlarged, on a sheet of paper; then place two white counters on thepoints 1 and 2, and two red counters on 9 and 10, The puzzle is to makethe red and white change places. You may move the counters one at a timein any order you like, along the lines from point to point, with theonly restriction that a red and a white counter may never stand at onceon the same straight line. Thus the first move can only be from 1 or 2to 3, or from 9 or 10 to 7.

[Illustration:

4 8 / \ / \ 2 6 10 \ / \ / 3 7 / \ / \ 1 5 9

]

327.--A NEW BISHOP'S PUZZLE.

[Illustration:

+---+---+---+---+ | b | b | b | b | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | B | B | B | B | +---+---+---+---+

]

This is quite a fascinating little puzzle. Place eight bishops (fourblack and four white) on the reduced chessboard, as shown in theillustration. The problem is to make the black bishops change placeswith the white ones, no bishop ever attacking another of the oppositecolour. They must move alternately--first a white, then a black, then awhite, and so on. When you have succeeded in doing it at all, try tofind the fewest possible moves.

If you leave out the bishops standing on black squares, and only play onthe white squares, you will discover my last puzzle turned on its side.

328.--THE QUEEN'S TOUR.

The puzzle of making a complete tour of the chessboard with the queen inthe fewest possible moves (in which squares may be visited more thanonce) was first given by the late Sam Loyd in his _Chess Strategy_. Butthe solution shown below is the one he gave in _American Chess-Nuts_ in1868. I have recorded at least six different solutions in the minimumnumber of moves--fourteen--but this one is the best of all, for reasonsI will explain.

[Illustration:

+---+---+---+---+---+---+---+---+ | | | | | | | | | | ............................. | | . | | | | | | | . | +-.-+---+---+---+---+---+---+-.-+ | . | | | | | | | . | | . | ..........................| | . | .| | | | | | . | +-.-+---.---+---+---+---+---+..-+ | . | |. | | | | . . | | . | ................. | .| . | | . | .| .| | |. | . | . | +-.-+---.---.---+---.---+.--+-.-+ | . | |. |. | .| . | . | | . | . | . | . | . | .| . | . | | . | ..| .| .|. | . |.. | . | +-.-+-.-.---.---.---+.--.-.-+-.-+ | . | . |. |. .|. . .| . | . | | . | . | . | . | ..| . | . | . | | . | . | .|. .| ..|. | . | . | +-.-+-.-+---.---..--.---+-.-+-.-+ | . | . | .|. .. .|. | . | . | | . | . | . | ..| . | . | . | . | | . | . |. | ..|. .| .| . | . | +-.-+-.-.---+.--.---.---.-.-+-.-+ | . | ..| . .|. |. |.. | . | | . | . | .| . | . | . | . | . | | . |.. | . |. | .| .| ..| . | +-.-.-.-+.--.---+---.---.-.-.-.-+ | ..| . . .| | |. |.. |.. | | . | ..| ............. | . | . | | | . | | | | | | | +---+---+---+---+---+---+---+---+

]

If you will look at the lettered square you will understand that thereare only ten really differently placed squares on a chessboard--thoseenclosed by a dark line--all the others are mere reversals orreflections. For example, every A is a corner square, and every J acentral square. Consequently, as the solution shown has a turning-pointat the enclosed D square, we can obtain a solution starting from andending at any square marked D--by just turning the board about. Now,this scheme will give you a tour starting from any A, B, C, D, E, F, orH, while no other route that I know can be adapted to more than fivedifferent starting-points. There is no Queen's Tour in fourteen moves(remember a tour must be re-entrant) that may start from a G, I, or J.But we can have a non-re-entrant path over the whole board in fourteenmoves, starting from any given square. Hence the following puzzle:--

[Illustration:

+---+---+---+---*---+---+---+---+ | A | B | C | G " G | C | B | A | *===*---+---+---*---+---+---+---+ | B " D | E | H " H | E | D | B | +---*===*---+---*---+---+---+---+ | C | E " F | I " I | F | E | C | +---+---*===*---*---+---+---+---+ | G | H | I " J " J | I | H | G | +---+---+---*===*---+---+---+---+ | G | H | I | J | J | I | H | G | +---+---+---+---+---+---+---+---+ | C | E | F | I | I | F | E | C | +---+---+---+---+---+---+---+---+ | B | D | E | H | H | E | D | B | +---+---+---+---+---+---+---+---+ | A | B | C | G | G | C | B | A | +---+---+---+---+---+---+---+---+

]

Start from the J in the enclosed part of the lettered diagram and visitevery square of the board in fourteen moves, ending wherever you like.

329.--THE STAR PUZZLE.

[Illustration:

+---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | ¤ | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | ¤ | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+

]

Put the point of your pencil on one of the white stars and (without everlifting your pencil from the paper) strike out all the stars in fourteencontinuous straight strokes, ending at the second white star. Yourstraight strokes may be in any direction you like, only every turningmust be made on a star. There is no objection to striking out any starmore than once.

In this case, where both your starting and ending squares are fixedinconveniently, you cannot obtain a solution by breaking a Queen's Tour,or in any other way by queen moves alone. But you are allowed to useoblique straight lines--such as from the upper white star direct to acorner star.

330.--THE YACHT RACE.

Now then, ye land-lubbers, hoist your baby-jib-topsails, break out yourspinnakers, ease off your balloon sheets, and get your head-sails set!

Our race consists in starting from the point at which the yacht is lyingin the illustration and touching every one of the sixty-four buoys infourteen straight courses, returning in the final tack to the buoy fromwhich we start. The seventh course must finish at the buoy from which aflag is flying.

This puzzle will call for a lot of skilful seamanship on account of thesharp angles at which it will occasionally be necessary to tack. Thepoint of a lead pencil and a good nautical eye are all the outfit thatwe require.

[Illustration]

This is difficult, because of the condition as to the flag-buoy, andbecause it is a re-entrant tour. But again we are allowed those obliquelines.

331.--THE SCIENTIFIC SKATER.

[Illustration]

It will be seen that this skater has marked on the ice sixty-four pointsor stars, and he proposes to start _from his present position_ near thecorner and enter every one of the points in fourteen straight lines. Howwill he do it? Of course there is no objection to his passing over anypoint more than once, but his last straight stroke must bring him backto the position from which he started.

It is merely a matter of taking your pencil and starting from the spoton which the skater's foot is at present resting, and striking out allthe stars in fourteen continuous straight lines, returning to the pointfrom which you set out.

332.--THE FORTY-NINE STARS.

[Illustration]

The puzzle in this case is simply to take your pencil and, starting fromone black star, strike out all the stars in twelve straight strokes,ending at the other black star. It will be seen that the attempt shownin the illustration requires fifteen strokes. Can you do it in twelve?Every turning must be made on a star, and the lines must be parallel tothe sides and diagonals of the square, as shown. In this case we aredealing with a chessboard of reduced dimensions, but only queen moves(without going outside the boundary as in the last case) are required.

333.--THE QUEEN'S JOURNEY.

[Illustration]

Place the queen on her own square, as shown in the illustration, andthen try to discover the greatest distance that she can travel over theboard in five queen's moves without passing over any square a secondtime. Mark the queen's path on the board, and note carefully also thatshe must never cross her own track. It seems simple enough, but thereader may find that he has tripped.

334.--ST. GEORGE AND THE DRAGON.

[Illustration]

Here is a little puzzle on a reduced chessboard of forty-nine squares.St. George wishes to kill the dragon. Killing dragons was a well-knownpastime of his, and, being a knight, it was only natural that he shoulddesire to perform the feat in a series of knight's moves. Can you showhow, starting from that central square, he may visit once, and onlyonce, every square of the board in a chain of chess knight's moves, andend by capturing the dragon on his last move? Of course a variety ofdifferent ways are open to him, so try to discover a route that formssome pretty design when you have marked each successive leap by astraight line from square to square.

335.--FARMER LAWRENCE'S CORNFIELDS.

One of the most beautiful districts within easy distance of London for asummer ramble is that part of Buckinghamshire known as the Valley of theChess--at least, it was a few years ago, before it was discovered by thespeculative builder. At the beginning of the present century therelived, not far from Latimers, a worthy but eccentric farmer namedLawrence. One of his queer notions was that every person who lived nearthe banks of the river Chess ought to be in some way acquainted with thenoble game of the same name, and in order to impress this fact on hismen and his neighbours he adopted at times strange terminology. Forexample, when one of his ewes presented him with a lamb, he would saythat it had "queened a pawn"; when he put up a new barn against thehighway, he called it "castling on the king's side"; and when he sent aman with a gun to keep his neighbour's birds off his fields, he spoke ofit as "attacking his opponent's rooks." Everybody in the neighbourhoodused to be amused at Farmer Lawrence's little jokes, and one boy (thewag of the village) who got his ears pulled by the old gentleman forstealing his "chestnuts" went so far as to call him "a silly oldchess-protector!"

One year he had a large square field divided into forty-nine squareplots, as shown in the illustration. The white squares were sown withwheat and the black squares with barley. When the harvest time cameround he gave orders that his men were first to cut the corn in thepatch marked 1, and that each successive cutting should be exactly aknight's move from the last one, the thirteenth cutting being in thepatch marked 13, the twenty-fifth in the patch marked 25, thethirty-seventh in the one marked 37, and the last, or forty-ninthcutting, in the patch marked 49. This was too much for poor Hodge, andeach day Farmer Lawrence had to go down to the field and show whichpiece had to be operated upon. But the problem will perhaps present nodifficulty to my readers.

[Illustration]

336.--THE GREYHOUND PUZZLE.

In this puzzle the twenty kennels do not communicate with one another bydoors, but are divided off by a low wall. The solitary occupant is thegreyhound which lives in the kennel in the top left-hand corner. When heis allowed his liberty he has to obtain it by visiting every kennel onceand only once in a series of knight's moves, ending at the bottomright-hand corner, which is open to the world. The lines in the abovediagram show one solution. The puzzle is to discover in how manydifferent ways the greyhound may thus make his exit from his cornerkennel.

[Illustration]

337.--THE FOUR KANGAROOS.

[Illustration]

In introducing a little Commonwealth problem, I must first explain thatthe diagram represents the sixty-four fields, all properly fenced offfrom one another, of an Australian settlement, though I need hardly saythat our kith and kin "down under" always _do_ set out their land inthis methodical and exact manner. It will be seen that in every one ofthe four corners is a kangaroo. Why kangaroos have a marked preferencefor corner plots has never been satisfactorily explained, and it wouldbe out of place to discuss the point here. I should also add thatkangaroos, as is well known, always leap in what we call "knight'smoves." In fact, chess players would probably have adopted the betterterm "kangaroo's move" had not chess been invented before kangaroos.

The puzzle is simply this. One morning each kangaroo went for hismorning hop, and in sixteen consecutive knight's leaps visited justfifteen different fields and jumped back to his corner. No field wasvisited by more than one of the kangaroos. The diagram shows how theyarranged matters. What you are asked to do is to show how they mighthave performed the feat without any kangaroo ever crossing thehorizontal line in the middle of the square that divides the board intotwo equal parts.

338.--THE BOARD IN COMPARTMENTS.

[Illustration]

We cannot divide the ordinary chessboard into four equal squarecompartments, and describe a complete tour, or even path, in eachcompartment. But we may divide it into four compartments, as in theillustration, two containing each twenty squares, and the other two eachtwelve squares, and so obtain an interesting puzzle. You are asked todescribe a complete re-entrant tour on this board, starting where youlike, but visiting every square in each successive compartment beforepassing into another one, and making the final leap back to the squarefrom which the knight set out. It is not difficult, but will be foundvery entertaining and not uninstructive.

Whether a re-entrant "tour" or a complete knight's "path" is possible ornot on a rectangular board of given dimensions depends not only on itsdimensions, but also on its shape. A tour is obviously not possible on aboard containing an odd number of cells, such as 5 by 5 or 7 by 7, forthis reason: Every successive leap of the knight must be from a whitesquare to a black and a black to a white alternately. But if there be anodd number of cells or squares there must be one more square of onecolour than of the other, therefore the path must begin from a square ofthe colour that is in excess, and end on a similar colour, and as aknight's move from one colour to a similar colour is impossible thepath cannot be re-entrant. But a perfect tour may be made on arectangular board of any dimensions provided the number of squares beeven, and that the number of squares on one side be not less than 6 andon the other not less than 5. In other words, the smallest rectangularboard on which a re-entrant tour is possible is one that is 6 by 5.

A complete knight's path (not re-entrant) over all the squares of aboard is never possible if there be only two squares on one side; nor isit possible on a square board of smaller dimensions than 5 by 5. So thaton a board 4 by 4 we can neither describe a knight's tour nor a completeknight's path; we must leave one square unvisited. Yet on a board 4 by 3(containing four squares fewer) a complete path may be described insixteen different ways. It may interest the reader to discover allthese. Every path that starts from and ends at different squares is herecounted as a different solution, and even reverse routes are calleddifferent.

339.--THE FOUR KNIGHTS' TOURS.

[Illustration]

I will repeat that if a chessboard be cut into four equal parts, asindicated by the dark lines in the illustration, it is not possible toperform a knight's tour, either re-entrant or not, on one of the parts.The best re-entrant attempt is shown, in which each knight has totrespass twice on other parts. The puzzle is to cut the boarddifferently into four parts, each of the same size and shape, so that are-entrant knight's tour may be made on each part. Cuts along the dottedlines will not do, as the four central squares of the board would beeither detached or hanging on by a mere thread.

340.--THE CUBIC KNIGHT'S TOUR.

Some few years ago I happened to read somewhere that Abnit Vandermonde,a clever mathematician, who was born in 1736 and died in 1793, haddevoted a good deal of study to the question of knight's tours. Beyondwhat may be gathered from a few fragmentary references, I am not awareof the exact nature or results of his investigations, but one thingattracted my attention, and that was the statement that he had proposedthe question of a tour of the knight over the six surfaces of a cube,each surface being a chessboard. Whether he obtained a solution or not Ido not know, but I have never seen one published. So I at once set towork to master this interesting problem. Perhaps the reader may like toattempt it.

341.--THE FOUR FROGS.

[Illustration]

In the illustration we have eight toadstools, with white frogs on 1 and3 and black frogs on 6 and 8. The puzzle is to move one frog at a time,in any order, along one of the straight lines from toadstool totoadstool, until they have exchanged places, the white frogs being lefton 6 and 8 and the black ones on 1 and 3. If you use four counters on asimple diagram, you will find this quite easy, but it is a little morepuzzling to do it in only seven plays, any number of successive moves byone frog counting as one play. Of course, more than one frog cannot beon a toadstool at the same time.

342.--THE MANDARIN'S PUZZLE.

The following puzzle has an added interest from the circumstance that acorrect solution of it secured for a certain young Chinaman the hand ofhis charming bride. The wealthiest mandarin within a radius of a hundredmiles of Peking was Hi-Chum-Chop, and his beautiful daughter, Peeky-Bo,had innumerable admirers. One of her most ardent lovers was Winky-Hi,and when he asked the old mandarin for his consent to their marriage,Hi-Chum-Chop presented him with the following puzzle and promised hisconsent if the youth brought him the correct answer within a week.Winky-Hi, following a habit which obtains among certain solvers to thisday, gave it to all his friends, and when he had compared theirsolutions he handed in the best one as his own. Luckily it was quiteright. The mandarin thereupon fulfilled his promise. The fatted pup waskilled for the wedding feast, and when Hi-Chum-Chop passed Winky-Hi theliver wing all present knew that it was a token of eternal goodwill, inaccordance with Chinese custom from time immemorial.

The mandarin had a table divided into twenty-five squares, as shown inthe diagram. On each of twenty-four of these squares was placed anumbered counter, just as I have indicated. The puzzle is to get thecounters in numerical order by moving them one at a time in what we call"knight's moves." Counter 1 should be where 16 is, 2 where 11 is, 4where 13 now is, and so on. It will be seen that all the counters onshaded squares are in their proper positions. Of course, two countersmay never be on a square at the same time. Can you perform the feat inthe fewest possible moves?

[Illustration]

In order to make the manner of moving perfectly clear I will point outthat the first knight's move can only be made by 1 or by 2 or by 10.Supposing 1 moves, then the next move must be by 23, 4, 8, or 21. Asthere is never more than one square vacant, the order in which thecounters move may be written out as follows: 1--21--14--18--22, etc. Arough diagram should be made on a larger scale for practice, andnumbered counters or pieces of cardboard used.

343.--EXERCISE FOR PRISONERS.

The following is the plan of the north wing of a certain gaol, showingthe sixteen cells all communicating by open doorways. Fifteen prisonerswere numbered and arranged in the cells as shown. They were allowed tochange their cells as much as they liked, but if two prisoners were everin the same cell together there was a severe punishment promised them.

