Part 36

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 Mathématique au jeu des Échecs (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 × 2) + (4 × 1) + (8 × 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 × 64 and 732 × 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. × 10 ft. × 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. × 12.6 ft. × 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.