Can you place the numbers 1 to 8 on the outer squares of a 3x3 grid, so on every side the middle number is the sum of the corners next to it?

Can you manage to place the same numbers so the middle number of each side is the average of the corners next to it?

First, the lowest two values (1 and 2) must be on the corners, and the highest value (8) must be in the middle of an edge.

If 1 and 2 are adjacent to each other, the 3 must be between them. That would mean that the 4 cannot be on an edge (1 and 3 are taken); it must be on a corner adjacent to either 1 (making 5) or 2 (making 6). In either case, it would be impossible to place the last corner, as any remaining value (5, 6, 7, or 8) added to four would be greater than 8.

The 1 and the 2 are on opposite corners. Thus, the pairs of adjacent edges must differ by one; with 7 and 8 on one side and 4 and 5 on the other:

1 7 6
4   8
3 5 2

To answer the second part, first note that the highest and lowest values, 1 and 8, cannot be the average of two other values. So, they must be on the corners. Also, for two numbers to have an integral average value, they must add up to an even number; that is, they must be both odd or both even. Since each corner is obviously adjacent to two other corners, all four corners must be both odd or both even for this to work. That is not possible with both 1 and 8 on the corners, so no, the second scenario is not possible.

Posted by DJ on 2004-07-30 09:29:42