[Illustration]

Now, in order to reduce their growing obesity, and to combine physicalexercise with mental recreation, the prisoners decided, on thesuggestion of one of their number who was interested in knight's tours,to try to form themselves into a perfect knight's path without breakingthe prison regulations, and leaving the bottom right-hand corner cellvacant, as originally. The joke of the matter is that the arrangement atwhich they arrived was as follows:--

8 3 12 1 11 14 9 6 4 7 2 13 15 10 5

The warders failed to detect the important fact that the men could notpossibly get into this position without two of them having been at sometime in the same cell together. Make the attempt with counters on aruled diagram, and you will find that this is so. Otherwise the solutionis correct enough, each member being, as required, a knight's move fromthe preceding number, and the original corner cell vacant.

The puzzle is to start with the men placed as in the illustration andshow how it might have been done in the fewest moves, while giving acomplete rest to as many prisoners as possible.

As there is never more than one vacant cell for a man to enter, it isonly necessary to write down the numbers of the men in the order inwhich they move. It is clear that very few men can be left throughout intheir cells undisturbed, but I will leave the solver to discover justhow many, as this is a very essential part of the puzzle.

344.--THE KENNEL PUZZLE.

[Illustration]

A man has twenty-five dog kennels all communicating with each other bydoorways, as shown in the illustration. He wishes to arrange his twentydogs so that they shall form a knight's string from dog No. 1 to dog No.20, the bottom row of five kennels to be left empty, as at present. Thisis to be done by moving one dog at a time into a vacant kennel. The dogsare well trained to obedience, and may be trusted to remain in thekennels in which they are placed, except that if two are placed in thesame kennel together they will fight it out to the death. How is thepuzzle to be solved in the fewest possible moves without two dogs everbeing together?

345.--THE TWO PAWNS.

[Illustration]

Here is a neat little puzzle in counting. In how many different ways maythe two pawns advance to the eighth square? You may move them in anyorder you like to form a different sequence. For example, you may movethe Q R P (one or two squares) first, or the K R P first, or one pawn asfar as you like before touching the other. Any sequence is permissible,only in this puzzle as soon as a pawn reaches the eighth square it isdead, and remains there unconverted. Can you count the number ofdifferent sequences? At first it will strike you as being verydifficult, but I will show that it is really quite simple when properlyattacked.

VARIOUS CHESS PUZZLES.

"Chesse-play is a good and wittie exercise of the minde for some kinde of men." Burton's _Anatomy of Melancholy_.

346.--SETTING THE BOARD.

I have a single chessboard and a single set of chessmen. In how manydifferent ways may the men be correctly set up for the beginning of agame? I find that most people slip at a particular point in making thecalculation.

347.--COUNTING THE RECTANGLES.

Can you say correctly just how many squares and other rectangles thechessboard contains? In other words, in how great a number of differentways is it possible to indicate a square or other rectangle enclosed bylines that separate the squares of the board?

348.--THE ROOKERY.

[Illustration]

The White rooks cannot move outside the little square in which they areenclosed except on the final move, in giving checkmate. The puzzle ishow to checkmate Black in the fewest possible moves with No. 8 rook, theother rooks being left in numerical order round the sides of theirsquare with the break between 1 and 7.

349.--STALEMATE.

Some years ago the puzzle was proposed to construct an imaginary game ofchess, in which White shall be stalemated in the fewest possible moveswith all the thirty-two pieces on the board. Can you build up such aposition in fewer than twenty moves?

350.--THE FORSAKEN KING.

[Illustration]

Set up the position shown in the diagram. Then the condition of thepuzzle is--White to play and checkmate in six moves. Notwithstanding thecomplexities, I will show how the manner of play may be condensed intoquite a few lines, merely stating here that the first two moves of Whitecannot be varied.

351.--THE CRUSADER.

The following is a prize puzzle propounded by me some years ago. Producea game of chess which, after sixteen moves, shall leave White with allhis sixteen men on their original squares and Black in possession of hisking alone (not necessarily on his own square). White is then to _force_mate in three moves.

352.--IMMOVABLE PAWNS.

Starting from the ordinary arrangement of the pieces as for a game, whatis the smallest possible number of moves necessary in order to arrive atthe following position? The moves for both sides must, of course, beplayed strictly in accordance with the rules of the game, though theresult will necessarily be a very weird kind of chess.

[Illustration]

353.--THIRTY-SIX MATES.

[Illustration]

Place the remaining eight White pieces in such a position that Whiteshall have the choice of thirty-six different mates on the move. Everymove that checkmates and leaves a different position is a differentmate. The pieces already placed must not be moved.

354.--AN AMAZING DILEMMA.

In a game of chess between Mr. Black and Mr. White, Black was indifficulties, and as usual was obliged to catch a train. So he proposedthat White should complete the game in his absence on condition that nomoves whatever should be made for Black, but only with the White pieces.Mr. White accepted, but to his dismay found it utterly impossible to winthe game under such conditions. Try as he would, he could not checkmatehis opponent. On which square did Mr. Black leave his king? The otherpieces are in their proper positions in the diagram. White may leaveBlack in check as often as he likes, for it makes no difference, as hecan never arrive at a checkmate position.

[Illustration]

355.--CHECKMATE!

[Illustration]

Strolling into one of the rooms of a London club, I noticed a positionleft by two players who had gone. This position is shown in the diagram.It is evident that White has checkmated Black. But how did he do it?That is the puzzle.

356.--QUEER CHESS.

Can you place two White rooks and a White knight on the board so thatthe Black king (who must be on one of the four squares in the middle ofthe board) shall be in check with no possible move open to him? "Inother words," the reader will say, "the king is to be shown checkmated."Well, you can use the term if you wish, though I intentionally do notemploy it myself. The mere fact that there is no White king on the boardwould be a sufficient reason for my not doing so.

357.--ANCIENT CHINESE PUZZLE.

[Illustration]

My next puzzle is supposed to be Chinese, many hundreds of years old,and never fails to interest. White to play and mate, moving each of thethree pieces once, and once only.

358.--THE SIX PAWNS.

In how many different ways may I place six pawns on the chessboard sothat there shall be an even number of unoccupied squares in every rowand every column? We are not here considering the diagonals at all, andevery different six squares occupied makes a different solution, so wehave not to exclude reversals or reflections.

359.--COUNTER SOLITAIRE.

Here is a little game of solitaire that is quite easy, but not so easyas to be uninteresting. You can either rule out the squares on a sheetof cardboard or paper, or you can use a portion of your chessboard. Ihave shown numbered counters in the illustration so as to make thesolution easy and intelligible to all, but chess pawns or draughts willserve just as well in practice.

[Illustration]

The puzzle is to remove all the counters except one, and this one thatis left must be No. 1. You remove a counter by jumping over anothercounter to the next space beyond, if that square is vacant, but youcannot make a leap in a diagonal direction. The following moves willmake the play quite clear: 1-9, 2-10, 1-2, and so on. Here 1 jumps over9, and you remove 9 from the board; then 2 jumps over 10, and you remove10; then 1 jumps over 2, and you remove 2. Every move is thus a capture,until the last capture of all is made by No. 1.

360.--CHESSBOARD SOLITAIRE.

[Illustration]

Here is an extension of the last game of solitaire. All you need is achessboard and the thirty-two pieces, or the same number of draughts orcounters. In the illustration numbered counters are used. The puzzle isto remove all the counters except two, and these two must haveoriginally been on the same side of the board; that is, the two leftmust either belong to the group 1 to 16 or to the other group, 17 to 32.You remove a counter by jumping over it with another counter to the nextsquare beyond, if that square is vacant, but you cannot make a leap in adiagonal direction. The following moves will make the play quite clear:3-11, 4-12, 3-4, 13-3. Here 3 jumps over 11, and you remove 11; 4 jumpsover 12, and you remove 12; and so on. It will be found a fascinatinglittle game of patience, and the solution requires the exercise of someingenuity.

361.--THE MONSTROSITY.

One Christmas Eve I was travelling by rail to a little place in one ofthe southern counties. The compartment was very full, and the passengerswere wedged in very tightly. My neighbour in one of the corner seats wasclosely studying a position set up on one of those little foldingchessboards that can be carried conveniently in the pocket, and I couldscarcely avoid looking at it myself. Here is the position:--

[Illustration]

My fellow-passenger suddenly turned his head and caught the look ofbewilderment on my face.

"Do you play chess?" he asked.

"Yes, a little. What is that? A problem?"

"Problem? No; a game."

"Impossible!" I exclaimed rather rudely. "The position is a perfectmonstrosity!"

He took from his pocket a postcard and handed it to me. It bore anaddress at one side and on the other the words "43. K to Kt 8."

"It is a correspondence game." he exclaimed. "That is my friend's lastmove, and I am considering my reply."

"But you really must excuse me; the position seems utterly impossible.How on earth, for example--"

"Ah!" he broke in smilingly. "I see; you are a beginner; you play towin."

"Of course you wouldn't play to lose or draw!"

He laughed aloud.

"You have much to learn. My friend and myself do not play for results ofthat antiquated kind. We seek in chess the wonderful, the whimsical, theweird. Did you ever see a position like that?"

I inwardly congratulated myself that I never had.

"That position, sir, materializes the sinuous evolvements and syncretic,synthetic, and synchronous concatenations of two cerebralindividualities. It is the product of an amphoteric and intercalatoryinterchange of--"

"Have you seen the evening paper, sir?" interrupted the man opposite,holding out a newspaper. I noticed on the margin beside his thumb somepencilled writing. Thanking him, I took the paper and read--"Insane, butquite harmless. He is in my charge."

After that I let the poor fellow run on in his wild way until both gotout at the next station.

But that queer position became fixed indelibly in my mind, with Black'slast move 43. K to Kt 8; and a short time afterwards I found it actuallypossible to arrive at such a position in forty-three moves. Can thereader construct such a sequence? How did White get his rooks and king'sbishop into their present positions, considering Black can never havemoved his king's bishop? No odds were given, and every move wasperfectly legitimate.

MEASURING, WEIGHING, AND PACKING PUZZLES.

"Measure still for measure." _Measure for Measure_, v. 1.

Apparently the first printed puzzle involving the measuring of a givenquantity of liquid by pouring from one vessel to others of knowncapacity was that propounded by Niccola Fontana, better known as"Tartaglia" (the stammerer), 1500-1559. It consists in dividing 24 oz.of valuable balsam into three equal parts, the only measures availablebeing vessels holding 5, 11, and 13 ounces respectively. There are manydifferent solutions to this puzzle in six manipulations, or pouringsfrom one vessel to another. Bachet de Meziriac reprinted this and otherof Tartaglia's puzzles in his _Problemes plaisans et delectables_(1612). It is the general opinion that puzzles of this class can only besolved by trial, but I think formulae can be constructed for the solutiongenerally of certain related cases. It is a practically unexplored fieldfor investigation.

The classic weighing problem is, of course, that proposed by Bachet. Itentails the determination of the least number of weights that wouldserve to weigh any integral number of pounds from 1 lb. to 40 lbs.inclusive, when we are allowed to put a weight in either of the twopans. The answer is 1, 3, 9, and 27 lbs. Tartaglia had previouslypropounded the same puzzle with the condition that the weights may onlybe placed in one pan. The answer in that case is 1, 2, 4, 8, 16, 32 lbs.Major MacMahon has solved the problem quite generally. A full accountwill be found in Ball's _Mathematical Recreations_ (5th edition).

Packing puzzles, in which we are required to pack a maximum number ofarticles of given dimensions into a box of known dimensions, are, Ibelieve, of quite recent introduction. At least I cannot recall anyexample in the books of the old writers. One would rather expect to findin the toy shops the idea presented as a mechanical puzzle, but I do notthink I have ever seen such a thing. The nearest approach to it wouldappear to be the puzzles of the jig-saw character, where there is onlyone depth of the pieces to be adjusted.

362.--THE WASSAIL BOWL.

One Christmas Eve three Weary Willies came into possession of what wasto them a veritable wassail bowl, in the form of a small barrel,containing exactly six quarts of fine ale. One of the men possessed afive-pint jug and another a three-pint jug, and the problem for them wasto divide the liquor equally amongst them without waste. Of course, theyare not to use any other vessels or measures. If you can show how it wasto be done at all, then try to find the way that requires the fewestpossible manipulations, every separate pouring from one vessel toanother, or down a man's throat, counting as a manipulation.

363.--THE DOCTOR'S QUERY.

"A curious little point occurred to me in my dispensary this morning,"said a doctor. "I had a bottle containing ten ounces of spirits of wine,and another bottle containing ten ounces of water. I poured a quarter ofan ounce of spirits into the water and shook them up together. Themixture was then clearly forty to one. Then I poured back aquarter-ounce of the mixture, so that the two bottles should again eachcontain the same quantity of fluid. What proportion of spirits to waterdid the spirits of wine bottle then contain?"

364.--THE BARREL PUZZLE.

The men in the illustration are disputing over the liquid contents of abarrel. What the particular liquid is it is impossible to say, for weare unable to look into the barrel; so we will call it water. One mansays that the barrel is more than half full, while the other insiststhat it is not half full. What is their easiest way of settling thepoint? It is not necessary to use stick, string, or implement of anykind for measuring. I give this merely as one of the simplest possibleexamples of the value of ordinary sagacity in the solving of puzzles.What are apparently very difficult problems may frequently be solved ina similarly easy manner if we only use a little common sense.

[Illustration]

365.--NEW MEASURING PUZZLE.

Here is a new poser in measuring liquids that will be found interesting.A man has two ten-quart vessels full of wine, and a five-quart and afour-quart measure. He wants to put exactly three quarts into each ofthe two measures. How is he to do it? And how many manipulations(pourings from one vessel to another) do you require? Of course, wasteof wine, tilting, and other tricks are not allowed.

366.--THE HONEST DAIRYMAN.

An honest dairyman in preparing his milk for public consumption employeda can marked B, containing milk, and a can marked A, containing water.From can A he poured enough to double the contents of can B. Then hepoured from can B into can A enough to double its contents. Then hefinally poured from can A into can B until their contents were exactlyequal. After these operations he would send the can A to London, and thepuzzle is to discover what are the relative proportions of milk andwater that he provides for the Londoners' breakfast-tables. Do they getequal proportions of milk and water--or two parts of milk and one ofwater--or what? It is an interesting question, though, curiously enough,we are not told how much milk or water he puts into the cans at thestart of his operations.

367.--WINE AND WATER.

Mr. Goodfellow has adopted a capital idea of late. When he gives alittle dinner party and the time arrives to smoke, after the departureof the ladies, he sometimes finds that the conversation is apt to becometoo political, too personal, too slow, or too scandalous. Then he alwaysmanages to introduce to the company some new poser that he has secretedup his sleeve for the occasion. This invariably results in no end ofinteresting discussion and debate, and puts everybody in a good humour.

Here is a little puzzle that he propounded the other night, and it isextraordinary how the company differed in their answers. He filled awine-glass half full of wine, and another glass twice the size one-thirdfull of wine. Then he filled up each glass with water and emptied thecontents of both into a tumbler. "Now," he said, "what part of themixture is wine and what part water?" Can you give the correct answer?

368.--THE KEG OF WINE.

Here is a curious little problem. A man had a ten-gallon keg full ofwine and a jug. One day he drew off a jugful of wine and filled up thekeg with water. Later on, when the wine and water had got thoroughlymixed, he drew off another jugful and again filled up the keg withwater. It was then found that the keg contained equal proportions ofwine and water. Can you find from these facts the capacity of the jug?

369.--MIXING THE TEA.

"Mrs. Spooner called this morning," said the honest grocer to hisassistant. "She wants twenty pounds of tea at 2s. 41/2d. per lb. Ofcourse we have a good 2s. 6d. tea, a slightly inferior at 2s. 3d., and acheap Indian at 1s. 9d., but she is very particular always about herprices."

"What do you propose to do?" asked the innocent assistant.

"Do?" exclaimed the grocer. "Why, just mix up the three teas indifferent proportions so that the twenty pounds will work out fairly atthe lady's price. Only don't put in more of the best tea than you canhelp, as we make less profit on that, and of course you will use onlyour complete pound packets. Don't do any weighing."

How was the poor fellow to mix the three teas? Could you have shown himhow to do it?

370.--A PACKING PUZZLE.

As we all know by experience, considerable ingenuity is often requiredin packing articles into a box if space is not to be unduly wasted. Aman once told me that he had a large number of iron balls, all exactlytwo inches in diameter, and he wished to pack as many of these aspossible into a rectangular box 24+9/10 inches long, 22+4/5 incheswide, and 14 inches deep. Now, what is the greatest number of theballs that he could pack into that box?

371.--GOLD PACKING IN RUSSIA.

The editor of the _Times_ newspaper was invited by a high Russianofficial to inspect the gold stored in reserve at St. Petersburg, inorder that he might satisfy himself that it was not another "Humbertsafe." He replied that it would be of no use whatever, for although thegold might appear to be there, he would be quite unable from a mereinspection to declare that what he saw was really gold. A correspondentof the _Daily Mail_ thereupon took up the challenge, but, although hewas greatly impressed by what he saw, he was compelled to confess hisincompetence (without emptying and counting the contents of every boxand sack, and assaying every piece of gold) to give any assurance on thesubject. In presenting the following little puzzle, I wish it to be alsounderstood that I do not guarantee the real existence of the gold, andthe point is not at all material to our purpose. Moreover, if the readersays that gold is not usually "put up" in slabs of the dimensions that Igive, I can only claim problematic licence.

Russian officials were engaged in packing 800 gold slabs, each measuring121/2 inches long, 11 inches wide, and 1 inch deep. What are theinterior dimensions of a box of equal length and width, and necessarydepth, that will exactly contain them without any space being left over?Not more than twelve slabs may be laid on edge, according to the rulesof the government. It is an interesting little problem in packing, andnot at all difficult.

372.--THE BARRELS OF HONEY.

[Illustration]

Once upon a time there was an aged merchant of Bagdad who was muchrespected by all who knew him. He had three sons, and it was a rule ofhis life to treat them all exactly alike. Whenever one received apresent, the other two were each given one of equal value. One day thisworthy man fell sick and died, bequeathing all his possessions to histhree sons in equal shares.

The only difficulty that arose was over the stock of honey. There wereexactly twenty-one barrels. The old man had left instructions that notonly should every son receive an equal quantity of honey, but shouldreceive exactly the same number of barrels, and that no honey should betransferred from barrel to barrel on account of the waste involved. Now,as seven of these barrels were full of honey, seven were half-full, andseven were empty, this was found to be quite a puzzle, especially aseach brother objected to taking more than four barrels of, the samedescription--full, half-full, or empty. Can you show how they succeededin making a correct division of the property?

CROSSING RIVER PROBLEMS

"My boat is on the shore." BYRON.

This is another mediaeval class of puzzles. Probably the earliest examplewas by Abbot Alcuin, who was born in Yorkshire in 735 and died at Toursin 804. And everybody knows the story of the man with the wolf, goat,and basket of cabbages whose boat would only take one of the three at atime with the man himself. His difficulties arose from his being unableto leave the wolf alone with the goat, or the goat alone with thecabbages. These puzzles were considered by Tartaglia and Bachet, andhave been later investigated by Lucas, De Fonteney, Delannoy, Tarry, andothers. In the puzzles I give there will be found one or two newconditions which add to the complexity somewhat. I also include a pulleyproblem that practically involves the same principles.

[Illustration]

373.--CROSSING THE STREAM.

During a country ramble Mr. and Mrs. Softleigh found themselves in apretty little dilemma. They had to cross a stream in a small boat whichwas capable of carrying only 150 lbs. weight. But Mr. Softleigh and hiswife each weighed exactly 150 lbs., and each of their sons weighed 75lbs. And then there was the dog, who could not be induced on any termsto swim. On the principle of "ladies first," they at once sent Mrs.Softleigh over; but this was a stupid oversight, because she had to comeback again with the boat, so nothing was gained by that operation. Howdid they all succeed in getting across? The reader will find it mucheasier than the Softleigh family did, for their greatest enemy could nothave truthfully called them a brilliant quartette--while the dog was aperfect fool.

374--CROSSING THE RIVER AXE.

Many years ago, in the days of the smuggler known as "Rob Roy of theWest," a piratical band buried on the coast of South Devon a quantity oftreasure which was, of course, abandoned by them in the usualinexplicable way. Some time afterwards its whereabouts was discovered bythree countrymen, who visited the spot one night and divided the spoilbetween them, Giles taking treasure to the value of L800, Jasper L500worth, and Timothy L300 worth. In returning they had to cross the riverAxe at a point where they had left a small boat in readiness. Here,however, was a difficulty they had not anticipated. The boat would onlycarry two men, or one man and a sack, and they had so little confidencein one another that no person could be left alone on the land or in theboat with more than his share of the spoil, though two persons (being acheck on each other) might be left with more than their shares. Thepuzzle is to show how they got over the river in the fewest possiblecrossings, taking their treasure with them. No tricks, such as ropes,"flying bridges," currents, swimming, or similar dodges, may beemployed.

375.--FIVE JEALOUS HUSBANDS.

During certain local floods five married couples found themselvessurrounded by water, and had to escape from their unpleasant position ina boat that would only hold three persons at a time. Every husband wasso jealous that he would not allow his wife to be in the boat or oneither bank with another man (or with other men) unless he was himselfpresent. Show the quickest way of getting these five men and their wivesacross into safety.

Call the men A, B, C, D, E, and their respective wives a, b, c, d, e. Togo over and return counts as two crossings. No tricks such as ropes,swimming, currents, etc., are permitted.

376.--THE FOUR ELOPEMENTS.

Colonel B---- was a widower of a very taciturn disposition. Histreatment of his four daughters was unusually severe, almost cruel, andthey not unnaturally felt disposed to resent it. Being charming girlswith every virtue and many accomplishments, it is not surprising thateach had a fond admirer. But the father forbade the young men to call athis house, intercepted all letters, and placed his daughters understricter supervision than ever. But love, which scorns locks and keysand garden walls, was equal to the occasion, and the four youthsconspired together and planned a general elopement.

At the foot of the tennis lawn at the bottom of the garden ran thesilver Thames, and one night, after the four girls had been safelyconducted from a dormitory window to _terra firma_, they all creptsoftly down to the bank of the river, where a small boat belonging tothe Colonel was moored. With this they proposed to cross to the oppositeside and make their way to a lane where conveyances were waiting tocarry them in their flight. Alas! here at the water's brink theirdifficulties already began.

The young men were so extremely jealous that not one of them would allowhis prospective bride to remain at any time in the company of anotherman, or men, unless he himself were present also. Now, the boat wouldonly hold two persons, though it could, of course, be rowed by one, andit seemed impossible that the four couples would ever get across. Butmidway in the stream was a small island, and this seemed to present away out of the difficulty, because a person or persons could be leftthere while the boat was rowed back or to the opposite shore. If theyhad been prepared for their difficulty they could have easily worked outa solution to the little poser at any other time. But they were now sohurried and excited in their flight that the confusion they soon gotinto was exceedingly amusing--or would have been to any one exceptthemselves.

As a consequence they took twice as long and crossed the river twice asoften as was really necessary. Meanwhile, the Colonel, who was a verylight sleeper, thought he heard a splash of oars. He quickly raised thealarm among his household, and the young ladies were found to bemissing. Somebody was sent to the police-station, and a number ofofficers soon aided in the pursuit of the fugitives, who, in consequenceof that delay in crossing the river, were quickly overtaken. The fourgirls returned sadly to their homes, and afterwards broke off theirengagements in disgust.

For a considerable time it was a mystery how the party of eight managedto cross the river in that little boat without any girl being ever leftwith a man, unless her betrothed was also present. The favourite methodis to take eight counters or pieces of cardboard and mark them A, B, C,D, a, b, c, d, to represent the four men and their prospective brides,and carry them from one side of a table to the other in a matchbox (torepresent the boat), a penny being placed in the middle of the table asthe island.

Readers are now asked to find the quickest method of getting the partyacross the river. How many passages are necessary from land to land? By"land" is understood either shore or island. Though the boat would notnecessarily call at the island every time of crossing, the possibilityof its doing so must be provided for. For example, it would not do for aman to be alone in the boat (though it were understood that he intendedmerely to cross from one bank to the opposite one) if there happened tobe a girl alone on the island other than the one to whom he was engaged.

377.--STEALING THE CASTLE TREASURE.

The ingenious manner in which a box of treasure, consisting principallyof jewels and precious stones, was stolen from Gloomhurst Castle hasbeen handed down as a tradition in the De Gourney family. The thievesconsisted of a man, a youth, and a small boy, whose only mode of escapewith the box of treasure was by means of a high window. Outside thewindow was fixed a pulley, over which ran a rope with a basket at eachend. When one basket was on the ground the other was at the window. Therope was so disposed that the persons in the basket could neither helpthemselves by means of it nor receive help from others. In short, theonly way the baskets could be used was by placing a heavier weight inone than in the other.

Now, the man weighed 195 lbs., the youth 105 lbs., the boy 90 lbs., andthe box of treasure 75 lbs. The weight in the descending basket couldnot exceed that in the other by more than 15 lbs. without causing adescent so rapid as to be most dangerous to a human being, though itwould not injure the stolen property. Only two persons, or one personand the treasure, could be placed in the same basket at one time. Howdid they all manage to escape and take the box of treasure with them?

The puzzle is to find the shortest way of performing the feat, which initself is not difficult. Remember, a person cannot help himself byhanging on to the rope, the only way being to go down "with a bump,"with the weight in the other basket as a counterpoise.

PROBLEMS CONCERNING GAMES.

"The little pleasure of the game." MATTHEW PRIOR.

Every game lends itself to the propounding of a variety of puzzles. Theycan be made, as we have seen, out of the chessboard and the peculiarmoves of the chess pieces. I will now give just a few examples ofpuzzles with playing cards and dominoes, and also go out of doors andconsider one or two little posers in the cricket field, at the footballmatch, and the horse race and motor-car race.

378.--DOMINOES IN PROGRESSION.

[Illustration]

It will be seen that I have played six dominoes, in the illustration, inaccordance with the ordinary rules of the game, 4 against 4, 1 against1, and so on, and yet the sum of the spots on the successive dominoes,4, 5, 6, 7, 8, 9, are in arithmetical progression; that is, the numberstaken in order have a common difference of 1. In how many different waysmay we play six dominoes, from an ordinary box of twenty-eight, so thatthe numbers on them may lie in arithmetical progression? We must alwaysplay from left to right, and numbers in decreasing arithmeticalprogression (such as 9, 8, 7, 6, 5, 4) are not admissible.

379.--THE FIVE DOMINOES.

[Illustration]

Here is a new little puzzle that is not difficult, but will probably befound entertaining by my readers. It will be seen that the five dominoesare so arranged in proper sequence (that is, with 1 against 1, 2 against2, and so on), that the total number of pips on the two end dominoes isfive, and the sum of the pips on the three dominoes in the middle isalso five. There are just three other arrangements giving five for theadditions. They are: --

(1--0) (0--0) (0--2) (2--1) (1--3) (4--0) (0--0) (0--2) (2--1) (1--0) (2--0) (0--0) (0--1) (1--3) (3--0)

Now, how many similar arrangements are there of five dominoes that shallgive six instead of five in the two additions?

380.--THE DOMINO FRAME PUZZLE.

[Illustration]

It will be seen in the illustration that the full set of twenty-eightdominoes is arranged in the form of a square frame, with 6 against 6, 2against 2, blank against blank, and so on, as in the game. It will befound that the pips in the top row and left-hand column both add up 44.The pips in the other two sides sum to 59 and 32 respectively. Thepuzzle is to rearrange the dominoes in the same form so that all of thefour sides shall sum to 44. Remember that the dominoes must be correctlyplaced one against another as in the game.

381.--THE CARD FRAME PUZZLE.

In the illustration we have a frame constructed from the ten playingcards, ace to ten of diamonds. The children who made it wanted the pipson all four sides to add up alike, but they failed in their attempt andgave it up as impossible. It will be seen that the pips in the top row,the bottom row, and the left-hand side all add up 14, but the right-handside sums to 23. Now, what they were trying to do is quite possible. Canyou rearrange the ten cards in the same formation so that all four sidesshall add up alike? Of course they need not add up 14, but any numberyou choose to select.

[Illustration]

382.--THE CROSS OF CARDS.

[Illustration]

In this case we use only nine cards--the ace to nine of diamonds. Thepuzzle is to arrange them in the form of a cross, exactly in the wayshown in the illustration, so that the pips in the vertical bar and inthe horizontal bar add up alike. In the example given it will be foundthat both directions add up 23. What I want to know is, how manydifferent ways are there of rearranging the cards in order to bringabout this result? It will be seen that, without affecting the solution,we may exchange the 5 with the 6, the 5 with the 7, the 8 with the 3,and so on. Also we may make the horizontal and the vertical bars changeplaces. But such obvious manipulations as these are not to be regardedas different solutions. They are all mere variations of one fundamentalsolution. Now, how many of these fundamentally different solutions arethere? The pips need not, of course, always add up 23.

383.--THE "T" CARD PUZZLE.

[Illustration]

An entertaining little puzzle with cards is to take the nine cards of asuit, from ace to nine inclusive, and arrange them in the form of theletter "T," as shown in the illustration, so that the pips in thehorizontal line shall count the same as those in the column. In theexample given they add up twenty-three both ways. Now, it is quite easyto get a single correct arrangement. The puzzle is to discover in justhow many different ways it may be done. Though the number is high, thesolution is not really difficult if we attack the puzzle in the rightmanner. The reverse way obtained by reflecting the illustration in amirror we will not count as different, but all other changes in therelative positions of the cards will here count. How many different waysare there?

384.--CARD TRIANGLES.

Here you pick out the nine cards, ace to nine of diamonds, and arrangethem in the form of a triangle, exactly as shown in the illustration, sothat the pips add up the same on the three sides. In the example givenit will be seen that they sum to 20 on each side, but the particularnumber is of no importance so long as it is the same on all three sides.The puzzle is to find out in just how many different ways this can bedone.

If you simply turn the cards round so that one of the other two sides isnearest to you this will not count as different, for the order will bethe same. Also, if you make the 4, 9, 5 change places with the 7, 3, 8,and at the same time exchange the 1 and the 6, it will not be different.But if you only change the 1 and the 6 it will be different, because theorder round the triangle is not the same. This explanation will preventany doubt arising as to the conditions.

[Illustration]

385.--"STRAND" PATIENCE.

The idea for this came to me when considering the game of Patience thatI gave in the _Strand Magazine_ for December, 1910, which has beenreprinted in Ernest Bergholt's _Second Book of Patience Games_, underthe new name of "King Albert."

Make two piles of cards as follows: 9 D, 8 S, 7 D, 6 S, 5 D, 4 S, 3 D, 2S, 1 D, and 9 H, 8 C, 7 H, 6 C, 5 H, 4 C, 3 H, 2 C, 1 H, with the 9 ofdiamonds at the bottom of one pile and the 9 of hearts at the bottom ofthe other. The point is to exchange the spades with the clubs, so thatthe diamonds and clubs are still in numerical order in one pile and thehearts and spades in the other. There are four vacant spaces in additionto the two spaces occupied by the piles, and any card may be laid on aspace, but a card can only be laid on another of the next highervalue--an ace on a two, a two on a three, and so on. Patience isrequired to discover the shortest way of doing this. When there are fourvacant spaces you can pile four cards in seven moves, with only threespaces you can pile them in nine moves, and with two spaces you cannotpile more than two cards. When you have a grasp of these and similarfacts you will be able to remove a number of cards bodily and write down7, 9, or whatever the number of moves may be. The gradual shortening ofplay is fascinating, and first attempts are surprisingly lengthy.

386.--A TRICK WITH DICE.

[Illustration]

Here is a neat little trick with three dice. I ask you to throw the dicewithout my seeing them. Then I tell you to multiply the points of thefirst die by 2 and add 5; then multiply the result by 5 and add thepoints of the second die; then multiply the result by 10 and add thepoints of the third die. You then give me the total, and I can at oncetell you the points thrown with the three dice. How do I do it? As anexample, if you threw 1, 3, and 6, as in the illustration, the resultyou would give me would be 386, from which I could at once say what youhad thrown.

387.--THE VILLAGE CRICKET MATCH.

In a cricket match, Dingley Dell v. All Muggleton, the latter had thefirst innings. Mr. Dumkins and Mr. Podder were at the wickets, when thewary Dumkins made a splendid late cut, and Mr. Podder called on him torun. Four runs were apparently completed, but the vigilant umpires ateach end called, "three short," making six short runs in all. Whatnumber did Mr. Dumkins score? When Dingley Dell took their turn at thewickets their champions were Mr. Luffey and Mr. Struggles. The lattermade a magnificent off-drive, and invited his colleague to "come along,"with the result that the observant spectators applauded them for whatwas supposed to have been three sharp runs. But the umpires declaredthat there had been two short runs at each end--four in all. To whatextent, if any, did this manoeuvre increase Mr. Struggles's total?

388.--SLOW CRICKET.

