68.—The Postage Stamps Puzzles.

The first of these puzzles is based on a similar principle, though it is really much easier, because the condition that nine of the stamps must be of different values makes their selection a simple matter, though how they are to be placed requires a little thought or trial until one knows the rule respecting putting the fractions in the corners. I give the solution.

[4-1/2 d] [1 d] [3 d]

[2 d] [3 d] [4 d]

[2-1/2 d] [5 d] [1-1/2 d] ]

[1/2 d] [3 d] [1-1/2 d] [9 d]

[10 d] [6 d] [2 d]

[1 d] [1 s.] [5 d] ]

I also show the solution to the second stamp puzzle. All the columns, rows, and diagonals add up 1s. 6d. There is no stamp on one square, and the conditions did not forbid this omission. The stamps at present in circulation are these:—1/2d., 1d., 1-1/2d., 2d., 2-1/2d., 3d., 4d., 5d., 6d., 9d., 10d., 1s., 2s. 6d., 5s., 10s., L1, and L5. In the first solution the numbers are in arithmetical progression—1, 1-1/2, 2, 2-1/2, 3, 3-1/2, 4, 4-1/2, 5. But any nine numbers will form a magic square if we can write them thus:—

1 2 3 7 8 9 13 14 15

where the horizontal differences are all alike and the vertical differences all alike, but not necessarily the same as the horizontal. This happens in the case of the second solution, the numbers of which may be written:—

0 1 2 5 6 7 10 11 12

Also in the case of the solution to No. 67, the Coinage Puzzle, the numbers are, in shillings:—

2 2-1/2 3 4-1/2 5 5-1/2 7 7-1/2 8

If there are to be nine different numbers, 0 may occur once (as in the solution to No. 22). Yet one might construct squares with negative numbers, as follows:—

-2 -1 0 5 6 7 12 13 14



69.—The Frogs and Tumblers.

It is perfectly true, as the Professor said, that there is only one solution (not counting a reversal) to this puzzle. The frogs that jump are George in the third horizontal row; Chang, the artful-looking batrachian at the end of the fourth row; and Wilhelmina, the fair creature in the seventh row. George jumps downwards to the second tumbler in the seventh row; Chang, who can only leap short distances in consequence of chronic rheumatism, removes somewhat unwillingly to the glass just above him—the eighth in the third row; while Wilhelmina, with all the sprightliness of her youth and sex, performs the very creditable saltatory feat of leaping to the fourth tumbler in the fourth row. In their new positions, as shown in the accompanying diagram, it will be found that of the eight frogs no two are in line vertically, horizontally, or diagonally.



70.—Romeo and Juliet.

This is rather a difficult puzzle, though, as the Professor remarked when Hawkhurst hit on the solution, it is "just one of those puzzles that a person might solve at a glance" by pure luck. Yet when the solution, with its pretty, symmetrical arrangement, is seen, it looks ridiculously simple.

It will be found that Romeo reaches Juliet's balcony after visiting every house once and only once, and making fourteen turnings, not counting the turn he makes at starting. These are the fewest turnings possible, and the problem can only be solved by the route shown or its reversal.



71.—Romeo's Second Journey.



In order to take his trip through all the white squares only with the fewest possible turnings, Romeo would do well to adopt the route I have shown, by means of which only sixteen turnings are required to perform the feat. The Professor informs me that the Helix Aspersa, or common or garden snail, has a peculiar aversion to making turnings—so much so that one specimen with which he made experiments went off in a straight line one night and has never come back since.



72.—The Frogs who would a-wooing go.

This is one of those puzzles in which a plurality of solutions is practically unavoidable. There are two or three positions into which four frogs may jump so as to form five rows with four in each row, but the case I have given is the most satisfactory arrangement.



The frogs that have jumped have left their astral bodies behind, in order to show the reader the positions which they originally occupied. Chang, the frog in the middle of the upper row, suffering from rheumatism, as explained above in the Frogs and Tumblers solution, makes the shortest jump of all—a little distance between the two rows; George and Wilhelmina leap from the ends of the lower row to some distance N. by N.W. and N. by N.E. respectively; while the frog in the middle of the lower row, whose name the Professor forgot to state, goes direct S.



73.—The Game of Kayles.

To win at this game you must, sooner or later, leave your opponent an even number of similar groups. Then whatever he does in one group you repeat in a similar group. Suppose, for example, that you leave him these groups: o.o.ooo.ooo. Now, if he knocks down a single, you knock down a single; if he knocks down two in one triplet, you knock down two in the other triplet; if he knocks down the central kayle in a triplet, you knock down the central one in the other triplet. In this way you must eventually win. As the game is started with the arrangement o.ooooooooooo, the first player can always win, but only by knocking down the sixth or tenth kayle (counting the one already fallen as the second), and this leaves in either case o.ooo.ooooooo, as the order of the groups is of no importance. Whatever the second player now does, this can always be resolved into an even number of equal groups. Let us suppose that he knocks down the single one; then we play to leave him oo.ooooooo. Now, whatever he does we can afterwards leave him either ooo.ooo or o.oo.ooo. We know why the former wins, and the latter wins also; because, however he may play, we can always leave him either o.o, or o.o.o.o, or oo.oo, as the case may be. The complete analysis I can now leave for the amusement of the reader.



74.—The Broken Chessboard.

The illustration will show how the thirteen pieces can be put together so as to construct the perfect board, and the reverse problem of cutting these particular pieces out will be found equally entertaining.



Compare with Nos. 293 and 294 in A. in M.



75.—The Spider and the Fly.

Though this problem was much discussed in the Daily Mail from 18th January to 7th February 1905, when it appeared to create great public interest, it was actually first propounded by me in the Weekly Dispatch of 14th June 1903.

Imagine the room to be a cardboard box. Then the box may be cut in various different ways, so that the cardboard may be laid flat on the table. I show four of these ways, and indicate in every case the relative positions of the spider and the fly, and the straightened course which the spider must take without going off the cardboard. These are the four most favourable cases, and it will be found that the shortest route is in No. 4, for it is only 40 feet in length (add the square of 32 to the square of 24 and extract the square root). It will be seen that the spider actually passes along five of the six sides of the room! Having marked the route, fold the box up (removing the side the spider does not use), and the appearance of the shortest course is rather surprising. If the spider had taken what most persons will consider obviously the shortest route (that shown in No. 1), he would have gone 42 feet! Route No. 2 is 43.174 feet in length, and Route No. 3 is 40.718 feet.



I will leave the reader to discover which are the shortest routes when the spider and the fly are 2, 3, 4, 5, and 6 feet from the ceiling and the floor respectively.



76.—The Perplexed Cellarman.

Brother John gave the first man three large bottles and one small bottleful of wine, and one large and three small empty bottles. To each of the other two men he gave two large and three small bottles of wine, and two large and one small empty bottle. Each of the three then receives the same quantity of wine, and the same number of each size of bottle.



