IMO 1982 Problem A1 The function f(n) is defined on the positive integers and takes non-negative integer values. f(2) = 0, f(3) > 0, f(9999) = 3333 and for all m, n: f(m+n) - f(m) - f(n) = 0 or 1. Determine f(1982).
We show that f(n) = [n/3] for n  0, so f(3) = 1. It follows by induction that f(3n) ≥ n. For f(3n+3) = f(3) + f(3n) + 0 or 1 = f(3n) + 1 or 2. Moreover if we ever get f(3n) > n, then the same argument shows that f(3m) > m for all m > n. But f(3.3333) = 3333, so f(3n) = n for all n <= 3333. Now f(3n+1) = f(3n) + f(1) + 0 or 1 = n or n + 1. But 3n+1 = f(9n+3) ≥ f(6n+2) + f(3n+1) ≥ 3f(3n+1), so f(3n+1) < n+1. Hence f(3n+1) = n. Similarly, f(3n+2) = n. In particular f(1982) = 660. s are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 23rd IMO 1982 © John Scholes jscholes@kalva.demon.co.uk 14 Oct 1998