CIRCLING THE SQUARES.
(
Money Puzzles)
The puzzle is to place a different number in each of the ten squares so
that the sum of the squares of any two adjacent numbers shall be equal
to the sum of the squares of the two numbers diametrically opposite to
them. The four numbers placed, as examples, must stand as they are. The
square of 16 is 256, and the square of 2 is 4. Add these together, and
the result is 260. Also--the square of 14 is 196, and the square of 8 is
64. These together also make 260. Now, in precisely the same way, B and
C should be equal to G and H (the sum will not necessarily be 260), A
and K to F and E, H and I to C and D, and so on, with any two adjoining
squares in the circle.
All you have to do is to fill in the remaining six numbers. Fractions
are not allowed, and I shall show that no number need contain more than
two figures.
Answer:
Though this problem might strike the novice as being rather difficult,
it is, as a matter of fact, quite easy, and is made still easier by
inserting four out of the ten numbers.
First, it will be found that squares that are diametrically opposite
have a common difference. For example, the difference between the square
of 14 and the square of 2, in the diagram, is 192; and the difference
between the square of 16 and the square of 8 is also 192. This must be
so in every case. Then it should be remembered that the difference
between squares of two consecutive numbers is always twice the smaller
number plus 1, and that the difference between the squares of any two
numbers can always be expressed as the difference of the numbers
multiplied by their sum. Thus the square of 5 (25) less the square of 4
(16) equals (2 x 4) + 1, or 9; also, the square of 7 (49) less the
square of 3 (9) equals (7 + 3) x (7 - 3), or 40.
Now, the number 192, referred to above, may be divided into five
different pairs of even factors: 2 x 96, 4 x 48, 6 x 32, 8 x 24, and 12
x 16, and these divided by 2 give us, 1 x 48, 2 x 24, 3 x 16, 4 x 12,
and 6 x 8. The difference and sum respectively of each of these pairs in
turn produce 47, 49; 22, 26; 13, 19; 8, 16; and 2, 14. These are the
required numbers, four of which are already placed. The six numbers that
have to be added may be placed in just six different ways, one of which
is as follows, reading round the circle clockwise: 16, 2, 49, 22, 19, 8,
14, 47, 26, 13.
I will just draw the reader's attention to one other little point. In
all circles of this kind, the difference between diametrically opposite
numbers increases by a certain ratio, the first numbers (with the
exception of a circle of 6) being 4 and 6, and the others formed by
doubling the next preceding but one. Thus, in the above case, the first
difference is 2, and then the numbers increase by 4, 6, 8, and 12. Of
course, an infinite number of solutions may be found if we admit
fractions. The number of squares in a circle of this kind must, however,
be of the form 4n + 6; that is, it must be a number composed of 6 plus a
multiple of 4.