77.—Making a Flag.

The diagram shows how the piece of bunting is to be cut into two pieces. Lower the piece on the right one "tooth," and they will form a perfect square, with the roses symmetrically placed.



It will be found interesting to compare this with No. 154 in A. in M.



78.—Catching the Hogs.

A very short examination of this puzzle game should convince the reader that Hendrick can never catch the black hog, and that the white hog can never be caught by Katruen.

Each hog merely runs in and out of one of the nearest corners and can never be captured. The fact is, curious as it must at first sight appear, a Dutchman cannot catch a black hog, and a Dutchwoman can never capture a white one! But each can, without difficulty, catch one of the other colour.

So if the first player just determines that he will send Hendrick after the white porker and Katruen after the black one, he will have no difficulty whatever in securing both in a very few moves.

It is, in fact, so easy that there is no necessity whatever to give the line of play. We thus, by means of the game, solve the puzzle in real life, why the Dutchman and his wife could not catch their pigs: in their simplicity and ignorance of the peculiarities of Dutch hogs, each went after the wrong animal.

The little principle involved in this puzzle is that known to chess-players as "getting the opposition." The rule, in the case of my puzzle (where the moves resemble rook moves in chess, with the added condition that the rook may only move to an adjoining square), is simply this. Where the number of squares on the same row, between the man or woman and the hog, is odd, the hog can never be captured; where the number of squares is even, a capture is possible. The number of squares between Hendrick and the black hog, and between Katruen and the white hog, is 1 (an odd number), therefore these individuals cannot catch the animals they are facing. But the number between Hendrick and the white hog, and between Katruen and the black one, is 6 (an even number), therefore they may easily capture those behind them.



79.—The Thirty-one Game.

By leading with a 5 the first player can always win. If your opponent plays another 5, you play a 2 and score 12. Then as often as he plays a 5 you play a 2, and if at any stage he drops out of the series, 3, 10, 17, 24, 31, you step in and win. If after your lead of 5 he plays anything but another 5, you make 10 or 17 and win. The first player may also win by leading a 1 or a 2, but the play is complicated. It is, however, well worth the reader's study.



80.—The Chinese Railways.

This puzzle was artfully devised by the yellow man. It is not a matter for wonder that the representatives of the five countries interested were bewildered. It would have puzzled the engineers a good deal to construct those circuitous routes so that the various trains might run with safety. Diagram 1 shows directions for the five systems of lines, so that no line shall ever cross another, and this appears to be the method that would require the shortest possible mileage.



The reader may wish to know how many different solutions there are to the puzzle. To this I should answer that the number is indeterminate, and I will explain why. If we simply consider the case of line A alone, then one route would be Diagram 2, another 3, another 4, and another 5. If 3 is different from 2, as it undoubtedly is, then we must regard 5 as different from 4. But a glance at the four diagrams, 2, 3, 4, 5, in succession will show that we may continue this "winding up" process for ever; and as there will always be an unobstructed way (however long and circuitous) from stations B and E to their respective main lines, it is evident that the number of routes for line A alone is infinite. Therefore the number of complete solutions must also be infinite, if railway lines, like other lines, have no breadth; and indeterminate, unless we are told the greatest number of parallel lines that it is possible to construct in certain places. If some clear condition, restricting these "windings up," were given, there would be no great difficulty in giving the number of solutions. With any reasonable limitation of the kind, the number would, I calculate, be little short of two thousand, surprising though it may appear.



81.—The Eight Clowns.

This is a little novelty in magic squares. These squares may be formed with numbers that are in arithmetical progression, or that are not in such progression. If a square be formed of the former class, one place may be left vacant, but only under particular conditions. In the case of our puzzle there would be no difficulty in making the magic square with 9 missing; but with 1 missing (that is, using 2, 3, 4, 5, 6, 7, 8, and 9) it is not possible. But a glance at the original illustration will show that the numbers we have to deal with are not actually those just mentioned. The clown that has a 9 on his body is portrayed just at the moment when two balls which he is juggling are in mid-air. The positions of these balls clearly convert his figure into the recurring decimal .[.9]. Now, since the recurring decimal .[.9] is equal to 9/9, and therefore to 1, it is evident that, although the clown who bears the figure 1 is absent, the man who bears the figure 9 by this simple artifice has for the occasion given his figure the value of the number 1. The troupe can consequently be grouped in the following manner:—

7 5 2 4 6 3 8 .[.9]

Every column, every row, and each of the two diagonals now add up to 12. This is the correct solution to the puzzle.



82.—The Wizard's Arithmetic.

This puzzle is both easy and difficult, for it is a very simple matter to find one of the multipliers, which is 86. If we multiply 8 by 86, all we need do is to place the 6 in front and the 8 behind in order to get the correct answer, 688. But the second number is not to be found by mere trial. It is 71, and the number to be multiplied is no less than 1639344262295081967213114754098360655737704918032787. If you want to multiply this by 71, all you have to do is to place another 1 at the beginning and another 7 at the end—a considerable saving of labour! These two, and the example shown by the wizard, are the only two-figure multipliers, but the number to be multiplied may always be increased. Thus, if you prefix to 41096 the number 41095890, repeated any number of times, the result may always be multiplied by 83 in the wizard's peculiar manner.

If we add the figures of any number together and then, if necessary, again add, we at last get a single-figure number. This I call the "digital root." Thus, the digital root of 521 is 8, and of 697 it is 4. This digital analysis is extensively dealt with in A. in M. Now, it is evident that the digital roots of the two numbers required by the puzzle must produce the same root in sum and product. This can only happen when the roots of the two numbers are 2 and 2, or 9 and 9, or 3 and 6, or 5 and 8. Therefore the two-figure multiplier must have a digital root of 2, 3, 5, 6, 8, or 9. There are ten such numbers in each case. I write out all the sixty, then I strike out all those numbers where the second figure is higher than the first, and where the two figures are alike (thirty-six numbers in all); also all remaining numbers where the first figure is odd and the second figure even (seven numbers); also all multiples of 5 (three more numbers). The numbers 21 and 62 I reject on inspection, for reasons that I will not enter into. I then have left, out of the original sixty, only the following twelve numbers: 83, 63, 81, 84, 93, 42, 51, 87, 41, 86, 53, and 71. These are the only possible multipliers that I have really to examine.

My process is now as curious as it is simple in working. First trying 83, I deduct 10 and call it 73. Adding 0's to the second figure, I say if 30000, etc., ever has a remainder 43 when divided by 73, the dividend will be the required multiplier for 83. I get the 43 in this way. The only multiplier of 3 that produces an 8 in the digits place is 6. I therefore multiply 73 by 6 and get 438, or 43 after rejecting the 8. Now, 300,000 divided by 73 leaves the remainder 43, and the dividend is 4,109. To this 1 add the 6 mentioned above and get 41,096 x 83, the example given on page 129.

