Determine all 3 digit numbers N which are divisible by 11 and where N/11 is equal to the sum of the squares of the digits of N.

**Answer**

550, 803.

**Solution**

So, put N/11 = 10a + b. If a + b ≤ 9, we have 2a^{2} + 2ab + 2b^{2} = 10a + b (*), so b is even. Put b = 2B, then B = a(a-5) + 2aB + 4B^{2}, which is even. So b must be a multiple of 4, so b = 0, 4 or 8. If b = 0, then (*) gives a = 5 and we get the solution 550. If b = 4, then (*) gives a^{2} - a + 14 = 0, which has no integral solutions. If b = 8, then (since a + b ≤ 9 and a > 0) a must be 1, but that does not satisfy (*).

If a + b > 9, we have (a+1)^{2} + (a+b-10)^{2} + b^{2} = 10a + b, or 2a^{2} + 2ab + 2b^{2} - 28a - 21b + 101 = 0 (**), so b is odd. Put b = 2B+1. Then a^{2} + 2aB + 4B^{2} - 13a - 17B + 41 = 0. But a(a-13) is even, so B is odd. Hence b = 3 or 7. If b = 3, then (**) gives a^{2} - 11a + 28 = 0, so a = 4 or 7. But a + b > 9, so a = 7. That gives the solution 803. If b = 7, then (**) gives a^{2} - 7a + 26 = 0, which has no integral solutions.

*Comment*

Personally, I hate this type of question. The fastest way to solve it is almost certainly to scan the 81 multiples of 11 from 110 to 990.

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

2nd IMO 1960

© John Scholes

jscholes@kalva.demon.co.uk

17 Sep 1998

Last corrected/updated 24 Sep 2003