In the recent county match between Wessex and Nincomshire the formerteam were at the wickets all day, the last man being put out a fewminutes before the time for drawing stumps. The play was so slow thatmost of the spectators were fast asleep, and, on being awakened by oneof the officials clearing the ground, we learnt that two men had beenput out leg-before-wicket for a combined score of 19 runs; four men werecaught for a combined score or 17 runs; one man was run out for a duck'segg; and the others were all bowled for 3 runs each. There were noextras. We were not told which of the men was the captain, but he madeexactly 15 more than the average of his team. What was the captain'sscore?

389.--THE FOOTBALL PLAYERS.

"It is a glorious game!" an enthusiast was heard to exclaim. "At theclose of last season, of the footballers of my acquaintance four hadbroken their left arm, five had broken their right arm, two had theright arm sound, and three had sound left arms." Can you discover fromthat statement what is the smallest number of players that the speakercould be acquainted with?

It does not at all follow that there were as many as fourteen men,because, for example, two of the men who had broken the left arm mightalso be the two who had sound right arms.

390.--THE HORSE-RACE PUZZLE.

There are no morals in puzzles. When we are solving the old puzzle ofthe captain who, having to throw half his crew overboard in a storm,arranged to draw lots, but so placed the men that only the Turks weresacrificed, and all the Christians left on board, we do not stop todiscuss the questionable morality of the proceeding. And when we aredealing with a measuring problem, in which certain thirsty pilgrims areto make an equitable division of a barrel of beer, we do not objectthat, as total abstainers, it is against our conscience to have anythingto do with intoxicating liquor. Therefore I make no apology forintroducing a puzzle that deals with betting.

Three horses--Acorn, Bluebottle, and Capsule--start in a race. The oddsare 4 to 1, Acorn; 3 to 1, Bluebottle; 2 to 1, Capsule. Now, how muchmust I invest on each horse in order to win L13, no matter which horsecomes in first? Supposing, as an example, that I betted L5 on eachhorse. Then, if Acorn won, I should receive L20 (four times L5), andhave to pay L5 each for the other two horses; thereby winning L10. Butit will be found that if Bluebottle was first I should only win L5, andif Capsule won I should gain nothing and lose nothing. This will makethe question perfectly clear to the novice, who, like myself, is notinterested in the calling of the fraternity who profess to be engaged inthe noble task of "improving the breed of horses."

391.--THE MOTOR-CAR RACE.

Sometimes a quite simple statement of fact, if worded in an unfamiliarmanner, will cause considerable perplexity. Here is an example, and itwill doubtless puzzle some of my more youthful readers just a little. Ihappened to be at a motor-car race at Brooklands, when one spectatorsaid to another, while a number of cars were whirling round and roundthe circular track:--

"There's Gogglesmith--that man in the white car!"

"Yes, I see," was the reply; "but how many cars are running in thisrace?"

Then came this curious rejoinder:--

"One-third of the cars in front of Gogglesmith added to three-quartersof those behind him will give you the answer."

Now, can you tell how many cars were running in the race?

PUZZLE GAMES.

"He that is beaten may be said To lie in honour's truckle bed." HUDIBRAS.

It may be said generally that a game is a contest of skill for two ormore persons, into which we enter either for amusement or to win aprize. A puzzle is something to be done or solved by the individual. Forexample, if it were possible for us so to master the complexities of thegame of chess that we could be assured of always winning with the firstor second move, as the case might be, or of always drawing, then itwould cease to be a game and would become a puzzle. Of course among theyoung and uninformed, when the correct winning play is not understood, apuzzle may well make a very good game. Thus there is no doubt childrenwill continue to play "Noughts and Crosses," though I have shown (No.109, "_Canterbury Puzzles_") that between two players who boththoroughly understand the play, every game should be drawn. Neitherplayer could ever win except through the blundering of his opponent. ButI am writing from the point of view of the student of these things.

The examples that I give in this class are apparently games, but, sinceI show in every case how one player may win if he only play correctly,they are in reality puzzles. Their interest, therefore, lies inattempting to discover the leading method of play.

392.--THE PEBBLE GAME.

Here is an interesting little puzzle game that I used to play with anacquaintance on the beach at Slocomb-on-Sea. Two players place an oddnumber of pebbles, we will say fifteen, between them. Then each takes inturn one, two, or three pebbles (as he chooses), and the winner is theone who gets the odd number. Thus, if you get seven and your opponenteight, you win. If you get six and he gets nine, he wins. Ought thefirst or second player to win, and how? When you have settled thequestion with fifteen pebbles try again with, say, thirteen.

393.--THE TWO ROOKS.

This is a puzzle game for two players. Each player has a single rook.The first player places his rook on any square of the board that he maychoose to select, and then the second player does the same. They nowplay in turn, the point of each play being to capture the opponent'srook. But in this game you cannot play through a line of attack withoutbeing captured. That is to say, if in the diagram it is Black's turn toplay, he cannot move his rook to his king's knight's square, or to hisking's rook's square, because he would enter the "line of fire" whenpassing his king's bishop's square. For the same reason he cannot moveto his queen's rook's seventh or eighth squares. Now, the game can neverend in a draw. Sooner or later one of the rooks must fall, unless, ofcourse, both players commit the absurdity of not trying to win. Thetrick of winning is ridiculously simple when you know it. Can you solvethe puzzle?

[Illustration]

394.--PUSS IN THE CORNER.

[Illustration]

This variation of the last puzzle is also played by two persons. Oneputs a counter on No. 6, and the other puts one on No. 55, and they playalternately by removing the counter to any other number in a line. Ifyour opponent moves at any time on to one of the lines you occupy, oreven crosses one of your lines, you immediately capture him and win. Wewill take an illustrative game.

A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes to15; A retreats to 26; B retreats to 13; A advances to 21; B retreats to2; A advances to 7; B goes to 3; A moves to 6; B must now go to 4; Aestablishes himself at 11, and B must be captured next move because heis compelled to cross a line on which A stands. Play this over and youwill understand the game directly. Now, the puzzle part of the game isthis: Which player should win, and how many moves are necessary?

395.--A WAR PUZZLE GAME.

[Illustration]

Here is another puzzle game. One player, representing the Britishgeneral, places a counter at B, and the other player, representing theenemy, places his counter at E. The Britisher makes the first advancealong one of the roads to the next town, then the enemy moves to one ofhis nearest towns, and so on in turns, until the British general getsinto the same town as the enemy and captures him. Although each mustalways move along a road to the next town only, and the second playermay do his utmost to avoid capture, the British general (as we shouldsuppose, from the analogy of real life) must infallibly win. But how?That is the question.

396.--A MATCH MYSTERY.

Here is a little game that is childishly simple in its conditions. Butit is well worth investigation.

Mr. Stubbs pulled a small table between himself and his friend, Mr.Wilson, and took a box of matches, from which he counted out thirty.

"Here are thirty matches," he said. "I divide them into three unequalheaps. Let me see. We have 14, 11, and 5, as it happens. Now, the twoplayers draw alternately any number from any one heap, and he who drawsthe last match loses the game. That's all! I will play with you, Wilson.I have formed the heaps, so you have the first draw."

"As I can draw any number," Mr. Wilson said, "suppose I exhibit my usualmoderation and take all the 14 heap."

"That is the worst you could do, for it loses right away. I take 6 fromthe 11, leaving two equal heaps of 5, and to leave two equal heaps is acertain win (with the single exception of 1, 1), because whatever you doin one heap I can repeat in the other. If you leave 4 in one heap, Ileave 4 in the other. If you then leave 2 in one heap, I leave 2 in theother. If you leave only 1 in one heap, then I take all the other heap.If you take all one heap, I take all but one in the other. No, you mustnever leave two heaps, unless they are equal heaps and more than 1, 1.Let's begin again."

"Very well, then," said Mr. Wilson. "I will take 6 from the 14, andleave you 8, 11, 5."

Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5, 3;Mr. Wilson,4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr.Stubbs, 2, 3, 1; Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1, 1,1.

"It is now quite clear that I must win," said Mr. Stubbs, because youmust take 1, and then I take 1, leaving you the last match. You neverhad a chance. There are just thirteen different ways in which thematches may be grouped at the start for a certain win. In fact, thegroups selected, 14, 11, 5, are a certain win, because for whatever youropponent may play there is another winning group you can secure, and soon and on down to the last match."

397.--THE MONTENEGRIN DICE GAME.

It is said that the inhabitants of Montenegro have a little dice gamethat is both ingenious and well worth investigation. The two playersfirst select two different pairs of odd numbers (always higher than 3)and then alternately toss three dice. Whichever first throws the dice sothat they add up to one of his selected numbers wins. If they are bothsuccessful in two successive throws it is a draw and they try again. Forexample, one player may select 7 and 15 and the other 5 and 13. Then ifthe first player throws so that the three dice add up 7 or 15 he wins,unless the second man gets either 5 or 13 on his throw.

The puzzle is to discover which two pairs of numbers should be selectedin order to give both players an exactly even chance.

398.--THE CIGAR PUZZLE.

I once propounded the following puzzle in a London club, and for aconsiderable period it absorbed the attention of the members. They couldmake nothing of it, and considered it quite impossible of solution. Andyet, as I shall show, the answer is remarkably simple.

Two men are seated at a square-topped table. One places an ordinarycigar (flat at one end, pointed at the other) on the table, then theother does the same, and so on alternately, a condition being that nocigar shall touch another. Which player should succeed in placing thelast cigar, assuming that they each will play in the best possiblemanner? The size of the table top and the size of the cigar are notgiven, but in order to exclude the ridiculous answer that the tablemight be so diminutive as only to take one cigar, we will say that thetable must not be less than 2 feet square and the cigar not more than 41/2inches long. With those restrictions you may take any dimensions youlike. Of course we assume that all the cigars are exactly alike inevery respect. Should the first player, or the second player, win?

MAGIC SQUARE PROBLEMS.

"By magic numbers." CONGREVE, _The Mourning Bride._

This is a very ancient branch of mathematical puzzledom, and it has animmense, though scattered, literature of its own. In their simple formof consecutive whole numbers arranged in a square so that every column,every row, and each of the two long diagonals shall add up alike, thesemagic squares offer three main lines of investigation: Construction,Enumeration, and Classification. Of recent years many ingenious methodshave been devised for the construction of magics, and the law of theirformation is so well understood that all the ancient mystery hasevaporated and there is no longer any difficulty in making squares ofany dimensions. Almost the last word has been said on this subject. Thequestion of the enumeration of all the possible squares of a given orderstands just where it did over two hundred years ago. Everybody knowsthat there is only one solution for the third order, three cells bythree; and Frenicle published in 1693 diagrams of all the arrangementsof the fourth order--880 in number--and his results have been verifiedover and over again. I may here refer to the general solution for thisorder, for numbers not necessarily consecutive, by E. Bergholt in_Nature_, May 26, 1910, as it is of the greatest importance to studentsof this subject. The enumeration of the examples of any higher order isa completely unsolved problem.

As to classification, it is largely a matter of individualtaste--perhaps an aesthetic question, for there is beauty in the law andorder of numbers. A man once said that he divided the human race intotwo great classes: those who take snuff and those who do not. I am notsure that some of our classifications of magic squares are not almost asvalueless. However, lovers of these things seem somewhat agreed thatNasik magic squares (so named by Mr. Frost, a student of them, after thetown in India where he lived, and also called Diabolique andPandiagonal) and Associated magic squares are of special interest, so Iwill just explain what these are for the benefit of the novice.

[Illustration: SIMPLE]

[Illustration: SEMI-NASIK]

[Illustration: ASSOCIATED]

[Illustration: NASIK]

I published in _The Queen_ for January 15, 1910, an article that wouldenable the reader to write out, if he so desired, all the 880 magics ofthe fourth order, and the following is the complete classification thatI gave. The first example is that of a Simple square that fulfils thesimple conditions and no more. The second example is a Semi-Nasik, whichhas the additional property that the opposite short diagonals of twocells each together sum to 34. Thus, 14 + 4 + 11 + 5 = 34 and 12 + 6 +13 + 3 = 34. The third example is not only Semi-Nasik but alsoAssociated, because in it every number, if added to the number that isequidistant, in a straight line, from the centre gives 17. Thus, 1 + 16,2 + 15, 3 + 14, etc. The fourth example, considered the most "perfect"of all, is a Nasik. Here all the broken diagonals sum to 34. Thus, forexample, 15 + 14 + 2 + 3, and 10 + 4 + 7 + 13, and 15 + 5 + 2 + 12. As aconsequence, its properties are such that if you repeat the square inall directions you may mark off a square, 4 x 4, wherever you please,and it will be magic.

The following table not only gives a complete enumeration under the fourforms described, but also a classification under the twelve graphictypes indicated in the diagrams. The dots at the end of each linerepresent the relative positions of those complementary pairs, 1 + 16, 2+ 15, etc., which sum to 17. For example, it will be seen that the firstand second magic squares given are of Type VI., that the third square isof Type III., and that the fourth is of Type I. Edouard Lucas indicatedthese types, but he dropped exactly half of them and did not attempt theclassification.

NASIK (Type I.) . . . . . 48 SEMI-NASIK (Type II., Transpositions of Nasik) . 48 " (Type III., Associated) 48 " (Type IV.) . . . 96 " (Type V.) . . . 96 192 ___ " (Type VI.) . . . 96 384 ___ SIMPLE. (Type VI.) . . . 208 " (Type VII.) . . . 56 " (Type VIII.). . . 56 " (Type IX.) . . . 56 " (Type X.) . . . 56 224 ___ " (Type XI.) . . . 8 " (Type XII.) . . . 8 16 448 ___ ___ ___ 880 ___

It is hardly necessary to say that every one of these squares willproduce seven others by mere reversals and reflections, which we do notcount as different. So that there are 7,040 squares of this order, 880of which are fundamentally different.

An infinite variety of puzzles may be made introducing new conditionsinto the magic square. In _The Canterbury Puzzles_ I have given examplesof such squares with coins, with postage stamps, with cutting-outconditions, and other tricks. I will now give a few variants involvingfurther novel conditions.

399.--THE TROUBLESOME EIGHT.

Nearly everybody knows that a "magic square" is an arrangement ofnumbers in the form of a square so that every row, every column, andeach of the two long diagonals adds up alike. For example, you wouldfind little difficulty in merely placing a different number in each ofthe nine cells in the illustration so that the rows, columns, anddiagonals shall all add up 15. And at your first attempt you willprobably find that you have an 8 in one of the corners. The puzzle is toconstruct the magic square, under the same conditions, with the 8 in theposition shown.

[Illustration]

400.--THE MAGIC STRIPS.

[Illustration]

I happened to have lying on my table a number of strips of cardboard,with numbers printed on them from 1 upwards in numerical order. The ideasuddenly came to me, as ideas have a way of unexpectedly coming, to makea little puzzle of this. I wonder whether many readers will arrive atthe same solution that I did.

Take seven strips of cardboard and lay them together as above. Thenwrite on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so thatthe numbers shall form seven rows and seven columns.

Now, the puzzle is to cut these strips into the fewest possible piecesso that they may be placed together and form a magic square, the sevenrows, seven columns, and two diagonals adding up the same number. Nofigures may be turned upside down or placed on their sides--that is, allthe strips must lie in their original direction.

Of course you could cut each strip into seven separate pieces, eachpiece containing a number, and the puzzle would then be very easy, but Ineed hardly say that forty-nine pieces is a long way from being thefewest possible.

401.--EIGHT JOLLY GAOL BIRDS.

[Illustration]

The illustration shows the plan of a prison of nine cells allcommunicating with one another by doorways. The eight prisoners havetheir numbers on their backs, and any one of them is allowed to exercisehimself in whichever cell may happen to be vacant, subject to the rulethat at no time shall two prisoners be in the same cell. The merrymonarch in whose dominions the prison was situated offered them specialcomforts one Christmas Eve if, without breaking that rule, they could soplace themselves that their numbers should form a magic square.

Now, prisoner No. 7 happened to know a good deal about magic squares, sohe worked out a scheme and naturally selected the method that was mostexpeditious--that is, one involving the fewest possible moves from cellto cell. But one man was a surly, obstinate fellow (quite unfit for thesociety of his jovial companions), and he refused to move out of hiscell or take any part in the proceedings. But No. 7 was quite equal tothe emergency, and found that he could still do what was required in thefewest possible moves without troubling the brute to leave his cell. Thepuzzle is to show how he did it and, incidentally, to discover whichprisoner was so stupidly obstinate. Can you find the fellow?

402.--NINE JOLLY GAOL BIRDS.

