This is a little novelty in magic squares. These squares may be formed
with numbers that are in arithmetical progression, or that are not in
such progression. If a square be formed of the former class, one place
may be left vacant, but only under particular conditions. In the case of
our puzzle there would be no difficulty in making the magic square with 9
missing; but with 1 missing (that is, using 2, 3, 4, 5, 6, 7, 8, and 9)
it is not possible. But a glance at the original illustration will show
that the numbers we have to deal with are not actually those just
mentioned. The clown that has a 9 on his body is portrayed just at the
moment when two balls which he is juggling are in mid-air. The positions
of these balls clearly convert his figure into the recurring decimal
.[.9]. Now, since the recurring decimal .[.9] is equal to 9/9, and
therefore to 1, it is evident that, although the clown who bears the
figure 1 is absent, the man who bears the figure 9 by this simple
artifice has for the occasion given his _figure_ the value of the
_number_ 1. The troupe can consequently be grouped in the following
manner:--
7 5
2 4 6
3 8 .[.9]
Every column, every row, and each of the two diagonals now add up to 12.
This is the correct solution to the puzzle.