Arrange the blocks so as to form the two multiplication sums 915 × 64 and 732 × 80, and the product in both cases will be the same: 58,560.
[94].—Foxes and Geese.
The smallest possible number of moves is twenty-two—that is, eleven for the foxes and eleven for the geese. Here is one way of solving the puzzle:
| 10—5 | 11—6 | 12—7 | 5—12 | 6—1 | 7—6 |
| —— | —— | —— | —— | —— | —— |
| 1—8 | 2—9 | 3—4 | 8—3 | 9—10 | 4—9 |
| 12—7 | 1—8 | 6—1 | 7—2 | 8—3 | |
| —— | —— | —— | —— | —— | |
| 3—4 | 10—5 | 9—10 | 4—11 | 5—12 |
Of course, the reader will play the first move in the top line, then the first move in the second line, then the second move in the top line, and so on alternately.
In A. in M., p. 230, I have explained fully my "buttons and string" method of solving puzzles on chequered boards. In Diagram A is shown the puzzle in the form in which it may be presented on a portion of the chessboard with six knights. A comparison with the illustration on page 141 will show that I have there dispensed with the necessity of explaining the knight's move to the uninstructed reader by lines that indicate those moves. The two puzzles are the same thing in different dress. Now compare page 141 with Diagram B, and it will be seen that by disentangling the strings I have obtained a simplified diagram without altering the essential relations between the buttons or discs. The reader will now satisfy himself without any difficulty that the puzzle requires eleven moves for the foxes and eleven for the geese. He will see that a goose on 1 or 3 must go to 8, to avoid being one move from a fox and to enable the fox on 11 to come on to the ring. If we play 1—8, then it is clearly best to play 10—5 and not 12—5 for the foxes. When they are all on the circle, then they simply promenade round it in a clockwise direction, taking care to reserve 8—3 and 5—12 for the final moves. It is thus rendered ridiculously easy by this method. See also notes on solutions to Nos. [13] and [85].