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(Money Puzzles)
I had the other day in my possession a label bearing the number 3 0 2 5
in large figures. This got accidentally torn in half, so that 3 0 was on
one piece and 2 5 on the other, as shown on the illustration. On looking
at these pieces I began to make a calculation, scarcely conscious of
what I was doing, when I discovered this little peculiarity. If we add
the 3 0 and the 2 5 together and square the sum we get as the result the
complete original number on the label! Thus, 30 added to 25 is 55, and
55 multiplied by 55 is 3025. Curious, is it not? Now, the puzzle is to
find another number, composed of four figures, all different, which may
be divided in the middle and produce the same result.


The other number that answers all the requirements of the puzzle is
9,801. If we divide this in the middle into two numbers and add them
together we get 99, which, multiplied by itself, produces 9,801. It is
true that 2,025 may be treated in the same way, only this number is
excluded by the condition which requires that no two figures should be
The general solution is curious. Call the number of figures in each half
of the torn label n. Then, if we add 1 to each of the exponents of the
prime factors (other than 3) of 10^n - 1 (1 being regarded as a factor
with the constant exponent, 1), their product will be the number of
solutions. Thus, for a label of six figures, n = 3. The factors of 10^n
- 1 are 1 x 37 (not considering the 3 cubed), and the product of 2 x 2 =
4, the number of solutions. This always includes the special cases 98 -
01, 00 - 01, 998 - 01, 000 - 001, etc. The solutions are obtained as
follows:--Factorize 10 cubed - 1 in all possible ways, always keeping the
powers of 3 together, thus, 37 x 27, 999 x 1. Then solve the equation
37x = 27y + 1. Here x = 19 and y = 26. Therefore, 19 x 37 = 703, the
square of which gives one label, 494,209. A complementary solution
(through 27x = 37x + 1) can at once be found by 10^n - 703 = 297, the
square of which gives 088,209 for second label. (These non-significant
noughts to the left must be included, though they lead to peculiar cases
like 00238 - 04641 = 4879 squared, where 0238 - 4641 would not work.) The
special case 999 x 1 we can write at once 998,001, according to the law
shown above, by adding nines on one half and noughts on the other, and
its complementary will be 1 preceded by five noughts, or 000001. Thus we
get the squares of 999 and 1. These are the four solutions.

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