The Canterbury Puzzles, and Other Curious Problems

78 (#pgepubid00094).—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 Katrün.

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 Katrün 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 Katrün 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 Katrün and the black one, is 6 (an even number), therefore they may easily capture those behind them.

79 (#pgepubid00095).—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 (#pgepubid00096).—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 (#pgepubid00097).—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:—

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

82 (#pgepubid00098).—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 3×6 and 8×6 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 × 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 (#pgepubid00099).—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.

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, ½, 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 (#pgepubid00100).—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 (#pgepubid00101).—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 (#pgepubid00102).—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 (#pgepubid00231).

87 (#pgepubid00103).—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:—

Twenty-six moves in all.

88 (#pgepubid00104).—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 (#pgepubid00105).—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 (#pgepubid00106).—The Round Table

Here is the way of arranging the seven men:—

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.