In trying the even numbers there are two cases to be considered. Thus, taking 86, we may say that if 60000, etc., when divided by 76 leaves either 22 or 60 (because 3x6 and 8x6 both produce 8), we get a solution. But I reject the former on inspection, and see that 60 divided by 76 is 0, leaving a remainder 60. Therefore 8 x 86 = 688, the other example. It will be found in the case of 71 that 100000, etc., divided by 61 gives a remainder 42, (7 x 61 = 427) after producing the long dividend at the beginning of this article, with the 7 added.

The other multipliers fail to produce a solution, so 83, 86, and 71 are the only three possible multipliers. Those who are familiar with the principle of recurring decimals (as somewhat explained in my next note on No. 83, "The Ribbon Problem") will understand the conditions under which the remainders repeat themselves after certain periods, and will only find it necessary in two or three cases to make any lengthy divisions. It clearly follows that there is an unlimited number of multiplicands for each multiplier.



83.—The Ribbon Problem.

The solution is as follows: Place this rather lengthy number on the ribbon, 0212765957446808510638297872340425531914393617. It may be multiplied by any number up to 46 inclusive to give the same order of figures in the ring. The number previously given can be multiplied by any number up to 16. I made the limit 9 in order to put readers off the scent. The fact is these two numbers are simply the recurring decimals that equal 1/17 and 1/47 respectively. Multiply the one by seventeen and the other by forty-seven, and you will get all nines in each case.

In transforming a vulgar fraction, say 1/17, to a decimal fraction, we proceed as below, adding as many noughts to the dividend as we like until there is no remainder, or until we get a recurring series of figures, or until we have carried it as far as we require, since every additional figure in a never-ending decimal carries us nearer and nearer to exactitude.

17) 100 (.058823 85 —— 150 136 —— 140 136 —— 40 34 —— 60 51 —— 9

Now, since all powers of 10 can only contain factors of the powers of 2 and 5, it clearly follows that your decimal never will come to an end if any other factor than these occurs in the denominator of your vulgar fraction. Thus, 1/2, 1/4, and 1/8 give us the exact decimals, .5, .25, and .125; 1/5 and 1/25 give us .2 and .04; 1/10 and 1/20 give us .1 and .05: because the denominators are all composed of 2 and 5 factors. But if you wish to convert 1/3, 1/6, or 1/7, your division sum will never end, but you will get these decimals, .33333, etc., .166666, etc., and .142857142857142857, etc., where, in the first case, the 3 keeps on repeating for ever and ever; in the second case the 6 is the repeater, and in the last case we get the recurring period of 142857. In the case of 1/17 (in "The Ribbon Problem") we find the circulating period to be .0588235294117647.

Now, in the division sum above, the successive remainders are 1, 10, 15, 14, 4, 6, 9, etc., and these numbers I have inserted around the inner ring of the diagram. It will be seen that every number from 1 to 16 occurs once, and that if we multiply our ribbon number by any one of the numbers in the inner ring its position indicates exactly the point at which the product will begin. Thus, if we multiply by 4, the product will be 235, etc.; if we multiply by 6, 352, etc. We can therefore multiply by any number from 1 to 16 and get the desired result.



The kernel of the puzzle is this: Any prime number, with the exception of 2 and 5, which are the factors of 10, will exactly divide without remainder a number consisting of as many nines as the number itself, less one. Thus 999999 (six 9's) is divisible by 7, sixteen 9's are divisible by 17, eighteen 9's by 19, and so on. This is always the case, though frequently fewer 9's will suffice; for one 9 is divisible by 3, two by 11, six by 13, when our ribbon rule for consecutive multipliers breaks down and another law comes in. Therefore, since the 0 and 7 at the ends of the ribbon may not be removed, we must seek a fraction with a prime denominator ending in 7 that gives a full period circulator. We try 37, and find that it gives a short period decimal, .027, because 37 exactly divides 999; it, therefore, will not do. We next examine 47, and find that it gives us the full period circulator, in 46 figures, at the beginning of this article.

If you cut any of these full period circulators in half and place one half under the other, you will find that they will add up all 9's; so you need only work out one half and then write down the complements. Thus, in the ribbon above, if you add 05882352 to 94117647 the result is 99999999, and so with our long solution number. Note also in the diagram above that not only are the opposite numbers on the outer ring complementary, always making 9 when added, but that opposite numbers in the inner ring, our remainders, are also complementary, adding to 17 in every case. I ought perhaps to point out that in limiting our multipliers to the first nine numbers it seems just possible that a short period circulator might give a solution in fewer figures, but there are reasons for thinking it improbable.



84.—The Japanese Ladies and the Carpet.

If the squares had not to be all the same size, the carpet could be cut in four pieces in any one of the three manners shown. In each case the two pieces marked A will fit together and form one of the three squares, the other two squares being entire. But in order to have the squares exactly equal in size, we shall require six pieces, as shown in the larger diagram. No. 1 is a complete square, pieces 4 and 5 will form a second square, and pieces 2, 3, and 6 will form the third—all of exactly the same size.



If with the three equal squares we form the rectangle IDBA, then the mean proportional of the two sides of the rectangle will be the side of a square of equal area. Produce AB to C, making BC equal to BD. Then place the point of the compasses at E (midway between A and C) and describe the arc AC. I am showing the quite general method for converting rectangles to squares, but in this particular case we may, of course, at once place our compasses at E, which requires no finding. Produce the line BD, cutting the arc in F, and BF will be the required side of the square. Now mark off AG and DH, each equal to BF, and make the cut IG, and also the cut HK from H, perpendicular to ID. The six pieces produced are numbered as in the diagram on last page.

It will be seen that I have here given the reverse method first: to cut the three small squares into six pieces to form a large square. In the case of our puzzle we can proceed as follows:—

Make LM equal to half the diagonal ON. Draw the line NM and drop from L a perpendicular on NM. Then LP will be the side of all the three squares of combined area equal to the large square QNLO. The reader can now cut out without difficulty the six pieces, as shown in the numbered square on the last page.



85.—Captain Longbow and the Bears.



It might have struck the reader that the story of the bear impaled on the North Pole had no connection with the problem that followed. As a matter of fact it is essential to a solution. Eleven bears cannot possibly be arranged to form of themselves seven rows of bears with four bears in every row. But it is a different matter when Captain Longbow informs us that "they had so placed themselves that there were" seven rows of four bears. For if they were grouped as shown in the diagram, so that three of the bears, as indicated, were in line with the North Pole, that impaled animal would complete the seventh row of four, which cannot be obtained in any other way. It obviously does not affect the problem whether this seventh row is a hundred miles long or a hundred feet, so long as they were really in a straight line—a point that might perhaps be settled by the captain's pocket compass.



86.—The English Tour.

It was required to show how a resident at the town marked A might visit every one of the towns once, and only once, and finish up his tour at Z. This puzzle conceals a little trick. After the solver has demonstrated to his satisfaction that it cannot be done in accordance with the conditions as he at first understood them, he should carefully examine the wording in order to find some flaw. It was said, "This would be easy enough if he were able to cut across country by road, as well as by rail, but he is not."



Now, although he is prohibited from cutting across country by road, nothing is said about his going by sea! If, therefore, we carefully look again at the map, we shall find that two towns, and two only, lie on the sea coast. When he reaches one of these towns he takes his departure on board a coasting vessel and sails to the other port. The annexed illustration shows, by a dark line, the complete route.

This problem should be compared with No. 250, "The Grand Tour," in A. in M. It can be simplified in practically an identical manner, but as there is here no choice on the first stage from A, the solutions are necessarily quite different. See also solution to No. 94.



87.—The Chifu-Chemulpo Puzzle.

The solution is as follows. You may accept the invitation to "try to do it in twenty moves," but you will never succeed in performing the feat. The fewest possible moves are twenty-six. Play the cars so as to reach the following positions:—

E5678 ———— = 10 moves. 1234

E56 ———— = 2 moves. 123 87 4

56 ———— = 5 moves. E312 87 4

E ———— = 9 moves. 87654321

Twenty-six moves in all.



88.—The Eccentric Market-woman.

The smallest possible number of eggs that Mrs. Covey could have taken to market is 719. After selling half the number and giving half an egg over she would have 359 left; after the second transaction she would have 239 left; after the third deal, 179; and after the fourth, 143. This last number she could divide equally among her thirteen friends, giving each 11, and she would not have broken an egg.



89.—The Primrose Puzzle.

The two words that solve this puzzle are BLUEBELL and PEARTREE. Place the letters as follows: B 3-1, L 6-8, U 5-3, E 4-6, B 7-5, E 2-4, L 9-7, L 9-2. This means that you take B, jump from 3 to 1, and write it down on 1; and so on. The second word can be inserted in the same order. The solution depends on finding those words in which the second and eighth letters are the same, and also the fourth and sixth the same, because these letters interchange without destroying the words. MARITIMA (or sea-pink) would also solve the puzzle if it were an English word.

Compare with No. 226 in A. in M.



90.—The Round Table.

Here is the way of arranging the seven men:—

A B C D E F G A C D B G E F A D B C F G E A G B F E C D A F C E G D B A E D G F B C A C E B G F D A D G C F E B A B F D E G C A E F D C G B A G E B D F C A F G C B E D A E B F C D G A G C E D B F A F D G B C E

Of course, at a circular table, A will be next to the man at the end of the line.

I first gave this problem for six persons on ten days, in the Daily Mail for the 13th and 16th October 1905, and it has since been discussed in various periodicals by mathematicians. Of course, it is easily seen that the maximum number of sittings for n persons is (n - 1)(n -2)/2 ways. The comparatively easy method for solving all cases where n is a prime+1 was first discovered by Ernest Bergholt. I then pointed out the form and construction of a solution that I had obtained for 10 persons, from which E. D. Bewley found a general method for all even numbers. The odd numbers, however, are extremely difficult, and for a long time no progress could be made with their solution, the only numbers that could be worked being 7 (given above) and 5, 9, 17, and 33, these last four being all powers of 2+1. At last, however (though not without much difficulty), I discovered a subtle method for solving all cases, and have written out schedules for every number up to 25 inclusive. The case of 11 has been solved also by W. Nash. Perhaps the reader will like to try his hand at 13. He will find it an extraordinarily hard nut.

The solutions for all cases up to 12 inclusive are given in A. in M., pp. 205, 206.



91.—The Five Tea Tins.

There are twelve ways of arranging the boxes without considering the pictures. If the thirty pictures were all different the answer would be 93,312. But the necessary deductions for cases where changes of boxes may be made without affecting the order of pictures amount to 1,728, and the boxes may therefore be arranged, in accordance with the conditions, in 91,584 different ways. I will leave my readers to discover for themselves how the figures are to be arrived at.



92.—The Four Porkers.

The number of ways in which the four pigs may be placed in the thirty-six sties in accordance with the conditions is seventeen, including the example that I gave, not counting the reversals and reflections of these arrangements as different. Jaenisch, in his Analyse Mathematique au jeu des Echecs (1862), quotes the statement that there are just twenty-one solutions to the little problem on which this puzzle is based. As I had myself only recorded seventeen, I examined the matter again, and found that he was in error, and, doubtless, had mistaken reversals for different arrangements.

Here are the seventeen answers. The figures indicate the rows, and their positions show the columns. Thus, 104603 means that we place a pig in the first row of the first column, in no row of the second column, in the fourth row of the third column, in the sixth row of the fourth column, in no row of the fifth column, and in the third row of the sixth column. The arrangement E is that which I gave in diagram form:—

A. 104603 B. 136002 C. 140502 D. 140520 E. 160025 F. 160304 G. 201405 H. 201605 I. 205104 J. 206104 K. 241005 L. 250014 M. 250630 N. 260015 O. 261005 P. 261040 Q. 306104

It will be found that forms N and Q are semi-symmetrical with regard to the centre, and therefore give only two arrangements each by reversal and reflection; that form H is quarter-symmetrical, and gives only four arrangements; while all the fourteen others yield by reversal and reflection eight arrangements each. Therefore the pigs may be placed in (2 x 2) + (4 x 1) + (8 x 14) = 120 different ways by reversing and reflecting all the seventeen forms.

Three pigs alone may be placed so that every sty is in line with a pig, provided that the pigs are not forbidden to be in line with one another; but there is only one way of doing it (if we do not count reversals as different), as follows: 105030.



93.—The Number Blocks.

Arrange the blocks so as to form the two multiplication sums 915 x 64 and 732 x 80, and the product in both cases will be the same: 58,560.



94.—Foxes and Geese.

The smallest possible number of moves is twenty-two—that is, eleven for the foxes and eleven for the geese. Here is one way of solving the puzzle:

10—5 11—6 12—7 5—12 6—1 7—6 —— —— —— —— —— —— 1—8 2—9 3—4 8—3 9—10 4—9



12—7 1—8 6—1 7—2 8—3 —— —— —— —— —— 3—4 10—5 9—10 4—11 5—12

Of course, the reader will play the first move in the top line, then the first move in the second line, then the second move in the top line, and so on alternately.



In A. in M., p. 230, I have explained fully my "buttons and string" method of solving puzzles on chequered boards. In Diagram A is shown the puzzle in the form in which it may be presented on a portion of the chessboard with six knights. A comparison with the illustration on page 141 will show that I have there dispensed with the necessity of explaining the knight's move to the uninstructed reader by lines that indicate those moves. The two puzzles are the same thing in different dress. Now compare page 141 with Diagram B, and it will be seen that by disentangling the strings I have obtained a simplified diagram without altering the essential relations between the buttons or discs. The reader will now satisfy himself without any difficulty that the puzzle requires eleven moves for the foxes and eleven for the geese. He will see that a goose on 1 or 3 must go to 8, to avoid being one move from a fox and to enable the fox on 11 to come on to the ring. If we play 1—8, then it is clearly best to play 10—5 and not 12—5 for the foxes. When they are all on the circle, then they simply promenade round it in a clockwise direction, taking care to reserve 8—3 and 5—12 for the final moves. It is thus rendered ridiculously easy by this method. See also notes on solutions to Nos. 13 and 85.



95.—Robinson Crusoe's Table.

The diagram shows how the piece of wood should be cut in two pieces to form the square table-top. A, B, C, D are the corners of the table. The way in which the piece E fits into the piece F will be obvious to the eye of the reader. The shaded part is the wood that is discarded.



96.—The Fifteen Orchards.

The number must be the least common multiple of 1, 2, 3, etc., up to 15, that, when divided by 7, leaves the remainder 1, by 9 leaves 3, by 11 leaves 10, by 13 leaves 3, and by 14 leaves 8. Such a number is 120. The next number is 360,480, but as we have no record of a tree—especially a very young one—bearing anything like such a large number of apples, we may take 120 to be the only answer that is acceptable.



97.—The Perplexed Plumber.

The rectangular closed cistern that shall hold a given quantity of water and yet have the smallest possible surface of metal must be a perfect cube—that is, a cistern every side of which is a square. For 1,000 cubic feet of water the internal dimensions will be 10 ft. x 10 ft. x 10 ft., and the zinc required will be 600 square feet. In the case of a cistern without a top the proportions will be exactly half a cube. These are the "exact proportions" asked for in the second case. The exact dimensions cannot be given, but 12.6 ft. x 12.6 ft. x 6.3 ft. is a close approximation. The cistern will hold a little too much water, at which the buyer will not complain, and it will involve the plumber in a trifling loss not worth considering.



98.—The Nelson Column.

If you take a sheet of paper and mark it with a diagonal line, as in Figure A, you will find that when you roll it into cylindrical form, with the line outside, it will appear as in Figure B.



It will be seen that the spiral (in one complete turn) is merely the hypotenuse of a right-angled triangle, of which the length and width of the paper are the other two sides. In the puzzle given, the lengths of the two sides of the triangle are 40 ft. (one-fifth of 200 ft.) and 16 ft. 8 in. Therefore the hypotenuse is 43 ft. 4 in. The length of the garland is therefore five times as long—216 ft. 8 in. A curious feature of the puzzle is the fact that with the dimensions given the result is exactly the sum of the height and the circumference.



99.—The Two Errand Boys.

All that is necessary is to add the two distances at which they meet to twice their difference. Thus 720 + 400 + 640 = 1760 yards, or one mile, which is the distance required. Or, put another way, three times the first distance less the second distance will always give the answer, only the first distance should be more than two-thirds of the second.



100.—On the Ramsgate Sands.

Just six different rings may be formed without breaking the conditions. Here is one way of effecting the arrangements.

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

Join the ends and you have the six rings.

Lucas devised a simple mechanical method for obtaining the n rings that may be formed under the conditions by 2n+1 children.



101.—The Three Motor-Cars.

The only set of three numbers, of two, three, and five figures respectively, that will fulfil the required conditions is 27 x 594 = 16,038. These three numbers contain all the nine digits and 0, without repetition; the first two numbers multiplied together make the third, and the second is exactly twenty-two times the first. If the numbers might contain one, four, and five figures respectively, there would be many correct answers, such as 3 x 5,694 = 17,082; but it is a curious fact that there is only one answer to the problem as propounded, though it is no easy matter to prove that this is the case.



102.—A Reversible Magic Square.



It will be seen that in the arrangement given every number is different, and all the columns, all the rows, and each of the two diagonals, add up 179, whether you turn the page upside down or not. The reader will notice that I have not used the figures 3, 4, 5, 8, or 0.



103.—The Tube Railway.

There are 640 different routes. A general formula for puzzles of this kind is not practicable. We have obviously only to consider the variations of route between B and E. Here there are nine sections or "lines," but it is impossible for a train, under the conditions, to traverse more than seven of these lines in any route. In the following table by "directions" is meant the order of stations irrespective of "routes." Thus, the "direction" BCDE gives nine "routes," because there are three ways of getting from B to C, and three ways of getting from D to E. But the "direction" BDCE admits of no variation; therefore yields only one route.

2 two-line directions of 3 routes — 6 1 three-line " " 1 " — 1 1 " " " 9 " — 9 2 four-line " " 6 " — 12 2 " " " 18 " — 36 6 five-line " " 6 " — 36 2 " " " 18 " — 36 2 six-line " " 36 " — 72 12 seven-line " " 36 " — 432 —— Total 640

We thus see that there are just 640 different routes in all, which is the correct answer to the puzzle.



104.—The Skipper and the Sea-Serpent.

Each of the three pieces was clearly three cables long. But Simon persisted in assuming that the cuts were made transversely, or across, and that therefore the complete length was nine cables. The skipper, however, explained (and the point is quite as veracious as the rest of his yarn) that his cuts were made longitudinally—straight from the tip of the nose to the tip of the tail! The complete length was therefore only three cables, the same as each piece. Simon was not asked the exact length of the serpent, but how long it must have been. It must have been at least three cables long, though it might have been (the skipper's statement apart) anything from that up to nine cables, according to the direction of the cuts.



105.—The Dorcas Society.

If there were twelve ladies in all, there would be 132 kisses among the ladies alone, leaving twelve more to be exchanged with the curate—six to be given by him and six to be received. Therefore, of the twelve ladies, six would be his sisters. Consequently, if twelve could do the work in four and a half months, six ladies would do it in twice the time—four and a half months longer—which is the correct answer.

At first sight there might appear to be some ambiguity about the words, "Everybody kissed everybody else, except, of course, the bashful young man himself." Might this not be held to imply that all the ladies immodestly kissed the curate, although they were not (except the sisters) kissed by him in return? No; because, in that case, it would be found that there must have been twelve girls, not one of whom was a sister, which is contrary to the conditions. If, again, it should be held that the sisters might not, according to the wording, have kissed their brother, although he kissed them, I reply that in that case there must have been twelve girls, all of whom must have been his sisters. And the reference to the ladies who might have worked exclusively of the sisters shuts out the possibility of this.



106.—The Adventurous Snail.