[Illustration]

Shortly after the episode recorded in the last puzzle occurred, a ninthprisoner was placed in the vacant cell, and the merry monarch thenoffered them all complete liberty on the following strange conditions.They were required so to rearrange themselves in the cells that theirnumbers formed a magic square without their movements causing any two ofthem ever to be in the same cell together, except that at the start oneman was allowed to be placed on the shoulders of another man, and thusadd their numbers together, and move as one man. For example, No. 8might be placed on the shoulders of No. 2, and then they would moveabout together as 10. The reader should seek first to solve the puzzlein the fewest possible moves, and then see that the man who is burdenedhas the least possible amount of work to do.

403.--THE SPANISH DUNGEON.

Not fifty miles from Cadiz stood in the middle ages a castle, all tracesof which have for centuries disappeared. Among other interestingfeatures, this castle contained a particularly unpleasant dungeondivided into sixteen cells, all communicating with one another, as shownin the illustration.

Now, the governor was a merry wight, and very fond of puzzles withal.One day he went to the dungeon and said to the prisoners, "By myhalidame!" (or its equivalent in Spanish) "you shall all be set free ifyou can solve this puzzle. You must so arrange yourselves in the sixteencells that the numbers on your backs shall form a magic square in whichevery column, every row, and each of the two diagonals shall add up thesame. Only remember this: that in no case may two of you ever betogether in the same cell."

One of the prisoners, after working at the problem for two or threedays, with a piece of chalk, undertook to obtain the liberty of himselfand his fellow-prisoners if they would follow his directions and movethrough the doorway from cell to cell in the order in which he shouldcall out their numbers.

[Illustration]

He succeeded in his attempt, and, what is more remarkable, it would seemfrom the account of his method recorded in the ancient manuscript lyingbefore me, that he did so in the fewest possible moves. The reader isasked to show what these moves were.

404.--THE SIBERIAN DUNGEONS.

[Illustration]

The above is a trustworthy plan of a certain Russian prison in Siberia.All the cells are numbered, and the prisoners are numbered the same asthe cells they occupy. The prison diet is so fattening that thesepolitical prisoners are in perpetual fear lest, should their pardonarrive, they might not be able to squeeze themselves through the narrowdoorways and get out. And of course it would be an unreasonable thing toask any government to pull down the walls of a prison just to liberatethe prisoners, however innocent they might be. Therefore these men takeall the healthy exercise they can in order to retard their increasingobesity, and one of their recreations will serve to furnish us with thefollowing puzzle.

Show, in the fewest possible moves, how the sixteen men may formthemselves into a magic square, so that the numbers on their backs shalladd up the same in each of the four columns, four rows, and twodiagonals without two prisoners having been at any time in the same celltogether. I had better say, for the information of those who have notyet been made acquainted with these places, that it is a peculiarity ofprisons that you are not allowed to go outside their walls. Any prisonermay go any distance that is possible in a single move.

405.--CARD MAGIC SQUARES.

[Illustration]

Take an ordinary pack of cards and throw out the twelve court cards.Now, with nine of the remainder (different suits are of no consequence)form the above magic square. It will be seen that the pips add upfifteen in every row in every column, and in each of the two longdiagonals. The puzzle is with the remaining cards (without disturbingthis arrangement) to form three more such magic squares, so that each ofthe four shall add up to a different sum. There will, of course, be fourcards in the reduced pack that will not be used. These four may be anythat you choose. It is not a difficult puzzle, but requires just alittle thought.

406.--THE EIGHTEEN DOMINOES.

The illustration shows eighteen dominoes arranged in the form of asquare so that the pips in every one of the six columns, six rows, andtwo long diagonals add up 13. This is the smallest summation possiblewith any selection of dominoes from an ordinary box of twenty-eight. Thegreatest possible summation is 23, and a solution for this number may beeasily obtained by substituting for every number its complement to 6.Thus for every blank substitute a 6, for every 1 a 5, for every 2 a 4,for 3 a 3, for 4 a 2, for 5 a 1, and for 6 a blank. But the puzzle is tomake a selection of eighteen dominoes and arrange them (in exactly theform shown) so that the summations shall be 18 in all the fourteendirections mentioned.

[Illustration]

SUBTRACTING, MULTIPLYING, AND DIVIDING MAGICS.

Although the adding magic square is of such great antiquity, curiouslyenough the multiplying magic does not appear to have been mentioneduntil the end of the eighteenth century, when it was referred toslightly by one writer and then forgotten until I revived it in_Tit-Bits_ in 1897. The dividing magic was apparently first discussed byme in _The Weekly Dispatch_ in June 1898. The subtracting magic is hereintroduced for the first time. It will now be convenient to deal withall four kinds of magic squares together.

[Illustration: ADDING SUBTRACTING MULTIPLYING DIVIDING]

In these four diagrams we have examples in the third order of adding,subtracting, multiplying, and dividing squares. In the first theconstant, 15, is obtained by the addition of the rows, columns, and twodiagonals. In the second case you get the constant, 5, by subtractingthe first number in a line from the second, and the result from thethird. You can, of course, perform the operation in either direction;but, in order to avoid negative numbers, it is more convenient simply todeduct the middle number from the sum of the two extreme numbers. Thisis, in effect, the same thing. It will be seen that the constant of theadding square is n times that of the subtracting square derived fromit, where n is the number of cells in the side of square. And themanner of derivation here is simply to reverse the two diagonals. Bothsquares are "associated"--a term I have explained in the introductoryarticle to this department.

The third square is a multiplying magic. The constant, 216, is obtainedby multiplying together the three numbers in any line. It is"associated" by multiplication, instead of by addition. It is herenecessary to remark that in an adding square it is not essential thatthe nine numbers should be consecutive. Write down any nine numbers inthis way--

1 3 5 4 6 8 7 9 11

so that the horizontal differences are all alike and the verticaldifferences also alike (here 2 and 3), and these numbers will form anadding magic square. By making the differences 1 and 3 we, of course,get consecutive numbers--a particular case, and nothing more. Now, inthe case of the multiplying square we must take these numbers ingeometrical instead of arithmetical progression, thus--

1 3 9 2 6 18 4 12 36

Here each successive number in the rows is multiplied by 3, and in thecolumns by 2. Had we multiplied by 2 and 8 we should get the regulargeometrical progression, 1, 2, 4, 8, 16, 32, 64, 128, and 256, but Iwish to avoid high numbers. The numbers are arranged in the square inthe same order as in the adding square.

The fourth diagram is a dividing magic square. The constant 6 is hereobtained by dividing the second number in a line by the first (in eitherdirection) and the third number by the quotient. But, again, the processis simplified by dividing the product of the two extreme numbers by themiddle number. This square is also "associated" by multiplication. It isderived from the multiplying square by merely reversing the diagonals,and the constant of the multiplying square is the cube of that of thedividing square derived from it.

The next set of diagrams shows the solutions for the fifth order ofsquare. They are all "associated" in the same way as before. Thesubtracting square is derived from the adding square by reversing thediagonals and exchanging opposite numbers in the centres of the borders,and the constant of one is again n times that of the other. Thedividing square is derived from the multiplying square in the same way,and the constant of the latter is the 5th power (that is the nth) ofthat of the former.

[Illustration]

These squares are thus quite easy for odd orders. But the reader willprobably find some difficulty over the even orders, concerning which Iwill leave him to make his own researches, merely propounding two littleproblems.

407.--TWO NEW MAGIC SQUARES.

Construct a subtracting magic square with the first sixteen wholenumbers that shall be "associated" by _subtraction_. The constant is, ofcourse, obtained by subtracting the first number from the second inline, the result from the third, and the result again from the fourth.Also construct a dividing magic square of the same order that shall be"associated" by _division_. The constant is obtained by dividing thesecond number in a line by the first, the third by the quotient, and thefourth by the next quotient.

408.--MAGIC SQUARES OF TWO DEGREES.

While reading a French mathematical work I happened to come across, thefollowing statement: "A very remarkable magic square of 8, in twodegrees, has been constructed by M. Pfeffermann. In other words, he hasmanaged to dispose the sixty-four first numbers on the squares of achessboard in such a way that the sum of the numbers in every line,every column, and in each of the two diagonals, shall be the same; andmore, that if one substitutes for all the numbers their squares, thesquare still remains magic." I at once set to work to solve thisproblem, and, although it proved a very hard nut, one was rewarded bythe discovery of some curious and beautiful laws that govern it. Thereader may like to try his hand at the puzzle.

MAGIC SQUARES OF PRIMES.

The problem of constructing magic squares with prime numbers only wasfirst discussed by myself in _The Weekly Dispatch_ for 22nd July and 5thAugust 1900; but during the last three or four years it has receivedgreat attention from American mathematicians. First, they have sought toform these squares with the lowest possible constants. Thus, the firstnine prime numbers, 1 to 23 inclusive, sum to 99, which (being divisibleby 3) is theoretically a suitable series; yet it has been demonstratedthat the lowest possible constant is 111, and the required series asfollows: 1, 7, 13, 31, 37, 43, 61, 67, and 73. Similarly, in the case ofthe fourth order, the lowest series of primes that are "theoreticallysuitable" will not serve. But in every other order, up to the 12thinclusive, magic squares have been constructed with the lowest series ofprimes theoretically possible. And the 12th is the lowest order in whicha straight series of prime numbers, unbroken, from 1 upwards has beenmade to work. In other words, the first 144 odd prime numbers haveactually been arranged in magic form. The following summary is takenfrom _The Monist_ (Chicago) for October 1913:--

Order of Totals of Lowest Squares Square. Series. Constants. made by-- (Henry E. 3rd 333 111 { Dudeney ( (1900).

(Ernest Bergholt 4th 408 102 { and C. D. ( Shuldham.

5th 1065 213 H. A. Sayles.

(C. D. Shuldham 6th 2448 408 { and J. ( N. Muncey.

7th 4893 699 do. 8th 8912 1114 do. 9th 15129 1681 do. 10th 24160 2416 J. N. Muncey. 11th 36095 3355 do. 12th 54168 4514 do.

For further details the reader should consult the article itself, by W.S. Andrews and H. A. Sayles.

These same investigators have also performed notable feats inconstructing associated and bordered prime magics, and Mr. Shuldham hassent me a remarkable paper in which he gives examples of Nasik squaresconstructed with primes for all orders from the 4th to the 10th, withthe exception of the 3rd (which is clearly impossible) and the 9th,which, up to the time of writing, has baffled all attempts.

409.--THE BASKETS OF PLUMS.

[Illustration]

This is the form in which I first introduced the question of magicsquares with prime numbers. I will here warn the reader that there is alittle trap.

A fruit merchant had nine baskets. Every basket contained plums (allsound and ripe), and the number in every basket was different. Whenplaced as shown in the illustration they formed a magic square, so thatif he took any three baskets in a line in the eight possible directionsthere would always be the same number of plums. This part of the puzzleis easy enough to understand. But what follows seems at first sight alittle queer.

The merchant told one of his men to distribute the contents of anybasket he chose among some children, giving plums to every child so thateach should receive an equal number. But the man found it quiteimpossible, no matter which basket he selected and no matter how manychildren he included in the treat. Show, by giving contents of the ninebaskets, how this could come about.

410.--THE MANDARIN'S "T" PUZZLE.

[Illustration]

Before Mr. Beauchamp Cholmondely Marjoribanks set out on his tour in theFar East, he prided himself on his knowledge of magic squares, a subjectthat he had made his special hobby; but he soon discovered that he hadnever really touched more than the fringe of the subject, and that thewily Chinee could beat him easily. I present a little problem that onelearned mandarin propounded to our traveller, as depicted on the lastpage.

The Chinaman, after remarking that the construction of the ordinarymagic square of twenty-five cells is "too velly muchee easy," asked ourcountryman so to place the numbers 1 to 25 in the square that everycolumn, every row, and each of the two diagonals should add up 65, withonly prime numbers on the shaded "T." Of course the prime numbersavailable are 1, 2, 3, 5, 7, 11, 13, 17, 19, and 23, so you are atliberty to select any nine of these that will serve your purpose. Canyou construct this curious little magic square?

411.--A MAGIC SQUARE OF COMPOSITES.

As we have just discussed the construction of magic squares with primenumbers, the following forms an interesting companion problem. Make amagic square with nine consecutive composite numbers--the smallestpossible.

412.--THE MAGIC KNIGHT'S TOUR.

Here is a problem that has never yet been solved, nor has itsimpossibility been demonstrated. Play the knight once to every square ofthe chessboard in a complete tour, numbering the squares in the ordervisited, so that when completed the square shall be "magic," adding upto 260 in every column, every row, and each of the two long diagonals. Ishall give the best answer that I have been able to obtain, in whichthere is a slight error in the diagonals alone. Can a perfect solutionbe found? I am convinced that it cannot, but it is only a "piousopinion."

MAZES AND HOW TO THREAD THEM.

"In wandering mazes lost." _Paradise Lost._

The Old English word "maze," signifying a labyrinth, probably comes fromthe Scandinavian, but its origin is somewhat uncertain. The lateProfessor Skeat thought that the substantive was derived from the verb,and as in old times to be mazed or amazed was to be "lost in thought,"the transition to a maze in whose tortuous windings we are lost isnatural and easy.

The word "labyrinth" is derived from a Greek word signifying thepassages of a mine. The ancient mines of Greece and elsewhere inspiredfear and awe on account of their darkness and the danger of getting lostin their intricate passages. Legend was afterwards built round thesemazes. The most familiar instance is the labyrinth made by Daedalus inCrete for King Minos. In the centre was placed the Minotaur, and no onewho entered could find his way out again, but became the prey of themonster. Seven youths and seven maidens were sent regularly by theAthenians, and were duly devoured, until Theseus slew the monster andescaped from the maze by aid of the clue of thread provided by Ariadne;which accounts for our using to-day the expression "threading a maze."

The various forms of construction of mazes include complicated ranges ofcaverns, architectural labyrinths, or sepulchral buildings, tortuousdevices indicated by coloured marbles and tiled pavements, winding pathscut in the turf, and topiary mazes formed by clipped hedges. As a matterof fact, they may be said to have descended to us in precisely thisorder of variety.

Mazes were used as ornaments on the state robes of Christian emperorsbefore the ninth century, and were soon adopted in the decoration ofcathedrals and other churches. The original idea was doubtless to employthem as symbols of the complicated folds of sin by which man issurrounded. They began to abound in the early part of the twelfthcentury, and I give an illustration of one of this period in the parishchurch at St. Quentin (Fig. 1). It formed a pavement of the nave, andits diameter is 341/2 feet. The path here is the line itself. If you placeyour pencil at the point A and ignore the enclosing line, the line leadsyou to the centre by a long route over the entire area; but you neverhave any option as to direction during your course. As we shall find insimilar cases, these early ecclesiastical mazes were generally not of apuzzle nature, but simply long, winding paths that took you overpractically all the ground enclosed.

[Illustration: FIG. 1.--Maze at St. Quentin.]

[Illustration: FIG. 2.--Maze in Chartres Cathedral.]

In the abbey church of St. Berlin, at St. Omer, is another of thesecurious floors, representing the Temple of Jerusalem, with stations forpilgrims. These mazes were actually visited and traversed by them as acompromise for not going to the Holy Land in fulfilment of a vow. Theywere also used as a means of penance, the penitent frequently beingdirected to go the whole course of the maze on hands and knees.

[Illustration: FIG. 3.--Maze in Lucca Cathedral.]

The maze in Chartres Cathedral, of which I give an illustration (Fig.2), is 40 feet across, and was used by penitents following theprocession of Calvary. A labyrinth in Amiens Cathedral was octagonal,similar to that at St. Quentin, measuring 42 feet across. It bore thedate 1288, but was destroyed in 1708. In the chapter-house at Bayeux isa labyrinth formed of tiles, red, black, and encaustic, with a patternof brown and yellow. Dr. Ducarel, in his "_Tour through Part ofNormandy_" (printed in 1767), mentions the floor of the greatguard-chamber in the abbey of St. Stephen, at Caen, "the middle whereofrepresents a maze or labyrinth about 10 feet diameter, and so artfullycontrived that, were we to suppose a man following all the intricatemeanders of its volutes, he could not travel less than a mile before hegot from one end to the other."

[Illustration: FIG. 4.--Maze at Saffron Walden, Essex.]

Then these mazes were sometimes reduced in size and represented on asingle tile (Fig. 3). I give an example from Lucca Cathedral. It is onone of the porch piers, and is 191/2 inches in diameter. A writer in1858 says that, "from the continual attrition it has received fromthousands of tracing fingers, a central group of Theseus and theMinotaur has now been very nearly effaced." Other examples were, andperhaps still are, to be found in the Abbey of Toussarts, atChalons-sur-Marne, in the very ancient church of St. Michele at Pavia,at Aix in Provence, in the cathedrals of Poitiers, Rheims, and Arras, inthe church of Santa Maria in Aquiro in Rome, in San Vitale at Ravenna,in the Roman mosaic pavement found at Salzburg, and elsewhere. Thesemazes were sometimes called "Chemins de Jerusalem," as beingemblematical of the difficulties attending a journey to the earthlyJerusalem and of those encountered by the Christian before he can reachthe heavenly Jerusalem--where the centre was frequently called "Ciel."

