employment of sixpences and
half-crowns. I give the arrangement requiring the fewest possible current
English coins--fifteen. It will be seen that the amount in each corner is
a fractional one, the sum required in the total being a whole number of
shillings.

[Illustration]

68.--_The Postage Stamps Puzzles._

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.

[Illustration:

[1/2 d]
[4-1/2 d] [1 d] [3 d]

[2 d] [3 d] [4 d]

[2-1/2 d] [5 d] [1-1/2 d] ]

[Illustration:

[4 d] [1/2 d]
[3 d] [1-1/2 d]
[9 d]

[10 d] [6 d] [2 d]

[1 d] [1 s.] [5 d] ]

I also show the solution to the second stamp puzzle. All the columns,
rows, and diagonals add up 1_s._ 6_d._ There is no stamp on one square,
and the conditions did not forbid this omission. The stamps at present
in circulation are these:--1/2_d._, 1_d._, 1-1/2_d._, 2_d._, 2-1/2_d._,
3_d._, 4_d._, 5_d._, 6_d._, 9_d._, 10_d._, 1_s._, 2_s._ 6_d._, 5_s._,
10_s._, L1, and L5. In the first solution the numbers are in arithmetical
progression--1, 1-1/2, 2, 2-1/2, 3, 3-1/2, 4, 4-1/2, 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 2-1/2 3
4-1/2 5 5-1/2
7 7-1/2 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:--

-2 -1 0
5 6 7
12 13 14

69.--_The Frogs and Tumblers._

It is perfectly true, as the Professor said, that there is only one
solution (not counting a reversal) to this puzzle. The frogs that jump
are George in the third horizo