4th APMO 1992 Problem 5 a 1 , a 2 , a 3 , ... a n is a sequence of non-zero integers such that the sum of any 7 consecutive terms is positive and the sum of any 11 consecutive terms is negative. What is the largest possible value for n? Solution
: 16. We cannot have 17 terms, because then: a 1 + a 2 + ... + a 11 < 0 a 2 + a 3 + ... + a 12 < 0 a 3 + a 4 + ... + a 13 < 0 ... a 7 + a 8 + ... + a 17 < 0 So if we add the inequalities we get that an expression is negative. But notice that each column is positive. Contradiction. On the other hand, a valid sequence of 16 terms is: -5, -5, 13, -5, -5, -5, 13, -5, -5, 13, -5, -5, -5, 13, -5, -5. Any run of 7 terms has two 13s and five -5s, so sums to 1. Any run of 11 terms has three 13s and eight -5s, so sums to -1. 4th APMO 1992 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002