Common as these mazes were upon the Continent, it is probable that noexample is to be found in any English church; at least I am not aware ofthe existence of any. But almost every county has, or has had, itsspecimens of mazes cut in the turf. Though these are frequently known as"miz-mazes" or "mize-mazes," it is not uncommon to find them locallycalled "Troy-towns," "shepherds' races," or "Julian's Bowers"--namesthat are misleading, as suggesting a false origin. From the facts alonethat many of these English turf mazes are clearly copied from those inthe Continental churches, and practically all are found close to someecclesiastical building or near the site of an ancient one, we mayregard it as certain that they were of church origin and not invented bythe shepherds or other rustics. And curiously enough, these turf mazesare apparently unknown on the Continent. They are distinctly mentionedby Shakespeare:--

"The nine men's morris is filled up with mud, And the quaint mazes in the wanton green For lack of tread are undistinguishable."

_A Midsummer Night's Dream_, ii. 1.

"My old bones ache: here's a maze trod indeed, Through forth-rights and meanders!"

_The Tempest_, iii. 3.

[Illustration: FIG. 5.--Maze at Sneinton, Nottinghamshire.]

There was such a maze at Comberton, in Cambridgeshire, and another,locally called the "miz-maze," at Leigh, in Dorset. The latter was onthe highest part of a field on the top of a hill, a quarter of a milefrom the village, and was slightly hollow in the middle and enclosed bya bank about 3 feet high. It was circular, and was thirty paces indiameter. In 1868 the turf had grown over the little trenches, and itwas then impossible to trace the paths of the maze. The Comberton onewas at the same date believed to be perfect, but whether either or bothhave now disappeared I cannot say. Nor have I been able to verify theexistence or non-existence of the other examples of which I am able togive illustrations. I shall therefore write of them all in the pasttense, retaining the hope that some are still preserved.

[Illustration: FIG. 6.--Maze at Alkborough, Lincolnshire.]

In the next two mazes given--that at Saffron Walden, Essex (110 feet indiameter, Fig. 4), and the one near St. Anne's Well, at Sneinton,Nottinghamshire (Fig. 5), which was ploughed up on February 27th, 1797(51 feet in diameter, with a path 535 yards long)--the paths must ineach case be understood to be on the lines, black or white, as the casemay be.

[Illustration: FIG. 7.--Maze at Boughton Green, Nottinghamshire.]

I give in Fig. 6 a maze that was at Alkborough, Lincolnshire,overlooking the Humber. This was 44 feet in diameter, and theresemblance between it and the mazes at Chartres and Lucca (Figs. 2 and3) will be at once perceived. A maze at Boughton Green, inNottinghamshire, a place celebrated at one time for its fair (Fig. 7),was 37 feet in diameter. I also include the plan (Fig. 8) of one thatused to be on the outskirts of the village of Wing, near Uppingham,Rutlandshire. This maze was 40 feet in diameter.

[Illustration: FIG. 8.--Maze at Wing, Rutlandshire.]

[Illustration: FIG. 9.--Maze on St. Catherine's Hill, Winchester.]

The maze that was on St. Catherine's Hill, Winchester, in the parish ofChilcombe, was a poor specimen (Fig. 9), since, as will be seen, therewas one short direct route to the centre, unless, as in Fig. 10 again,the path is the line itself from end to end. This maze was 86 feetsquare, cut in the turf, and was locally known as the "Mize-maze." Itbecame very indistinct about 1858, and was then recut by the Warden ofWinchester, with the aid of a plan possessed by a lady living in theneighbourhood.

[Illustration: FIG. 10.--Maze on Ripon Common.]

A maze formerly existed on Ripon Common, in Yorkshire (Fig. 10). It wasploughed up in 1827, but its plan was fortunately preserved. Thisexample was 20 yards in diameter, and its path is said to have been 407yards long.

[Illustration: FIG. 11.--Maze at Theobalds, Hertfordshire.]

In the case of the maze at Theobalds, Hertfordshire, after you havefound the entrance within the four enclosing hedges, the path is forced(Fig. 11). As further illustrations of this class of maze, I give onetaken from an Italian work on architecture by Serlio, published in 1537(Fig. 12), and one by London and Wise, the designers of the HamptonCourt maze, from their book, _The Retired Gard'ner_, published in 1706(Fig. 13). Also, I add a Dutch maze (Fig. 14).

[Illustration: FIG. 12.--Italian Maze of Sixteenth Century.]

[Illustration: FIG. 13.--By the Designers of Hampton Court Maze.]

[Illustration: FIG. 14.--A Dutch Maze.]

So far our mazes have been of historical interest, but they havepresented no difficulty in threading. After the Reformation period wefind mazes converted into mediums for recreation, and they generallyconsisted of labyrinthine paths enclosed by thick and carefully trimmedhedges. These topiary hedges were known to the Romans, with whom the_topiarius_ was the ornamental gardener. This type of maze has of lateyears degenerated into the seaside "Puzzle Gardens. Teas, sixpence,including admission to the Maze." The Hampton Court Maze, sometimescalled the "Wilderness," at the royal palace, was designed, as I havesaid, by London and Wise for William III., who had a liking for suchthings (Fig. 15). I have before me some three or four versions of it,all slightly different from one another; but the plan I select is takenfrom an old guide-book to the palace, and therefore ought to betrustworthy. The meaning of the dotted lines, etc., will be explainedlater on.

[Illustration: FIG. 15.--Maze at Hampton Court Palace.]

[Illustration: FIG. 16.--Maze at Hatfield House, Herts.]

[Illustration: FIG. 17.--Maze formerly at South Kensington.]

[Illustration: FIG. 18.--A German Maze.]

The maze at Hatfield House (Fig. 16), the seat of the Marquis ofSalisbury, like so many labyrinths, is not difficult on paper; but boththis and the Hampton Court Maze may prove very puzzling to actuallythread without knowing the plan. One reason is that one is so apt to godown the same blind alleys over and over again, if one proceeds withoutmethod. The maze planned by the desire of the Prince Consort for theRoyal Horticultural Society's Gardens at South Kensington was allowed togo to ruin, and was then destroyed--no great loss, for it was a feeblething. It will be seen that there were three entrances from the outside(Fig. 17), but the way to the centre is very easy to discover. I includea German maze that is curious, but not difficult to thread on paper(Fig. 18). The example of a labyrinth formerly existing at Pimperne, inDorset, is in a class by itself (Fig. 19). It was formed of small ridgesabout a foot high, and covered nearly an acre of ground; but it was,unfortunately, ploughed up in 1730.

[Illustration: FIG. 19.--Maze at Pimperne, Dorset.]

We will now pass to the interesting subject of how to thread any maze.While being necessarily brief, I will try to make the matter clear toreaders who have no knowledge of mathematics. And first of all we willassume that we are trying to enter a maze (that is, get to the "centre")of which we have no plan and about which we know nothing. The first ruleis this: If a maze has no parts of its hedges detached from the rest,then if we always keep in touch with the hedge with the right hand (oralways touch it with the left), going down to the stop in every blindalley and coming back on the other side, we shall pass through everypart of the maze and make our exit where we went in. Therefore we mustat one time or another enter the centre, and every alley will betraversed twice.

[Illustration: FIG. 20.--M. Tremaux's Method of Solution.]

[Illustration: FIG. 21.--How to thread the Hatfield Maze.]

Now look at the Hampton Court plan. Follow, say to the right, the pathindicated by the dotted line, and what I have said is clearly correct ifwe obliterate the two detached parts, or "islands," situated on eachside of the star. But as these islands are there, you cannot by thismethod traverse every part of the maze; and if it had been so plannedthat the "centre" was, like the star, between the two islands, you wouldnever pass through the "centre" at all. A glance at the Hatfield mazewill show that there are three of these detached hedges or islands atthe centre, so this method will never take you to the "centre" of thatone. But the rule will at least always bring you safely out again unlessyou blunder in the following way. Suppose, when you were going in thedirection of the arrow in the Hampton Court Maze, that you could notdistinctly see the turning at the bottom, that you imagined you were ina blind alley and, to save time, crossed at once to the opposite hedge,then you would go round and round that U-shaped island with your righthand still always on the hedge--for ever after!

[Illustration: FIG. 22. The Philadelphia Maze, and its Solution.]

This blunder happened to me a few years ago in a little maze on the isleof Caldy, South Wales. I knew the maze was a small one, but after a verylong walk I was amazed to find that I did not either reach the "centre"or get out again. So I threw a piece of paper on the ground, and sooncame round to it; from which I knew that I had blundered over a supposedblind alley and was going round and round an island. Crossing to theopposite hedge and using more care, I was quickly at the centre and outagain. Now, if I had made a similar mistake at Hampton Court, anddiscovered the error when at the star, I should merely have passed fromone island to another! And if I had again discovered that I was on adetached part, I might with ill luck have recrossed to the first islandagain! We thus see that this "touching the hedge" method should alwaysbring us safely out of a maze that we have entered; it may happen totake us through the "centre," and if we miss the centre we shall knowthere must be islands. But it has to be done with a little care, and inno case can we be sure that we have traversed every alley or that thereare no detached parts.

[Illustration: FIG. 23.--Simplified Diagram of Fig. 22.]

If the maze has many islands, the traversing of the whole of it may be amatter of considerable difficulty. Here is a method for solving anymaze, due to M. Tremaux, but it necessitates carefully marking in someway your entrances and exits where the galleries fork. I give a diagramof an imaginary maze of a very simple character that will serve ourpurpose just as well as something more complex (Fig. 20). The circles atthe regions where we have a choice of turnings we may call nodes. A"new" path or node is one that has not been entered before on the route;an "old" path or node is one that has already been entered, 1. No pathmay be traversed more than twice. 2. When you come to a new node, takeany path you like. 3. When by a new path you come to an old node or tothe stop of a blind alley, return by the path you came. 4. When by anold path you come to an old node, take a new path if there is one; ifnot, an old path. The route indicated by the dotted line in the diagramis taken in accordance with these simple rules, and it will be seenthat it leads us to the centre, although the maze consists of fourislands.

[Illustration: FIG. 24.--Can you find the Shortest Way to Centre?]

Neither of the methods I have given will disclose to us the shortest wayto the centre, nor the number of the different routes. But we can easilysettle these points with a plan. Let us take the Hatfield maze (Fig.21). It will be seen that I have suppressed all the blind alleys by theshading. I begin at the stop and work backwards until the path forks.These shaded parts, therefore, can never be entered without our havingto retrace our steps. Then it is very clearly seen that if we enter at Awe must come out at B; if we enter at C we must come out at D. Then wehave merely to determine whether A, B, E, or C, D, E, is the shorterroute. As a matter of fact, it will be found by rough measurement orcalculation that the shortest route to the centre is by way of C, D, E,F.

[Illustration: FIG. 25.--Rosamund's Bower.]

I will now give three mazes that are simply puzzles on paper, for, sofar as I know, they have never been constructed in any other way. Thefirst I will call the Philadelphia maze (Fig. 22). Fourteen years ago atravelling salesman, living in Philadelphia, U.S.A., developed acuriously unrestrained passion for puzzles. He neglected his business,and soon his position was taken from him. His days and nights were nowpassed with the subject that fascinated him, and this little maze seemsto have driven him into insanity. He had been puzzling over it for sometime, and finally it sent him mad and caused him to fire a bulletthrough his brain. Goodness knows what his difficulties could have been!But there can be little doubt that he had a disordered mind, and that ifthis little puzzle had not caused him to lose his mental balance someother more or less trivial thing would in time have done so. There is nomoral in the story, unless it be that of the Irish maxim, which appliesto every occupation of life as much as to the solving of puzzles: "Takethings aisy; if you can't take them aisy, take them as aisy as you can."And it is a bad and empirical way of solving any puzzle--by blowing yourbrains out.

Now, how many different routes are there from A to B in this maze if wemust never in any route go along the same passage twice? The four openspaces where four passages end are not reckoned as "passages." In thediagram (Fig. 22) it will be seen that I have again suppressed the blindalleys. It will be found that, in any case, we must go from A to C, andalso from F to B. But when we have arrived at C there are three ways,marked 1, 2, 3, of getting to D. Similarly, when we get to E there arethree ways, marked 4, 5, 6, of getting to F. We have also the dottedroute from C to E, the other dotted route from D to F, and the passagefrom D to E, indicated by stars. We can, therefore, express the positionof affairs by the little diagram annexed (Fig. 23). Here everycondition of route exactly corresponds to that in the circular maze,only it is much less confusing to the eye. Now, the number of routes,under the conditions, from A to B on this simplified diagram is 640, andthat is the required answer to the maze puzzle.

Finally, I will leave two easy maze puzzles (Figs. 24, 25) for myreaders to solve for themselves. The puzzle in each case is to find theshortest possible route to the centre. Everybody knows the story of FairRosamund and the Woodstock maze. What the maze was like or whether itever existed except in imagination is not known, many writers believingthat it was simply a badly-constructed house with a large number ofconfusing rooms and passages. At any rate, my sketch lacks the authorityof the other mazes in this article. My "Rosamund's Bower" is simplydesigned to show that where you have the plan before you it oftenhappens that the easiest way to find a route into a maze is by workingbackwards and first finding a way out.

THE PARADOX PARTY.

"Is not life itself a paradox?" C.L. DODGSON, _Pillow Problems_.

"It is a wonderful age!" said Mr. Allgood, and everybody at the tableturned towards him and assumed an attitude of expectancy.

This was an ordinary Christmas dinner of the Allgood family, with asprinkling of local friends. Nobody would have supposed that the aboveremark would lead, as it did, to a succession of curious puzzles andparadoxes, to which every member of the party contributed something ofinterest. The little symposium was quite unpremeditated, so we must notbe too critical respecting a few of the posers that were forthcoming.The varied character of the contributions is just what we would expecton such an occasion, for it was a gathering not of expert mathematiciansand logicians, but of quite ordinary folk.

"It is a wonderful age!" repeated Mr. Allgood. "A man has just designeda square house in such a cunning manner that all the windows on the foursides have a south aspect."

"That would appeal to me," said Mrs. Allgood, "for I cannot endure aroom with a north aspect."

"I cannot conceive how it is done," Uncle John confessed. "I suppose heputs bay windows on the east and west sides; but how on earth can becontrive to look south from the north side? Does he use mirrors, orsomething of that kind?"

"No," replied Mr. Allgood, "nothing of the sort. All the windows areflush with the walls, and yet you get a southerly prospect from everyone of them. You see, there is no real difficulty in designing the houseif you select the proper spot for its erection. Now, this house isdesigned for a gentleman who proposes to build it exactly at the NorthPole. If you think a moment you will realize that when you stand at theNorth Pole it is impossible, no matter which way you may turn, to lookelsewhere than due south! There are no such directions as north, east,or west when you are exactly at the North Pole. Everything is duesouth!"

"I am afraid, mother," said her son George, after the laughter hadsubsided, "that, however much you might like the aspect, the situationwould be a little too bracing for you."

"Ah, well!" she replied. "Your Uncle John fell also into the trap. I amno good at catches and puzzles. I suppose I haven't the right sort ofbrain. Perhaps some one will explain this to me. Only last week Iremarked to my hairdresser that it had been said that there are morepersons in the world than any one of them has hairs on his head. Hereplied, 'Then it follows, madam, that two persons, at least, must haveexactly the same number of hairs on their heads.' If this is a fact, Iconfess I cannot see it."

"How do the bald-headed affect the question?" asked Uncle John.

"If there are such persons in existence," replied Mrs. Allgood, "whohaven't a solitary hair on their heads discoverable under amagnifying-glass, we will leave them out of the question. Still, Idon't see how you are to prove that at least two persons have exactlythe same number to a hair."

"I think I can make it clear," said Mr. Filkins, who had dropped in forthe evening. "Assume the population of the world to be only one million.Any number will do as well as another. Then your statement was to theeffect that no person has more than nine hundred and ninety-ninethousand nine hundred and ninety-nine hairs on his head. Is that so?"

"Let me think," said Mrs. Allgood. "Yes--yes--that is correct."

"Very well, then. As there are only nine hundred and ninety-ninethousand nine hundred and ninety-nine _different_ ways of bearing hair,it is clear that the millionth person must repeat one of those ways. Doyou see?"

"Yes; I see that--at least I think I see it."

"Therefore two persons at least must have the same number of hairs ontheir heads; and as the number of people on the earth so greatly exceedsthe number of hairs on any one person's head, there must, of course, bean immense number of these repetitions."