At the end of seventeen days the snail will have climbed 17 ft., and at the end of its eighteenth day-time task it will be at the top. It instantly begins slipping while sleeping, and will be 2 ft. down the other side at the end of the eighteenth day of twenty-four hours. How long will it take over the remaining 18 ft.? If it slips 2 ft. at night it clearly overcomes the tendency to slip 2 ft. during the daytime, in climbing up. In rowing up a river we have the stream against us, but in coming down it is with us and helps us. If the snail can climb 3 ft. and overcome the tendency to slip 2 ft. in twelve hours' ascent, it could with the same exertion crawl 5 ft. a day on the level. Therefore, in going down, the same exertion carries it 7 ft. in twelve hours—that is, 5 ft. by personal exertion and 2 ft. by slip. This, with the night slip, gives it a descending progress of 9 ft. in the twenty-four hours. It can, therefore, do the remaining 18 ft. in exactly two days, and the whole journey, up and down, will take it exactly twenty days.



107.—The Four Princes.

When Montucla, in his edition of Ozanam's Recreations in Mathematics, declared that "No more than three right-angled triangles, equal to each other, can be found in whole numbers, but we may find as many as we choose in fractions," he curiously overlooked the obvious fact that if you give all your sides a common denominator and then cancel that denominator you have the required answer in integers!

Every reader should know that if we take any two numbers, m and n, then m^2 + n^2, m^2 - n^2, and 2mn will be the three sides of a rational right-angled triangle. Here m and n are called generating numbers. To form three such triangles of equal area, we use the following simple formula, where m is the greater number:—

mn + m^2 + n^2 = a m^2 - n^2 = b 2mn + n^2 = c

Now, if we form three triangles from the following pairs of generators, a and b, a and c, a and b + c, they will all be of equal area. This is the little problem respecting which Lewis Carroll says in his diary (see his Life and Letters by Collingwood, p. 343), "Sat up last night till 4 a.m., over a tempting problem, sent me from New York, 'to find three equal rational-sided right-angled triangles.' I found two ... but could not find three!"

The following is a subtle formula by means of which we may always find a R.A.T. equal in area to any given R.A.T. Let z = hypotenuse, b = base, h = height, a = area of the given triangle; then all we have to do is to form a R.A.T. from the generators z^2 and 4a, and give each side the denominator 2z (b^2 - h^2), and we get the required answer in fractions. If we multiply all three sides of the original triangle by the denominator, we shall get at once a solution in whole numbers.

The answer to our puzzle in smallest possible numbers is as follows:—

First Prince ... 518 1320 1418 Second Prince ... 280 2442 2458 Third Prince ... 231 2960 2969 Fourth Prince ... 111 6160 6161

The area in every case is 341,880 square furlongs. I must here refrain from showing fully how I get these figures. I will explain, however, that the first three triangles are obtained, in the manner shown, from the numbers 3 and 4, which give the generators 37, 7; 37, 33; 37, 40. These three pairs of numbers solve the indeterminate equation, a^3b -b^3a = 341,880. If we can find another pair of values, the thing is done. These values are 56, 55, which generators give the last triangle. The next best answer that I have found is derived from 5 and 6, which give the generators 91, 11; 91, 85; 91, 96. The fourth pair of values is 63, 42.

The reader will understand from what I have written above that there is no limit to the number of rational-sided R.A.T.'s of equal area that may be found in whole numbers.



108.—Plato and the Nines.

The following is the simple solution of the three nines puzzle:—

9 + 9 —— .9

To divide 18 by .9 (or nine-tenths) we, of course, multiply by 10 and divide by 9. The result is 20, as required.



109.—Noughts and Crosses.

The solution is as follows: Between two players who thoroughly understand the play every game should be drawn. Neither player could ever win except through the blundering of his opponent. If Nought (the first player) takes the centre, Cross must take a corner, or Nought may beat him with certainty. If Nought takes a corner on his first play, Cross must take the centre at once, or again be beaten with certainty. If Nought leads with a side, both players must be very careful to prevent a loss, as there are numerous pitfalls. But Nought may safely lead anything and secure a draw, and he can only win through Cross's blunders.



110.—Ovid's Game.

The solution here is: The first player can always win, provided he plays to the centre on his first move. But a good variation of the game is to bar the centre for the first move of the first player. In that case the second player should take the centre at once. This should always end in a draw, but to ensure it the first player must play to two adjoining corners (such as 1 and 3) on his first and second moves. The game then requires great care on both sides.



111.—The Farmer's Oxen.

Sir Isaac Newton has shown us, in his Universal Arithmetic, that we may divide the bullocks in each case in two parts—one part to eat the increase, and the other the accumulated grass. The first will vary directly as the size of the field, and will not depend on the time; the second part will also vary directly as the size of the field, and in addition inversely with the time. We find from the farmer's statements that 6 bullocks keep down the growth in a 10-acre field, and 6 bullocks eat the grass on 10 acres in 16 weeks. Therefore, if 6 bullocks keep down the growth on 10 acres, 24 will keep down the growth on 40 acres.

Again, we find that if 6 bullocks eat the accumulated grass on 10 acres in 16 weeks, then

12 eat the grass on 10 acres in 8 weeks, 48 " " 40 " 8 " 192 " " 40 " 2 " 64 " " 40 " 6 "

Add the two results together (24 + 64), and we find that 88 oxen may be fed on a 40-acre meadow for 6 weeks, the grass growing regularly all the time.



112.—The Great Grangemoor Mystery.

We were told that the bullet that killed Mr. Stanton Mowbray struck the very centre of the clock face and instantly welded together the hour, minute, and second hands, so that all revolved in one piece. The puzzle was to tell from the fixed relative positions of the three hands the exact time when the pistol was fired.

We were also told, and the illustration of the clock face bore out the statement, that the hour and minute hands were exactly twenty divisions apart, "the third of the circumference of the dial." Now, there are eleven times in twelve hours when the hour hand is exactly twenty divisions ahead of the minute hand, and eleven times when the minute hand is exactly twenty divisions ahead of the hour hand. The illustration showed that we had only to consider the former case. If we start at four o'clock, and keep on adding 1 h. 5 m. 27-3/11 sec., we shall get all these eleven times, the last being 2 h. 54 min. 32-8/11 sec. Another addition brings us back to four o'clock. If we now examine the clock face, we shall find that the seconds hand is nearly twenty-two divisions behind the minute hand, and if we look at all our eleven times we shall find that only in the last case given above is the seconds hand at this distance. Therefore the shot must have been fired at 2 h. 54 min. 32-8/11 sec. exactly, or, put the other way, at 5 min. 27-3/11 sec. to three o'clock. This is the correct and only possible answer to the puzzle.



113.—Cutting a Wood Block.

Though the cubic contents are sufficient for twenty-five pieces, only twenty-four can actually be cut from the block. First reduce the length of the block by half an inch. The smaller piece cut off constitutes the portion that cannot be used. Cut the larger piece into three slabs, each one and a quarter inch thick, and it will be found that eight blocks may easily be cut out of each slab without any further waste.



114.—The Tramps and the Biscuits.

