This puzzle may be solved in as few as four moves, in the following manner:  Move 5 over 8, 9, 3, 1.  Move 7 over 4.  Move 6 over 2 and 7.  Move 5 over 6, and all the counters are removed except 5, which is left in the central square that it originally occupied.

230.—­THE TWELVE PENNIES.

Here is one of several solutions.  Move 12 to 3, 7 to 4, 10 to 6, 8 to 1, 9 to 5, 11 to 2.

231.—­PLATES AND COINS.

Number the plates from 1 to 12 in the order that the boy is seen to be going in the illustration.  Starting from 1, proceed as follows, where “1 to 4” means that you take the coin from plate No. 1 and transfer it to plate No. 4:  1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and complete the last revolution to 1, making three revolutions in all.  Or you can proceed this way:  4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10 to 1.  It is easy to solve in four revolutions, but the solutions in three are more difficult to discover.

This is “The Riddle of the Fishpond” (No. 41, Canterbury Puzzles) in a different dress.

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 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.