"But, Mr. Filkins," said little Willie Allgood, "why could not themillionth man have, say, ten thousand hairs and a half?"

"That is mere hair-splitting, Willie, and does not come into thequestion."

"Here is a curious paradox," said George. "If a thousand soldiers aredrawn up in battle array on a plane"--they understood him to mean"plain"--"only one man will stand upright."

Nobody could see why. But George explained that, according to Euclid, aplane can touch a sphere only at one point, and that person only whostands at that point, with respect to the centre of the earth, willstand upright.

"In the same way," he remarked, "if a billiard-table were quitelevel--that is, a perfect plane--the balls ought to roll to the centre."

Though he tried to explain this by placing a visiting-card on an orangeand expounding the law of gravitation, Mrs. Allgood declined to acceptthe statement. She could not see that the top of a true billiard-tablemust, theoretically, be spherical, just like a portion of theorange-peel that George cut out. Of course, the table is so small inproportion to the surface of the earth that the curvature is notappreciable, but it is nevertheless true in theory. A surface that wecall level is not the same as our idea of a true geometrical plane.

"Uncle John," broke in Willie Allgood, "there is a certain islandsituated between England and France, and yet that island is farther fromFrance than England is. What is the island?"

"That seems absurd, my boy; because if I place this tumbler, torepresent the island, between these two plates, it seems impossible thatthe tumbler can be farther from either of the plates than they are fromeach other."

"But isn't Guernsey between England and France?" asked Willie.

"Yes, certainly."

"Well, then, I think you will find, uncle, that Guernsey is abouttwenty-six miles from France, and England is only twenty-one miles fromFrance, between Calais and Dover."

"My mathematical master," said George, "has been trying to induce me toaccept the axiom that 'if equals be multiplied by equals the productsare equal.'"

"It is self-evident," pointed out Mr. Filkins. "For example, if 3 feetequal 1 yard, then twice 3 feet will equal 2 yards. Do you see?"

"But, Mr. Filkins," asked George, "is this tumbler half full of waterequal to a similar glass half empty?"

"Certainly, George."

"Then it follows from the axiom that a glass full must equal a glassempty. Is that correct?"

"No, clearly not. I never thought of it in that light."

"Perhaps," suggested Mr. Allgood, "the rule does not apply to liquids."

"Just what I was thinking, Allgood. It would seem that we must make anexception in the case of liquids."

"But it would be awkward," said George, with a smile, "if we also had toexcept the case of solids. For instance, let us take the solid earth.One mile square equals one square mile. Therefore two miles square mustequal two square miles. Is this so?"

"Well, let me see! No, of course not," Mr. Filkins replied, "because twomiles square is four square miles."

"Then," said George, "if the axiom is not true in these cases, when isit true?"

Mr. Filkins promised to look into the matter, and perhaps the readerwill also like to give it consideration at leisure.

"Look here, George," said his cousin Reginald Woolley: "by whatfractional part does four-fourths exceed three-fourths?"

"By one-fourth!" shouted everybody at once.

"Try another one," George suggested.

"With pleasure, when you have answered that one correctly," wasReginald's reply.

"Do you mean to say that it isn't one-fourth?"

"Certainly I do."

Several members of the company failed to see that the correct answer is"one-third," although Reginald tried to explain that three of anything,if increased by one-third, becomes four.

"Uncle John, how do you pronounce 't-o-o'?" asked Willie.

"'Too," my boy."

"And how do you pronounce 't-w-o'?"

"That is also 'too.'"

"Then how do you pronounce the second day of the week?"

"Well, that I should pronounce 'Tuesday,' not 'Toosday.'"

"Would you really? I should pronounce it 'Monday.'"

"If you go on like this, Willie," said Uncle John, with mock severity,"you will soon be without a friend in the world."

"Can any of you write down quickly in figures 'twelve thousand twelvehundred and twelve pounds'?" asked Mr. Allgood.

His eldest daughter, Miss Mildred, was the only person who happened tohave a pencil at hand.

"It can't be done," she declared, after making an attempt on the whitetable-cloth; but Mr. Allgood showed her that it should be written,"L13,212."

"Now it is my turn," said Mildred. "I have been waiting to ask you all aquestion. In the Massacre of the Innocents under Herod, a number of poorlittle children were buried in the sand with only their feet stickingout. How might you distinguish the boys from the girls?"

"I suppose," said Mrs. Allgood, "it is a conundrum--something to do withtheir poor little 'souls.'"

But after everybody had given it up, Mildred reminded the company thatonly boys were put to death.

"Once upon a time," began George, "Achilles had a race with atortoise--"

"Stop, George!" interposed Mr. Allgood. "We won't have that one. I knewtwo men in my youth who were once the best of friends, but theyquarrelled over that infernal thing of Zeno's, and they never spoke toone another again for the rest of their lives. I draw the line at that,and the other stupid thing by Zeno about the flying arrow. I don'tbelieve anybody understands them, because I could never do so myself."

"Oh, very well, then, father. Here is another. The Post-Office peoplewere about to erect a line of telegraph-posts over a high hill fromTurmitville to Wurzleton; but as it was found that a railway company wasmaking a deep level cutting in the same direction, they arranged to putup the posts beside the line. Now, the posts were to be a hundred yardsapart, the length of the road over the hill being five miles, and thelength of the level cutting only four and a half miles. How many postsdid they save by erecting them on the level?"

"That is a very simple matter of calculation," said Mr. Filkins. "Findhow many times one hundred yards will go in five miles, and how manytimes in four and a half miles. Then deduct one from the other, and youhave the number of posts saved by the shorter route."

"Quite right," confirmed Mr. Allgood. "Nothing could be easier."

"That is just what the Post-Office people said," replied George, "but itis quite wrong. If you look at this sketch that I have just made, youwill see that there is no difference whatever. If the posts are ahundred yards apart, just the same number will be required on the levelas over the surface of the hill."

[Illustration]

"Surely you must be wrong, George," said Mrs. Allgood, "for if the postsare a hundred yards apart and it is half a mile farther over the hill,you have to put up posts on that extra half-mile."

"Look at the diagram, mother. You will see that the distance from postto post is not the distance from base to base measured along the ground.I am just the same distance from you if I stand on this spot on thecarpet or stand immediately above it on the chair."

But Mrs. Allgood was not convinced.

Mr. Smoothly, the curate, at the end of the table, said at this pointthat he had a little question to ask.

"Suppose the earth were a perfect sphere with a smooth surface, and agirdle of steel were placed round the Equator so that it touched atevery point."

"'I'll put a girdle round about the earth in forty minutes,'" mutteredGeorge, quoting the words of Puck in _A Midsummer Night's Dream_.

"Now, if six yards were added to the length of the girdle, what wouldthen be the distance between the girdle and the earth, supposing thatdistance to be equal all round?"

"In such a great length," said Mr. Allgood, "I do not suppose thedistance would be worth mentioning."

"What do you say, George?" asked Mr. Smoothly.

"Well, without calculating I should imagine it would be a very minutefraction of an inch."

Reginald and Mr. Filkins were of the same opinion.

"I think it will surprise you all," said the curate, "to learn thatthose extra six yards would make the distance from the earth all roundthe girdle very nearly a yard!"

"Very nearly a yard!" everybody exclaimed, with astonishment; but Mr.Smoothly was quite correct. The increase is independent of the originallength of the girdle, which may be round the earth or round an orange;in any case the additional six yards will give a distance of nearly ayard all round. This is apt to surprise the non-mathematical mind.

"Did you hear the story of the extraordinary precocity of Mrs. Perkins'sbaby that died last week?" asked Mrs. Allgood. "It was only three monthsold, and lying at the point of death, when the grief-stricken motherasked the doctor if nothing could save it. 'Absolutely nothing!' saidthe doctor. Then the infant looked up pitifully into its mother's faceand said--absolutely nothing!"

"Impossible!" insisted Mildred. "And only three months old!"

"There have been extraordinary cases of infantile precocity," said Mr.Filkins, "the truth of which has often been carefully attested. But areyou sure this really happened, Mrs. Allgood?"

"Positive," replied the lady. "But do you really think it astonishingthat a child of three months should say absolutely nothing? What wouldyou expect it to say?"

"Speaking of death," said Mr. Smoothly, solemnly, "I knew two men,father and son, who died in the same battle during the South AfricanWar. They were both named Andrew Johnson and buried side by side, butthere was some difficulty in distinguishing them on the headstones. Whatwould you have done?"

"Quite simple," said Mr. Allgood. "They should have described one as'Andrew Johnson, Senior,' and the other as 'Andrew Johnson, Junior.'"

"But I forgot to tell you that the father died first."

"What difference can that make?"

"Well, you see, they wanted to be absolutely exact, and that was thedifficulty."

"But I don't see any difficulty," said Mr. Allgood, nor could anybodyelse.

"Well," explained Mr. Smoothly, "it is like this. If the father diedfirst, the son was then no longer 'Junior.' Is that so?"

"To be strictly exact, yes."

"That is just what they wanted--to be strictly exact. Now, if he was nolonger 'Junior,' then he did not die 'Junior." Consequently it must beincorrect so to describe him on the headstone. Do you see the point?"

"Here is a rather curious thing," said Mr. Filkins, "that I have justremembered. A man wrote to me the other day that he had recentlydiscovered two old coins while digging in his garden. One was dated '51B.C.,' and the other one marked 'George I.' How do I know that he wasnot writing the truth?"

"Perhaps you know the man to be addicted to lying," said Reginald.

"But that would be no proof that he was not telling the truth in thisinstance."

"Perhaps," suggested Mildred, "you know that there were no coins made atthose dates.

"On the contrary, they were made at both periods."

"Were they silver or copper coins?" asked Willie.

"My friend did not state, and I really cannot see, Willie, that it makesany difference."

"I see it!" shouted Reginald. "The letters 'B.C.' would never be used ona coin made before the birth of Christ. They never anticipated the eventin that way. The letters were only adopted later to denote datesprevious to those which we call 'A.D.' That is very good; but I cannotsee why the other statement could not be correct."

"Reginald is quite right," said Mr. Filkins, "about the first coin. Thesecond one could not exist, because the first George would never bedescribed in his lifetime as 'George I.'"

"Why not?" asked Mrs. Allgood. "He _was_ George I."

"Yes; but they would not know it until there was a George II."

"Then there was no George II. until George III. came to the throne?"

"That does not follow. The second George becomes 'George II.' on accountof there having been a 'George I.'"

"Then the first George was 'George I.' on account of there having beenno king of that name before him."

"Don't you see, mother," said George Allgood, "we did not call QueenVictoria 'Victoria I.;' but if there is ever a 'Victoria II.,' then shewill be known that way."

"But there _have_ been several Georges, and therefore he was 'George I.'There _haven't_ been several Victorias, so the two cases are notsimilar."

They gave up the attempt to convince Mrs. Allgood, but the reader will,of course, see the point clearly.

"Here is a question," said Mildred Allgood, "that I should like some ofyou to settle for me. I am accustomed to buy from our greengrocerbundles of asparagus, each 12 inches in circumference. I always put atape measure round them to make sure I am getting the full quantity. Theother day the man had no large bundles in stock, but handed me insteadtwo small ones, each 6 inches in circumference. 'That is the samething,' I said, 'and, of course, the price will be the same;' but heinsisted that the two bundles together contained more than the largeone, and charged me a few pence extra. Now, what I want to know is,which of us was correct? Would the two small bundles contain the samequantity as the large one? Or would they contain more?"

"That is the ancient puzzle," said Reginald, laughing, "of the sack ofcorn that Sempronius borrowed from Caius, which your greengrocer,perhaps, had been reading about somewhere. He caught you beautifully."

"Then they were equal?"

"On the contrary, you were both wrong, and you were badly cheated. Youonly got half the quantity that would have been contained in a largebundle, and therefore ought to have been charged half the originalprice, instead of more."

Yes, it was a bad swindle, undoubtedly. A circle with a circumferencehalf that of another must have its area a quarter that of the other.Therefore the two small bundles contained together only half as muchasparagus as a large one.

"Mr. Filkins, can you answer this?" asked Willie. "There is a man in thenext village who eats two eggs for breakfast every morning."

"Nothing very extraordinary in that," George broke in. "If you told usthat the two eggs ate the man it would be interesting."

"Don't interrupt the boy, George," said his mother.

"Well," Willie continued, "this man neither buys, borrows, barters,begs, steals, nor finds the eggs. He doesn't keep hens, and the eggs arenot given to him. How does he get the eggs?"

"Does he take them in exchange for something else?" asked Mildred.

"That would be bartering them," Willie replied.

"Perhaps some friend sends them to him," suggested Mrs. Allgood.

"I said that they were not given to him."

"I know," said George, with confidence. "A strange hen comes into hisplace and lays them."

"But that would be finding them, wouldn't it?"

"Does he hire them?" asked Reginald.

"If so, he could not return them after they were eaten, so that would bestealing them."

"Perhaps it is a pun on the word 'lay,'" Mr. Filkins said. "Does he laythem on the table?"

"He would have to get them first, wouldn't he? The question was, Howdoes he get them?"

"Give it up!" said everybody. Then little Willie crept round to theprotection of his mother, for George was apt to be rough on suchoccasions.

"The man keeps ducks!" he cried, "and his servant collects the eggsevery morning."

"But you said he doesn't keep birds!" George protested.

"I didn't, did I, Mr. Filkins? I said he doesn't keep hens."

"But he finds them," said Reginald.

"No; I said his servant finds them."

"Well, then," Mildred interposed, "his servant gives them to him."

"You cannot give a man his own property, can you?"

All agreed that Willie's answer was quite satisfactory. Then Uncle Johnproduced a little fallacy that "brought the proceedings to a close," asthe newspapers say.

413.--A CHESSBOARD FALLACY.

[Illustration]

"Here is a diagram of a chessboard," he said. "You see there aresixty-four squares--eight by eight. Now I draw a straight line from thetop left-hand corner, where the first and second squares meet, to thebottom right-hand corner. I cut along this line with the scissors, slideup the piece that I have marked B, and then clip off the little corner Cby a cut along the first upright line. This little piece will exactlyfit into its place at the top, and we now have an oblong with sevensquares on one side and nine squares on the other. There are, therefore,now only sixty-three squares, because seven multiplied by nine makessixty-three. Where on earth does that lost square go to? I have triedover and over again to catch the little beggar, but he always eludes me.For the life of me I cannot discover where he hides himself."

"It seems to be like the other old chessboard fallacy, and perhaps theexplanation is the same," said Reginald--"that the pieces do not exactlyfit."

"But they _do_ fit," said Uncle John. "Try it, and you will see."

Later in the evening Reginald and George, were seen in a corner withtheir heads together, trying to catch that elusive little square, and itis only fair to record that before they retired for the night theysucceeded in securing their prey, though some others of the companyfailed to see it when captured. Can the reader solve the little mystery?

UNCLASSIFIED PROBLEMS.

"A snapper up of unconsidered trifles." _Winter's Tale_, iv. 2.

414.--WHO WAS FIRST?

Anderson, Biggs, and Carpenter were staying together at a place by theseaside. One day they went out in a boat and were a mile at sea when arifle was fired on shore in their direction. Why or by whom the shot wasfired fortunately does not concern us, as no information on these pointsis obtainable, but from the facts I picked up we can get material for acurious little puzzle for the novice.

It seems that Anderson only heard the report of the gun, Biggs only sawthe smoke, and Carpenter merely saw the bullet strike the water nearthem. Now, the question arises: Which of them first knew of thedischarge of the rifle?

415.--A WONDERFUL VILLAGE.

There is a certain village in Japan, situated in a very low valley, andyet the sun is nearer to the inhabitants every noon, by 3,000 miles andupwards, than when he either rises or sets to these people. In what partof the country is the village situated?

416.--A CALENDAR PUZZLE.

If the end of the world should come on the first day of a new century,can you say what are the chances that it will happen on a Sunday?

417.--THE TIRING IRONS.

[Illustration]

The illustration represents one of the most ancient of all mechanicalpuzzles. Its origin is unknown. Cardan, the mathematician, wrote aboutit in 1550, and Wallis in 1693; while it is said still to be found inobscure English villages (sometimes deposited in strange places, such asa church belfry), made of iron, and appropriately called "tiring-irons,"and to be used by the Norwegians to-day as a lock for boxes and bags. Inthe toyshops it is sometimes called the "Chinese rings," though thereseems to be no authority for the description, and it more frequentlygoes by the unsatisfactory name of "the puzzling rings." The French callit "Baguenaudier."