The smallest number of biscuits must have been 1021, from which it is evident that they were of that miniature description that finds favour in the nursery. The general solution is that for n men the number must be m (n^{n+1}) - (n - 1), where m is any integer. Each man will receive m (n - 1)^n - 1 biscuits at the final division, though in the case of two men, when m = 1, the final distribution only benefits the dog. Of course, in every case each man steals an nth of the number of biscuits, after giving the odd one to the dog.



INDEX

"Abracadabra," 64.

Age and Kinship Puzzles, 20, 28.

Albanna, Ibn, 198.

Ale, Measuring the, 29.

Algebraical Puzzles. See Arithmetical Puzzles.

Alkalacadi, 198.

Amulet, The, 64, 190.

Archery Butt, The, 60, 187.

Arithmetical Puzzles, 18, 26, 34, 36, 45, 46, 51, 56, 59, 61, 64, 65, 73, 74, 75, 88, 89, 91, 92, 103, 107, 122, 125, 128, 129, 130, 135, 137, 139, 143, 147, 148, 150, 151, 152, 153, 154, 157, 158, 161.

Arrows, The Nine, 32.

Astronomical Problem, 55.

Bags, Four Money, 46.

Ball, W. W. Rouse, 202.

Bandy-Ball, The Game of, 58, 185.

Bears, Capt. Longbow and the, 132, 233.

Bergholt, Ernest, 237.

Bewley, E. D., 237.

Biscuits, The Tramps and the, 160, 250.

Block, Cutting a Wood, 160, 250.

Blocks, The Number, 139, 238.

Bottles, Sharing the, 122.

Bottles, The Sixteen, 45.

Bridges, The Eight, 48.

Bridging the Ditch, 83, 204.

Brooch, Cutting the, 41.

Buried Treasure, 107, 212.

Buttons and String Method, 171, 239.

Canon's Yeoman, Puzzle of the, 55, 181.

Canterbury Pilgrim's Puzzle, 33.

Canterbury Puzzles, 23, 163.

Card Puzzle, 125.

Carpenter's Puzzle, The, 31, 168.

Carpet, Japanese Ladies and, 131, 231.

Carroll, Lewis, 246.

Casket, Lady Isabel's, 67, 191.

Cats and Mice, 75.

Cellarer, The Riddle of the, 73, 196.

Cellarman, The Perplexed, 122, 222.

Chalked Numbers, The, 89, 206.

Chaucer's Puzzle, 54, 181.

Cheeses on Stools, 24.

Chessboard Problems, 21, 25, 32, 51, 72, 82, 90, 113, 114, 116, 119, 124, 138, 141.

Chessboard, The Broken, 119, 220.

Chifu-Chemulpo Puzzle, 134, 235.

Chinese Railways, The, 127, 224.

Christmas Puzzle Party, The Squire's, 86, 205.

Cisterns, Making, 145.

Clerk of Oxenford's Puzzle, The, 29, 167.

Cliff Mystery, The Cornish, 99, 210.

Clock Puzzle, 158.

Cloth, Cutting the, 50.

Clowns, The Eight, 128, 226.

Club, Adventures of the Puzzle, 94, 210.

Coinage Puzzle, The, 111, 214.

Coin Magic Square, 111.

Column, The Nelson, 146, 241.

Combination and Group Problems, 38, 39, 61, 70, 122, 137, 147.

Cook's Puzzle, The, 36, 171.

Cornish Cliff Mystery, The, 99, 210.

Counter Problems, Moving, 24, 35, 69, 77, 124, 135, 136, 141.

Counting out Puzzle, 39.

Crescent and the Cross, The, 63, 189.

Crossing River Problems, 82, 83.

Crusaders, The Riddle of the, 74, 197.

Crusoe's Table, Robinson, 142, 240.

Cubes, Sums of Two, 174, 209.

Cubes, The Silver, 92, 209.

Daily Mail, 179, 221, 236.

Decimals, Recurring, 228, 229.

Demoiselle, The Noble, 59, 186.

Diamond Letter Puzzles, 181.

Digital Analysis, 228.

Digital Puzzles, 18, 26, 90, 103, 129, 139, 148, 154.

Dispute between Friar and Sompnour, 51, 180.

Dissection Puzzles, 29, 30, 35, 41, 49, 63, 123, 131, 142.

Ditch, Bridging the, 83, 204.

Divisors of Numbers, To Find, 170.

Doctor of Physic, The Puzzle of the, 42, 174.

Donjon Keep Window, The, 62, 188.

Dorcas Society, The, 151, 245.

Dormitory Puzzle, The, 70.

Dungeon, The Death's-head, 60.

Dungeons, The Nine, 35.

Dyer's Puzzle, The, 50, 180.

Edward, Portrait of King, 46.

Eggs, Selling the, 135.

Eleven Pennies, The, 88, 206.

Errand Boys, The Two, 147, 242.

Escape of King's Jester, The Strange, 78, 201.

Executioner, The, 78.

Fallacy of Square's Diagonal, 52.

Farmer's Oxen, The, 157, 248.

Fermat, P. de, 174.

Fish-pond, The Riddle of the, 69, 194.

Flag, Making a, 123, 223.

Fleurs-de-Lys, Sixty-four, 50.

Flour, The Nine Sacks of, 26.

Fly, The Spider and the, 121, 221.

Footprints Puzzles, 101, 105.

Four Princes, The, 153, 246.

Foxes and Geese, 140, 239.

Franklin's Puzzle, The, 44, 176.

Friar and Sompnour's Dispute, 51, 180.

Friar's Puzzle, The, 46, 177.

Frogs and Tumblers, The, 113, 216.

Frogs who would a-wooing go, 116, 219.

Frogs' Ring, The Riddle of the, 76, 199.

Games, Puzzle, 118, 125, 156, 157.

Gardens, The Royal, 82, 203.

Geese, The Christmas, 88, 206.

Geometrical Problems, 52, 62, 67, 121, 131, 144, 146.

Grangemoor Mystery, The, 158, 249.

Group Problems, Combination and. See Combination and Group Problems.

Haberdasher's Puzzle, The, 49, 178.

Hogs, Catching the, 124, 223.

Hoppe, Oscar, 198.

Host's Puzzle, The, 28, 166.

Isabel's Casket, Lady, 67, 191.

Jaenisch, 237.

Japanese Ladies and the Carpet, 131, 231.

Jester, Strange Escape of the King's, 78, 201.

Kayles, The Game of, 118, 220.

Kennels, The Nine, 39.

King's Jester, Strange Escape of the, 78, 201.

Knight's Puzzle, The, 26, 165.

Lady Isabel's Casket, 67, 191.

L'Arithmetique Amusante, 198.

Legendre, 175.

Letter Puzzles, 16.

Lock, The Secret, 80, 202.

Locomotive and Speed Puzzle, 147.

Longbow and the Bears, Capt., 132, 233.

Lucas, Edouard, 175, 198, 242.

M'Elroy, C. W., 179.

Magdalen, Chart of the, 41.

Magic Square, A Reversible, 149, 243.

