IMO 1978 Problem A3 The set of all positive integers is the union of two disjoint subsets {f(1), f(2), f(3), ... }, {g(1), g(2), g(3), ... }, where f(1) < f(2) < f(3) < ..., and g(1) < g(2) < g(3) < ... , and g(n) = f(f(n)) + 1 for n = 1, 2, 3, ... . Determine f(240).
Let F = {f(1), f(2), f(3), ... }, G = {g(1), g(2), g(3), ... }, N n = {1, 2, 3, ... , n}. f(1) ≥ 1, so f(f(1)) ≥ 1 and hence g(1) ≥ 2. So 1 is not in G, and hence must be in F. It must be the smallest element of F and so f(1) = 1. Hence g(1) = 2. We can never have two successive integers n and n+1 in G, because if g(m) = n+1, then f(something) = n and so n is in F and G. Contradiction. In particular, 3 must be in F, and so f(2) = 3. Suppose f(n) = k. Then g(n) = f(k) + 1. So |N f(k)+1 ∩ G| = n. But |N f(k)+1 ∩ F| = k, so n + k = f(k) + 1, or f(k) = n + k - 1. Hence g(n) = n + k. So n + k + 1 must be in F and hence f(k+1) = n + k + 1. This so given the value of f for n we can find it for k and k+1. Using k+1 each time, we get, successively, f(2) = 3, f(4) = 6, f(7) = 11, f(12) = 19, f(20) = 32, f(33) = 53, f(54) = 87, f(88) = 142, f(143) = 231, f(232) = 375, which is not much help. Trying again with k, we get: f(3) = 4, f(4) = 6, f(6) = 9, f(9) = 14, f(14) = 22, f(22) = 35, f(35) = 56, f(56) = 90, f(90) = 145, f(145) = 234. Still not right, but we can try backing up slightly and using k+1: f(146) = 236. Still not right, we need to back up further: f(91) = 147, f(148) = 239, f(240) = 388. 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. 20th IMO 1978 © John Scholes jscholes@kalva.demon.co.uk 12 Oct 1998