Answers 251- DIVIDING BY ELEVEN
To be divisible by 11, four of the alternate digits must sum to 17 and the
remaining five to 28, or four to 28 and five to 17. Thus, in the example I gave
(482539761),4,2,3, 7, 1 sum to 17, and 8, 5, 9, 6 to 28. Now four digits
will sum to 17 in 9 different ways and five to 17 in 2 ways, making 11 together.
In each of the 11 cases 4 may be permuted in 24 ways and 5 in 120 ways, or
together in 2880 ways. So that 2880 X 11 = 31,680 ways. As the nine digits
can be permuted in 362,880 ways, the chances are just 115 to 11 against
a haphazard arrangement being divisible by 11.- DIVIDING BY 37
Write beneath the number successively, from right to left, the numbers 1,
10, 11, as follows:4 9 293 082 3
10 11 10 11 10 1 11 10
Now, regarding the lower figures as multipliers, add together all the products
of 1 and 10 and deduct all the products by 11. This is the same as adding 13,
08,29, and 49 together (99) and deducting eleven times 2 plus 3 plus 1 (66).
The difference, 33, will be the remainder when the large number is divided
by 37.
Here is the key. If we divide 1, 10, 100, 1000, etc., by 37 we get successively
the remainders 1, 10, 26, but for convenience we deduct the 26 from 37 and
call it minus 11. If you try 49,629,708,213 you will find the minus or negative
total 165, or in excess of the positive 99. The difference is 66. Deduct 37 and
you get 29. But as the result is minus, deduct it from 37 and you have 8 as
correct answer. You can now find the method for other prime divisors. The
cases of 7 and 13 are easy. In the former case you write 1, 3, 2 (I, 3, 2), 1, 3,
2, etc., from right to left, the bracketed numbers being minus. In the latter
case, 1 (3, 4, 1), 3, 4, 1 (3,4, 1), etc.
117. ANOTHER 37 DMSION
Call the required numbers ABCABCABC. If the sum of the A digits, the B
digits, and the C digits respectively are: