10th Brazil 1988 Problem 3 Let N be the natural numbers and N' = N ∪ {0}. Find all functions f:N→N' such that f(xy) = f(x) + f(y), f(30) = 0 and f(x) = 0 for all x = 7 mod 10.
f(n) = 0 for all n Solution f(30) = f(2) + f(3) + f(5) = 0 and f(n) is non-negative, so f(2) = f(3) = f(5) = 0. For any positive integer n not divisible by 2 or 5 we can find a positive integer m such that mn = 7 mod 10 (because 1·7 = 3·9 = 7 mod 10, so if n = 1 mod 10 we take any m = 7 mod 10 and vice versa, whilst if n = 3 mod 10 we take any m = 9 mod 10 and vice versa). But then f(mn) = 0, so f(n) = 0. It is a trivial induction that f(2 a 5 b n) = f(5 b n) = f(n), so f is identically zero. Thanks to Suat Namli . 10th Brazil 1988 © John Scholes jscholes@kalva.demon.co.uk 5 February 2004 Last corrected/updated 5 Feb 04