The first of these puzzles is based on a similar
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
they are to be placed requires a little thought or trial until one
the rule respecting putting the fractions in the corners.
I give the
I also show the solution to the second stamp
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.,
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:
where the horizontal differences are all alike and
differences all alike, but not necessarily the same as the
This happens in the case of the second solution, the numbers of which
Also in the case of the solution to the Coinage
numbers are, in shillings:
If there are to be nine different
0 may occur once.
Yet one might construct squares with negative
numbers, as follows: