## THE TORN NUMBER.

(

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.

## Answer:

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

alike.

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 1¹ x 37¹ (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.