Magic Square Problems, 21, 29, 44, 111, 112, 128, 149.

Manciple's Puzzle, The, 56, 183.

Man of Law's Puzzle, The, 34, 170.

Market Woman, The Eccentric, 135, 235.

Marksford and the Lady, Lord, 96.

Maze, The Underground, 79, 201.

Measuring, Weighing, and Packing Puzzles, 29, 31, 55, 72, 73, 160.

Merchant's Puzzle, The, 33, 170.

Merry Monks of Riddlewell, 68, 194.

Miller's Puzzle, The, 26, 164.

Miscellaneous Puzzles, 118, 220.

Mistletoe Bough, Under the, 91, 208.

Moat, Crossing the, 81, 202.

Money, Dividing the, 57.

Monks of Riddlewell, The Merry, 68, 194.

Monk's Puzzle, The, 39, 172.

Montucla, 246.

Motor-Car, The Runaway, 103, 211.

Motor-Cars, The Three, 147, 242.

Moving Counter Problems. See Counter Problems, Moving.

Nash, W., 237.

Nelson Column, The, 146, 241.

Newton, Sir Isaac, 248.

Nines, Plato and the, 154, 247.

Noble Demoiselle, The, 59, 186.

Noughts and Crosses, 156, 248.

Number Blocks, The, 139, 238.

Numbers on Motor-Cars, 103, 148.

Numbers Partition of, 46.

Numbers The Chalked, 89, 206.

Nun's Puzzle, The, 32, 169.

Ones, Numbers composed only of, 18, 75, 198.

Opposition in Chess, 224.

Orchards, The Fifteen, 143, 241.

Ovid's Game, 156, 248.

Oxen, The Farmer's, 157, 248.

Ozanam's Recreations, 246.

Packing Puzzles, Measuring, Weighing, and. See Measuring.

Palindromes, 17.

Pardoner's Puzzle, The, 25, 164.

Parental Command, A, 28.

Park, Mystery of Ravensdene, 105, 211.

Parson's Puzzle, The, 47, 177.

Party, The Squire's Christmas Puzzle, 86, 205.

Pellian Equation, 197.

Pennies, The Eleven, 88, 206.

Phials, The Two, 42.

Photograph, The Ambiguous, 94, 210.

Pie and the Pasty, The, 36.

Pilgrimages, The Fifteen, 25.

Pilgrims' Manner of Riding, 34.

Pilgrims The Riddle of the, 70, 194.

Pillar, The Carved Wooden, 31.

Plato and the Nines, 154, 247.

Ploughman's Puzzle, The, 43, 175.

Plumber, The Perplexed, 144, 241.

Plum Puddings, Tasting the, 90, 207.

Points and Lines Problems, 43, 116, 133.

Porkers, The Four, 138, 237.

Postage Stamps Puzzle, The, 112, 214.

Primrose Puzzle, The, 136, 235.

Princes, The Four, 153, 246.

Prioress, The Puzzle of the, 41, 173.

Professor's Puzzles, The, 110, 214.

Puzzle Club, Adventures of the, 94, 210.

Puzzles, How to solve, 18.

Puzzles, How they are made, 14.

Puzzles, Sophistical, 15.

Puzzles, The exact conditions of, 18.

Puzzles, The mysterious charm of, 12.

Puzzles, The nature of, 11.

Puzzles, The utility of, 13.

Puzzles, The variety of, 13, 16.

Puzzles, Unsolved, 20.

Puzzling Times at Solvamhall Castle, 58, 184.

Pyramids, Triangular, 163.

Railway Puzzle, 134.

Railway The Tube, 149, 243.

Railways, The Chinese, 127, 224.

Ramsgate Sands, On the, 147, 242.

Rat-catcher's Riddle, The, 56.

Ravensdene Park, Mystery of, 105, 211.

Reve's Puzzle, The, 24, 163.

Ribbon Problem, The, 130, 228.

Riddles, old, 16.

Riddlewell, The Merry Monks of, 68, 194.

River Crossing Problems, 82, 83.

Robinson Crusoe's Table, 142, 240.

Romeo and Juliet, 114, 217.

Romeo's Second Journey, 116, 218.

Rook's Path, The, 207.

Rope, The Mysterious, 79, 201.

Round Table, The, 137, 236.

Route Problems, Unicursal and. See Unicursal.

Sack Wine, The Riddle of the, 72, 196.

St. Edmondsbury, The Riddle of, 75, 197.

Sands, On the Ramsgate, 147, 242.

Sea-Serpent, The Skipper and the, 150, 244.

Shield, Squares on a, 27.

Shipman's Puzzle, The, 40, 173.

Skipper and the Sea-Serpent, The, 150, 244.

Snail on the Flagstaff, The, 66, 190.

Snail The Adventurous, 152, 245.

Snails, The Two, 115, 217.

Solvamhall Castle, Puzzling Times at, 58, 184.

Sompnour's and Friar's Dispute, 51, 180.

Sompnour's Puzzle, The, 38, 172.

Spherical Surface of Water, 181.

Spider and the Fly, The, 121, 221.

Square and Triangle, The, 49.

Square Field, The, 107.

Squares, Problem of, 74.

Square, Three Squares from One, 131, 231.

Squire's Christmas Puzzle Party, The, 86, 205.

Squire's Puzzle, The, 45, 176.

Squire's Yeoman, The Puzzle of the, 31, 168.

Stamps, Counting Postage, 137.

Stamps, Magic Squares of, 112.

Stamps, Puzzle, The Postage, 112, 214.

Superposition, Problem on, 179.

Sylvester, 175.

Table, Robinson Crusoe's, 142, 240.

Table, The Round, 137, 236.

Talkhys, 198.

Tapestry, Cutting the, 30.

Tapiser's Puzzle, The, 30, 167.

Teacups, The Three, 87, 205.

Tea Tins, The Five, 137, 237.

Thirty-one Game, The, 125, 224.

Tiled Hearth, The Riddle of the, 71, 195.

Tilting at the Ring, 59, 185.

Tour, The English, 134, 233.

Towns, Visiting the, 134.

Tramps and the Biscuits, The, 160, 250.

Treasure, The Buried, 107, 212.

Trees, The Sixteen Oak, 44.

Triangle and Square, 49.

Triangles of Equal Area, 153, 246.

Triangular numbers, 163.

Tube Railway, The, 149, 243.

Unicursal and Route Problems, 40, 45, 48, 56, 60, 83, 90, 105, 127, 134, 149.

Weaver's Puzzle, The, 35, 171.

Weekly Dispatch, 179, 221.

Weighing, and Packing Puzzles, Measuring. See Measuring.

Wife of Bath's Riddles, The, 27, 166.

Window, The Donjon Keep, 62, 188.

Wine, Stealing the, 73.

Wizard's Arithmetic, The, 129, 226.

Wood Block, Cutting a, 160, 250.

Wreath on Column, 146, 241.



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THE END