The Canterbury Puzzles Part 31

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 quant.i.ty 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.

[Ill.u.s.tration]

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 circ.u.mference.

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 2_n_+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 27 594 = 16,038. These three numbers contain all the nine digits and 0, without repet.i.tion; 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 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._

[Ill.u.s.tration:

11 77 62 29 69 22 17 71 27 61 79 12 72 19 21 67 ]

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 a.s.suming 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._