1. The problem about the farmer and the animals reduces \0 the Diophantine equation llx + 5y = 200. Applying the ijpethod of continued fractions, three solutions in positive integers can be found:

Cows Pigs Sheep

5 29 66

15 7 78

2. L. H. Longley-Cook, in Fun with Brain Puzzlers (Fawcett, _,J65), Problem 87, solves the rectangle problem as follows. Let •and y be the sides of the large rectangle. The total number of is it contains is xy. The border, one cell wide, contains 2x + t—4 cells. Since we are told that the border contains xyl2 cells, can write the equation:

Double both sides and rearrange the terms:

bi1 Add 16 to each side: f

It is clear that (* — 4) and (x—y) must be positive integral factors of 8. The only pairs of such factors are 8, 1 and 4, 2. They provide two solutions: x=\2, y = 5, andx = 8, y = 6.

The problem is closely related to integral-sided right triangles. The width of the border is an integer only when the diagonal of the large rectangle cuts it into two such "Pythagorean triangles."

If we generalize the problem to allow nonintegral solutions for borders of any uniform width, keeping only the proviso that the area of the border be equal to the area of the rectangle within it, there is an unusually simple formula for the width of the border. (I am indebted to S. L. Porter for it.) Merely add two adjacent sides of the border, subtract the diagonal of the large rectangle and divide the result by four. This procedure gives the width of the border.

Several readers generalized this problem to three dimen-sio cis, seeking integral edges for a brick composed of unit cubes equal to the number of unit cubes required to cover it on alH sides with a one-unit layer of cubes. Daniel Sleator used a computer to find the complete solution, a total of 20 bricks. The smallest-volume brick has edges of 8, 10, 12; the largest, 5, 13, 132. This confirms a guess made by M. H. Greenblatt in Mathematical Entertainments (Crowell, 1965), page 11, that the problem has "about" 20 solutions.

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