THE CANTERBURY PUZZLES
81.—
The Eight Clowns.
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 re-
curring decimal 9 is equal to $, 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.
82.—
The Wizard's Arithmetic.
This puzzle is both easy and difficult, for it is a very simple
matter to find one of the multipliers, which is 86. If we multiply 8
by 86, all we need do is to place the 6 in front and the 8 behind
in order to get the correct answer, 688. But the second number is
not to be found by mere trial. It is 71, and the number to be
multiplied is no less than 16393442622950819672131147540983-
60655737704918032787. If you want to multiply this by 71, all
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