Amusements in Mathematics by Henry Ernest Dudeney (e books for reading txt) 📖

or other of the designs indicated by the edges of the cards, or a reflection thereof. Let us direct our attention to Fig. 1. Here the card is so placed that the star is at the town T; it therefore gives us (by following the edge of the card) one of the circular routes from London: L, S, R, T, M, A, E, P, O, J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we should get L, I, F, H, K, Q, etc., but these reverse routes were not to be counted. When we have written out this first route we revolve the card until the star is at M, when we get another different route, at A a third route, at E a fourth route, and at P a fifth route. We have thus obtained five different routes by revolving the card as it lies. But it is evident that if we now take up the card and replace it with the other side uppermost, we shall in the same manner get five other routes by revolution.

We therefore see how, by using the revolving card in Fig. 1, we may, without any difficulty, at once write out ten routes. And if we employ the cards in Figs. 2 and 3, we similarly obtain in each case ten other routes. These thirty routes are all that are possible. I do not give the actual proof that the three cards exhaust all the possible cases, but leave the reader to reason that out for himself. If he works out any route at haphazard, he will certainly find that it falls into one or other of the three categories.

255.—THE LEVEL PUZZLE.—solution

Let us confine our attention to the L in the top left-hand corner. Suppose we go by way of the E on the right: we must then go straight on to the V, from which letter the word may be completed in four ways, for there are four E's available through which we may reach an L. There are therefore four ways of reading through the right-hand E. It is also clear that there must be the same number of ways through the E that is immediately below our starting point. That makes eight. If, however, we take the third route through the E on the diagonal, we then have the option of any one of the three V's, by means of each of which we may complete the word in four ways. We can therefore spell LEVEL in twelve ways through the diagonal E. Twelve added to eight gives twenty readings, all emanating from the L in the top left-hand corner; and as the four corners are equal, the answer must be four times twenty, or eighty different ways.

256.—THE DIAMOND PUZZLE.—solution

There are 252 different ways. The general formula is that, for words of n letters (not palindromes, as in the case of the next puzzle), when grouped in this manner, there are always 2(n+1) - 4 different readings. This does not allow diagonal readings, such as you would get if you used instead such a word as DIGGING, where it would be possible to pass from one G to another G by a diagonal step.

257.—THE DEIFIED PUZZLE.—solution

The correct answer is 1,992 different ways. Every F is either a corner F or a side F—standing next to a corner in its own square of F's. Now, FIED may be read from a corner F in 16 ways; therefore DEIF may be read into a corner F also in 16 ways; hence DEIFIED may be read through a corner F in 16 × 16 = 256 ways. Consequently, the four corner F's give 4 × 256 = 1,024 ways. Then FIED may be read from a side F in 11 ways, and DEIFIED therefore in 121 ways. But there are eight side F's; consequently these give together 8 × 121 = 968 ways. Add 968 to 1,024 and we get the answer, 1,992.

In this form the solution will depend on whether the number of letters in the palindrome be odd or even. For example, if you apply the word NUN in precisely the same manner, you will get 64 different readings; but if you use the word NOON, you will only get 56, because you cannot use the same letter twice in immediate succession (since you must "always pass from one letter to another") or diagonal readings, and every reading must involve the use of the central N.

The reader may like to find for himself the general formula in this case, which is complex and difficult. I will merely add that for such a case as MADAM, dealt with in the same way as DEIFIED, the number of readings is 400.

258.—THE VOTERS' PUZZLE.—solution

THE number of readings here is 63,504, as in the case of "WAS IT A RAT I SAW" (No. 30, Canterbury Puzzles). The general formula is that for palindromic sentences containing 2n + 1 letters there are [4(2n - 1)]² readings.

259.—HANNAH'S PUZZLE.—solution

Starting from any one of the N's, there are 17 different readings of NAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 ways of spelling HAN. If we were allowed to use the same N twice in a spelling, the answer would be 68 times 68, or 4,624 ways. But the conditions were, "always passing from one letter to another." Therefore, for every one of the 17 ways of spelling HAN with a particular N, there would be 51 ways (3 times 17) of completing the NAH, or 867 (17 times 51) ways for the complete word. Hence, as there are four N's to use in HAN, the correct solution of the puzzle is 3,468 (4 times 867) different ways.

260.—THE HONEYCOMB PUZZLE.—solution

The required proverb is, "There is many a slip 'twixt the cup and the lip." Start at the T on the outside at the bottom right-hand corner, pass to the H above it, and the rest is easy.

261.—THE MONK AND THE BRIDGES.—solution