20th CanMO 1988 Problem 5 If S is a sequence of positive integers let p(S) be the product of the members of S. Let m(S) be the arithmetic mean of p(T) for all non-empty subsets T of S. S' is formed from S by appending an additional positive integer. If m(S) = 13 and m(S') = 49, find S'.
Suppose there are n elements in S, namely a 1 , a 2 , ... , a n . Then m(S) = (K - 1)/(2 n - 1), where K = (a 1 + 1) (a 2 + 1) ... (a n + 1). If the additional integer is h, then m(S') = (K(h + 1) - 1)/(2 n+1 - 1). So we have K = 13(2 n - 1) + 1, (h + 1)K = 49(2 n+1 - 1) + 1. Hence h + 1 = (49·2 n+1 - 48)/(13·2 n - 12). But 7(13·2 n - 1) = 91·2 n - 84  3, 8(13·2 n - 12) = (52·2 n+1 - 96) > (49·2 n+1 - 48). So we must have n = 1, 2, or 3. If n = 1, then h+1 = 148/14, which is not an integer. If n = 2, then h+1 = 344/40, which is not an integer. If n = 3, then h+1 = 736/92 = 8, which is ok. If n = 3, then if the elements of S are a, b, c, we must have (a + 1)(b + 1)(c + 1) = 92 = 2.2.23. So S = {2, 2, 23} and S' = {2, 2, 7, 23}. 20th CanMO 1988 © John Scholes jscholes@kalva.demon.co.uk 21 June 2002