This is a little novelty in magic
These squares may
with numbers that are in arithmetical progression, or that are not in
If a square be formed of the former class, one place
may be left vacant, but only under particular conditions.
In the case
our puzzle there would be no difficulty in making the magic square with
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
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.
of these balls clearly convert his figure into the recurring decimal
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
Every column, every row, and each of the two
diagonals now add
up to 12.
This is the correct solution to the puzzle.