by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
Owner:
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, LNote 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 number, 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.? solution
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.? solution
Make the following exchanges of pairs: H-K, H-E, H-C, H-A, l-L, l-F, l-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.? solution
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 maybe 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.? solution
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 successive row shows how they appear after one of the five manipulations. It will thus be seen that we first move hats 2 and 3, then 7 and 8, then 4 and 5, then 10 and 11, and, finally, 1 and 2, leaving the four silk hats together, the four felt hats together, and the two vacant pegs at one end of the row. The first three pairs moved are dissimilar hats, the last two pairs being similar. There are other ways of solving the puzzle.
Pg197
237.?BOYS AND GIRLS.? solution
There are a good many different solutions to this puzzle. Any contiguous pair, except 7-8, may be moved first, and after the first move there are variations. The following solution shows the position from the start right through each successive move to the end:?
. . 12 3 4 5 6 7 8
4 3 12. .5678
4 3 12 7 6 5. .8
4 3 12 7. .568
4 . .2713568
238.?ARRANGING THE JAMPOTS.? solution
