IMO 1977 Problem A2 In a finite sequence of real numbers the sum of any seven successive terms is negative, and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence. Solution
: 16. x 1 + ... + x 7  0, so x 1 + x 2 + x 3  0, so x 4 > 0. If there are 17 or more elements then the same argument shows that x 5 , x 6 , x 7 > 0. But x 1 + ... + x 7  0, so x 5 + x 6 + x 7 < 0. Contradiction. If we assume that there is a solution for n = 16 and that the sum of 7 consecutive terms is -1 and that the sum of 11 consecutive terms is 1, then we can easily solve the equations to get: 5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5 and we can check that this works for 16. Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 19th IMO 1977 © John Scholes jscholes@kalva.demon.co.uk 12 Oct 1998