The puzzle will be seen to consist of a simple _loop_ of wire fixed in ahandle to be held in the left hand, and a certain number of _rings_secured by _wires_ which pass through holes in the _bar_ and are keptthere by their blunted ends. The wires work freely in the bar, butcannot come apart from it, nor can the wires be removed from the rings.The general puzzle is to detach the loop completely from all the rings,and then to put them all on again.

Now, it will be seen at a glance that the first ring (to the right) canbe taken off at any time by sliding it over the end and dropping itthrough the loop; or it may be put on by reversing the operation. Withthis exception, the only ring that can ever be removed is the one thathappens to be a contiguous second on the loop at the right-hand end.Thus, with all the rings on, the second can be dropped at once; with thefirst ring down, you cannot drop the second, but may remove the third;with the first three rings down, you cannot drop the fourth, but mayremove the fifth; and so on. It will be found that the first and secondrings can be dropped together or put on together; but to preventconfusion we will throughout disallow this exceptional double move, andsay that only one ring may be put on or removed at a time.

We can thus take off one ring in 1 move; two rings in 2 moves; threerings in 5 moves; four rings in 10 moves; five rings in 21 moves; and ifwe keep on doubling (and adding one where the number of rings is odd) wemay easily ascertain the number of moves for completely removing anynumber of rings. To get off all the seven rings requires 85 moves. Letus look at the five moves made in removing the first three rings, thecircles above the line standing for rings on the loop and those underfor rings off the loop.

Drop the first ring; drop the third; put up the first; drop the second;and drop the first--5 moves, as shown clearly in the diagrams. The darkcircles show at each stage, from the starting position to the finish,which rings it is possible to drop. After move 2 it will be noticed thatno ring can be dropped until one has been put on, because the first andsecond rings from the right now on the loop are not together. After thefifth move, if we wish to remove all seven rings we must now drop thefifth. But before we can then remove the fourth it is necessary to puton the first three and remove the first two. We shall then have 7, 6, 4,3 on the loop, and may therefore drop the fourth. When we have put on 2and 1 and removed 3, 2, 1, we may drop the seventh ring. The nextoperation then will be to get 6, 5, 4, 3, 2, 1 on the loop and remove 4,3, 2, 1, when 6 will come off; then get 5, 4, 3, 2, 1 on the loop, andremove 3, 2, 1, when 5 will come off; then get 4, 3, 2, 1 on the loopand remove 2, 1, when 4 will come off; then get 3, 2, 1 on the loop andremove 1, when 3 will come off; then get 2, 1 on the loop, when 2 willcome off; and 1 will fall through on the 85th move, leaving the loopquite free. The reader should now be able to understand the puzzle,whether or not he has it in his hand in a practical form.

[Illustration]

[Illustration:

o o o o o * * {-------------

o o o o * o 1{------------- o

o o o o o 2{------------- o o

o o o o * * 3{------------- o

o o o o * 4{------------- o o

o o * o 5{------------- o o o

]

The particular problem I propose is simply this. Suppose there arealtogether fourteen rings on the tiring-irons, and we proceed to takethem all off in the correct way so as not to waste any moves. What willbe the position of the rings after the 9,999th move has been made?

418.--SUCH A GETTING UPSTAIRS.

In a suburban villa there is a small staircase with eight steps, notcounting the landing. The little puzzle with which Tommy Smart perplexedhis family is this. You are required to start from the bottom and landtwice on the floor above (stopping there at the finish), having returnedonce to the ground floor. But you must be careful to use every tread thesame number of times. In how few steps can you make the ascent? It seemsa very simple matter, but it is more than likely that at your firstattempt you will make a great many more steps than are necessary. Ofcourse you must not go more than one riser at a time.

Tommy knows the trick, and has shown it to his father, who professes tohave a contempt for such things; but when the children are in bed thepater will often take friends out into the hall and enjoy a good laughat their bewilderment. And yet it is all so very simple when you knowhow it is done.

419.--THE FIVE PENNIES.

Here is a really hard puzzle, and yet its conditions are so absurdlysimple. Every reader knows how to place four pennies so that they areequidistant from each other. All you have to do is to arrange three ofthem flat on the table so that they touch one another in the form of atriangle, and lay the fourth penny on top in the centre. Then, as everypenny touches every other penny, they are all at equal distances fromone another. Now try to do the same thing with five pennies--place themso that every penny shall touch every other penny--and you will find ita different matter altogether.

420.--THE INDUSTRIOUS BOOKWORM.

[Illustration]

Our friend Professor Rackbrane is seen in the illustration to bepropounding another of his little posers. He is explaining that since helast had occasion to take down those three volumes of a learned bookfrom their place on his shelves a bookworm has actually bored a holestraight through from the first page to the last. He says that theleaves are together three inches thick in each volume, and that everycover is exactly one-eighth of an inch thick, and he asks how long atunnel had the industrious worm to bore in preparing his new tuberailway. Can you tell him?

421.--A CHAIN PUZZLE.

[Illustration]

This is a puzzle based on a pretty little idea first dealt with by thelate Mr. Sam Loyd. A man had nine pieces of chain, as shown in theillustration. He wanted to join these fifty links into one endlesschain. It will cost a penny to open any link and twopence to weld a linktogether again, but he could buy a new endless chain of the samecharacter and quality for 2s. 2d. What was the cheapest course for himto adopt? Unless the reader is cunning he may find himself a good wayout in his answer.

422.--THE SABBATH PUZZLE.

I have come across the following little poser in an old book. I wonderhow many readers will see the author's intended solution to the riddle.

Christians the week's _first_ day for Sabbath hold; The Jews the _seventh_, as they did of old; The Turks the _sixth_, as we have oft been told. How can these three, in the same place and day, Have each his own true Sabbath? tell, I pray.

423.--THE RUBY BROOCH.

The annals of Scotland Yard contain some remarkable cases of jewelrobberies, but one of the most perplexing was the theft of LadyLittlewood's rubies. There have, of course, been many greater robberiesin point of value, but few so artfully conceived. Lady Littlewood, ofRomley Manor, had a beautiful but rather eccentric heirloom in the formof a ruby brooch. While staying at her town house early in the eightiesshe took the jewel to a shop in Brompton for some slight repairs.

"A fine collection of rubies, madam," said the shopkeeper, to whom herladyship was a stranger.

"Yes," she replied; "but curiously enough I have never actually countedthem. My mother once pointed out to me that if you start from the centreand count up one line, along the outside and down the next line, thereare always eight rubies. So I should always know if a stone weremissing."

[Illustration]

Six months later a brother of Lady Littlewood's, who had returned fromhis regiment in India, noticed that his sister was wearing the rubybrooch one night at a county ball, and on their return home asked tolook at it more closely. He immediately detected the fact that four ofthe stones were gone.

"How can that possibly be?" said Lady Littlewood. "If you count up oneline from the centre, along the edge, and down the next line, in anydirection, there are always eight stones. This was always so and is sonow. How, therefore, would it be possible to remove a stone without mydetecting it?"

"Nothing could be simpler," replied the brother. "I know the broochwell. It originally contained forty-five stones, and there are now onlyforty-one. Somebody has stolen four rubies, and then reset as small anumber of the others as possible in such a way that there shall alwaysbe eight in any of the directions you have mentioned."

There was not the slightest doubt that the Brompton jeweller was thethief, and the matter was placed in the hands of the police. But the manwas wanted for other robberies, and had left the neighbourhood some timebefore. To this day he has never been found.

The interesting little point that at first baffled the police, and whichforms the subject of our puzzle, is this: How were the forty-five rubiesoriginally arranged on the brooch? The illustration shows exactly howthe forty-one were arranged after it came back from the jeweller; butalthough they count eight correctly in any of the directions mentioned,there are four stones missing.

424.--THE DOVETAILED BLOCK.

[Illustration]

Here is a curious mechanical puzzle that was given to me some years ago,but I cannot say who first invented it. It consists of two solid blocksof wood securely dovetailed together. On the other two vertical sidesthat are not visible the appearance is precisely the same as on thoseshown. How were the pieces put together? When I published this littlepuzzle in a London newspaper I received (though they were unsolicited)quite a stack of models, in oak, in teak, in mahogany, rosewood,satinwood, elm, and deal; some half a foot in length, and others varyingin size right down to a delicate little model about half an inch square.It seemed to create considerable interest.

425.--JACK AND THE BEANSTALK.

[Illustration]

The illustration, by a British artist, is a sketch of Jack climbing thebeanstalk. Now, the artist has made a serious blunder in this drawing.Can you find out what it is?

426.--THE HYMN-BOARD POSER.

The worthy vicar of Chumpley St. Winifred is in great distress. A littlechurch difficulty has arisen that all the combined intelligence of theparish seems unable to surmount. What this difficulty is I will statehereafter, but it may add to the interest of the problem if I first givea short account of the curious position that has been brought about. Itall has to do with the church hymn-boards, the plates of which havebecome so damaged that they have ceased to fulfil the purpose for whichthey were devised. A generous parishioner has promised to pay for a newset of plates at a certain rate of cost; but strange as it may seem, noagreement can be come to as to what that cost should be. The proposedmaker of the plates has named a price which the donor declares to beabsurd. The good vicar thinks they are both wrong, so he asks theschoolmaster to work out the little sum. But this individual declaresthat he can find no rule bearing on the subject in any of his arithmeticbooks. An application having been made to the local medicalpractitioner, as a man of more than average intellect at Chumpley, hehas assured the vicar that his practice is so heavy that he has not hadtime even to look at it, though his assistant whispers that the doctorhas been sitting up unusually late for several nights past. Widow Wilsonhas a smart son, who is reputed to have once won a prize forpuzzle-solving. He asserts that as he cannot find any solution to theproblem it must have something to do with the squaring of the circle,the duplication of the cube, or the trisection of an angle; at any rate,he has never before seen a puzzle on the principle, and he gives it up.

[Illustration]

This was the state of affairs when the assistant curate (who, I shouldsay, had frankly confessed from the first that a profound study oftheology had knocked out of his head all the knowledge of mathematics heever possessed) kindly sent me the puzzle.

A church has three hymn-boards, each to indicate the numbers of fivedifferent hymns to be sung at a service. All the boards are in use atthe same service. The hymn-book contains 700 hymns. A new set of numbersis required, and a kind parishioner offers to present a set painted onmetal plates, but stipulates that only the smallest number of platesnecessary shall be purchased. The cost of each plate is to be 6d., andfor the painting of each plate the charges are to be: For one plate,1s.; for two plates alike, 113/4d. each; for three plates alike,111/2d. each, and so on, the charge being one farthing less per platefor each similarly painted plate. Now, what should be the lowest cost?

Readers will note that they are required to use every legitimate andpractical method of economy. The illustration will make clear the natureof the three hymn-boards and plates. The five hymns are here indicatedby means of twelve plates. These plates slide in separately at the back,and in the illustration there is room, of course, for three more plates.

427.--PHEASANT-SHOOTING.

A Cockney friend, who is very apt to draw the long bow, and is evidentlyless of a sportsman than he pretends to be, relates to me the followingnot very credible yarn:--

"I've just been pheasant-shooting with my friend the duke. We hadsplendid sport, and I made some wonderful shots. What do you think ofthis, for instance? Perhaps you can twist it into a puzzle. The duke andI were crossing a field when suddenly twenty-four pheasants rose on thewing right in front of us. I fired, and two-thirds of them dropped deadat my feet. Then the duke had a shot at what were left, and brought downthree-twenty-fourths of them, wounded in the wing. Now, out of thosetwenty-four birds, how many still remained?"

It seems a simple enough question, but can the reader give a correctanswer?

428.--THE GARDENER AND THE COOK.

A correspondent, signing himself "Simple Simon," suggested that I shouldgive a special catch puzzle in the issue of _The Weekly Dispatch_ forAll Fools' Day, 1900. So I gave the following, and it causedconsiderable amusement; for out of a very large body of competitors,many quite expert, not a single person solved it, though it ran fornearly a month.

[Illustration]

"The illustration is a fancy sketch of my correspondent, 'Simple Simon,'in the act of trying to solve the following innocent little arithmeticalpuzzle. A race between a man and a woman that I happened to witness oneAll Fools' Day has fixed itself indelibly on my memory. It happened at acountry-house, where the gardener and the cook decided to run a race toa point 100 feet straight away and return. I found that the gardener ran3 feet at every bound and the cook only 2 feet, but then she made threebounds to his two. Now, what was the result of the race?"

A fortnight after publication I added the following note: "It has beensuggested that perhaps there is a catch in the 'return,' but there isnot. The race is to a point 100 feet away and home again--that is, adistance of 200 feet. One correspondent asks whether they take exactlythe same time in turning, to which I reply that they do. Another seemsto suspect that it is really a conundrum, and that the answer is that'the result of the race was a (matrimonial) tie.' But I had no suchintention. The puzzle is an arithmetical one, as it purports to be."

429.--PLACING HALFPENNIES.

[Illustration]

Here is an interesting little puzzle suggested to me by Mr. W. T. Whyte.Mark off on a sheet of paper a rectangular space 5 inches by 3 inches,and then find the greatest number of halfpennies that can be placedwithin the enclosure under the following conditions. A halfpenny isexactly an inch in diameter. Place your first halfpenny where you like,then place your second coin at exactly the distance of an inch from thefirst, the third an inch distance from the second, and so on. Nohalfpenny may touch another halfpenny or cross the boundary. Ourillustration will make the matter perfectly clear. No. 2 coin is an inchfrom No. 1; No. 3 an inch from No. 2; No. 4 an inch from No. 3; butafter No. 10 is placed we can go no further in this attempt. Yet severalmore halfpennies might have been got in. How many can the reader place?

430.--FIND THE MAN'S WIFE.

[Illustration]

One summer day in 1903 I was loitering on the Brighton front, watchingthe people strolling about on the beach, when the friend who was with mesuddenly drew my attention to an individual who was standing alone, andsaid, "Can you point out that man's wife? They are stopping at the samehotel as I am, and the lady is one of those in view." After a fewminutes' observation, I was successful in indicating the lady correctly.My friend was curious to know by what method of reasoning I had arrivedat the result. This was my answer:--

"We may at once exclude that Sister of Mercy and the girl in the shortfrock; also the woman selling oranges. It cannot be the lady in widows'weeds. It is not the lady in the bath chair, because she is not stayingat your hotel, for I happened to see her come out of a private housethis morning assisted by her maid. The two ladies in red breakfasted atmy hotel this morning, and as they were not wearing outdoor dress Iconclude they are staying there. It therefore rests between the lady inblue and the one with the green parasol. But the left hand that holdsthe parasol is, you see, ungloved and bears no wedding-ring.Consequently I am driven to the conclusion that the lady in blue is theman's wife--and you say this is correct."

Now, as my friend was an artist, and as I thought an amusing puzzlemight be devised on the lines of his question, I asked him to make me adrawing according to some directions that I gave him, and I havepleasure in presenting his production to my readers. It will be seenthat the picture shows six men and six ladies: Nos. 1, 3, 5, 7, 9, and11 are ladies, and Nos. 2, 4, 6, 8, 10, and 12 are men. These twelveindividuals represent six married couples, all strangers to one another,who, in walking aimlessly about, have got mixed up. But we are onlyconcerned with the man that is wearing a straw hat--Number 10. Thepuzzle is to find this man's wife. Examine the six ladies carefully, andsee if you can determine which one of them it is.

I showed the picture at the time to a few friends, and they expressedvery different opinions on the matter. One said, "I don't believe hewould marry a girl like Number 7." Another said, "I am sure a nice girllike Number 3 would not marry such a fellow!" Another said, "It must beNumber 1, because she has got as far away as possible from the brute!"It was suggested, again, that it must be Number 11, because "he seems tobe looking towards her;" but a cynic retorted, "For that very reason, ifhe is really looking at her, I should say that she is not his wife!"

I now leave the question in the hands of my readers. Which is reallyNumber 10's wife?

The illustration is of necessity considerably reduced from the largescale on which it originally appeared in _The Weekly Dispatch_ (24th May1903), but it is hoped that the details will be sufficiently clear toallow the reader to derive entertainment from its examination. In anycase the solution given will enable him to follow the points withinterest.

SOLUTIONS.

1.--A POST-OFFICE PERPLEXITY.

The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8twopence-halfpenny stamps, which delivery exactly fulfils the conditionsand represents a cost of five shillings.

2.--YOUTHFUL PRECOCITY.

The price of the banana must have been one penny farthing. Thus, 960bananas would cost L5, and 480 sixpences would buy 2,304 bananas.

3.--AT A CATTLE MARKET.

Jakes must have taken 7 animals to market, Hodge must have taken 11, andDurrant must have taken 21. There were thus 39 animals altogether.

4.--THE BEANFEAST PUZZLE.

The cobblers spent 35s., the tailors spent also 35s., t