Part 37

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 × 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 × 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.

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 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.