Page:Amusements in mathematics.djvu/208

232.—CATCHING THE MICE.

In order that the cat should eat every thirteenth mouse, and the white mouse last of all, it is necessary that the count should begin at the seventh mouse (calling the white one the first)—that is, at the one nearest the tip of the cat's tail. In this case it is not at all necessary to try starting at all the mice in turn until you come to the right one, for you can just start anywhere and note how far distant the last one eaten is from the starting point. You will find it to be the eighth, and therefore must start at the eighth, counting backwards from the white mouse. This is the one I have indicated.

In the case of the second puzzle, where you have to find the smallest number with which the cat may start at the white mouse and eat this one last of all, unless you have mastered the general solution of the problem, which is very difficult, there is no better course open to you than to try every number in succession until you come to one that works correctly. The smallest number is twenty-one. If you have to proceed by trial, you will shorten your labour a great deal by only counting out the remainders when the number is divided successively by 13, 12, 11, 10, etc. Thus, in the case of 21, we have the remainders 8, 9, 10, 1, 3, 5, 7, 3, 1, 1, 3, 1, 1. Note that I do not give the remainders of 7, 3, and 1 as nought, but as 7, 3, and 1. Now, count round each of these numbers in turn, and you will find that the white mouse is killed last of all. Of course, if we wanted simply any niunber, not the smallest, the solution is very easy, for we merely take the least common multiple of 13, 12, 11, 10, etc. down to 2. This is 360360, and you will find that the first count kills the thirteenth mouse, the next the twelfth, the next the eleventh, and so on down to the first. But the most arithmetically inclined cat could not be expected to take such a big number when a small one like twenty-one would equally serve its purpose.

In the third case, the smallest number is 100. The number 1,000 would also do, and there are just seventy-two other numbers between these that the cat might employ with equal success.

233.—THE ECCENTRIC CHEESEMONGER.

To leave the three piles at the extreme ends of the rows, the cheeses may be moved as follows—the numbers refer to the cheeses and not to their positions in the row: 7—2, 8—7, 9—8, 10—15, 6—10, 5—6, 14—16, 13—14, 12—13, 3—1, 4—3, 11—4. This is probably the easiest solution of all to find. To get three of the piles on cheeses 13, 14, and 15, play thus: 9—4, 10—9, 11—10, 6—14, 5—6, 12—15, 8—12, 7—8, 16—5, 3—13, 2—3, 1—2. To leave the piles on cheeses 3, 5, 12, and 14, play thus: 8—3, 9—14, 16—12, 1—5, 10—9, 7—10, 11—8, 2—1, 4—16, 13—2, 6—11, 15—4.

234.—THE EXCHANGE PUZZLE.

Make the following exchanges of pairs; H—K, H—E, H—C, H—A, I—L, I—F, I—D, K—L, G—J, J—A, F—K, L—E, D—K, E—F, E—D, E—B, B—K. It will be found that, although the white counters can be moved to their proper places in 11 moves, if we omit all consideration of exchanges, yet the black cannot be so moved in fewer than 17 moves. So we have to introduce waste moves with the white counters to equal the minimum required by the black. Thus fewer than 17 moves must be impossible. Some of the moves are, of course, interchangeable.

235.—TORPEDO PRACTICE.

If the enemy's fleet be anchored in the formation shown in the illustration, it will be seen that as many as ten out of the sixteen ships may be blown up by discharging the torpedoes in the order indicated by the numbers and in the directions indicated by the arrows. As each torpedo in succession passes under three ships and sinks the fourth, strike out each vessel with the pencil as it is sunk.

236.—THE HAT PUZZLE.

I suggested that the reader should try this puzzle with counters, so I give my solution in that form. The silk hats are represented by black counters and the felt hats by white counters. The first row shows the hats in their original positions, and then each succes-