The first of these puzzles is based on a similar principle, though it is really much easier, because the condition that nine of the stamps must be of different values makes their selection a simple matter, though how they are to be placed requires a little thought or trial until one knows the rule respecting putting the fractions in the corners. I give the solution.
I also show the solution to the second stamp puzzle. All the columns, rows, and diagonals add up 1s. 6d. There is no stamp on one square, and the conditions did not forbid this omission. The stamps at present in circulation are these:—½d., 1d., 1-½d., 2d., 2-½d., 3d., 4d., 5d., 6d., 9d., 10d., 1s., 2s. 6d., 5s., 10s., £1, and £5. In the first solution the numbers are in arithmetical progression—1, 1-½, 2, 2-½, 3, 3-½, 4, 4-½, 5. But any nine numbers will form a magic square if we can write them thus:—
| 1 | 2 | 3 |
| 7 | 8 | 9 |
| 13 | 14 | 15 |
where the horizontal differences are all alike and the vertical differences all alike, but not necessarily the same as the horizontal. This happens in the case of the second solution, the numbers of which may be written:—
| 0 | 1 | 2 |
| 5 | 6 | 7 |
| 10 | 11 | 12 |
Also in the case of the solution to No. 67, the Coinage Puzzle, the numbers are, in shillings:—
| 2 | 2½ | 3 |
| 4½ | 5 | 5½ |
| 7 | 7½ | 8 |
If there are to be nine different numbers, 0 may occur once (as in the solution to No. 22). Yet one might construct squares with negative numbers, as follows:—