536 Puzzles and Curious Problems

252 Answers

A B C

18 19 8

15 15 15

12 11 22

19 8 18

22 12 11

8 18 19

11 22 12

then in the first three groups 11A - lOB = C. In the next two groups

llA - lOB - C = III (3 X 37); and in the last two groups lOB + C-

llA = Ill. It does not matter what the figures are, but if they comply with

these conditions we can always divide by 37. Here is an example of the first

case-

ABCABCABC,

984763251,

where the 3 A's sum to 18, the 3 B's to 19, and the 3 C's to 8.

You will find 22 cases with the first equation, 10 with the second, and 10 with

the third, making 42 fundamental cases in all. But in every case the A figures

may be permuted in 6 ways, and the B figures in 6 ways, and the C figures in

6 ways, making 6 X 6 X 6 = 216, which multiplied by 42 gives the answer

9072 ways divisible by 37. As the 9 digits may be permuted in 362880 ways

th e c h ances are 362880 9072 or 40 I or^39 to I agamst. d·· lVlSl ·bil· lty.

118. A DIGITAL DIFFICULTY

There are four solutions, as follows: 2,438,195,760; 3,785,942,160;
4,753,869,120; 4,876,391,520. The last figure must be zero. Any arrangement
with an even figure next to the zero will be divisible by 2, 3, 4, 5, 6, 9, 10, 12,
15, and 18. We have therefore only to consider 7, 11, 13, 16, and 17. To be di-
visible by 11 the odd digits must sum to 28 and the even to 17, or vice versa.
To be divisible by 7 X 11 X 13 = 1001, if we ignore the zero, the numbers
formed by the first three and the last three digits must sum to the middle
three. (Note that the third case above is really 474-1386-912, with the 1