8th CanMO 1976 Problem 3 n+2 students played a tournament. Each pair played each other once. A player scored 1 for a win, 1/2 for a draw and nil for a loss. Two students scored a total of 8 and the other players all had equal total scores. Find n.
There are (n+2)(n+1)/2 matches, so the total score is (n+2)(n+1)/2. Let the other players score k each. Then 8 + nk = (n+2)(n+1)/2, so n 2 - (2k-3) - 14 = 0. We know this equation has one root which is a positive integer. The product of the roots is -14, so the possibilities for the roots are: 1, -14; 2, -7; 7, -2; 14, -1. Hence the sum of the roots is -13, -5, 5, or 13 (respecively). Hence k = -5, -1, 4 or 8 (respectively). But k must be non-negative, so n = 7 or 14 is a necessary condition. We need to check that these values can be achieved. Take n = 7, so there are 9 players in total. If every match is a draw, then every player draws 8 matches and scores 4, which satisfies the conditions. Take n = 14, so there are 16 players in total. Suppose one player loses to everyone, and all the other games end in a draw. Then the first player scores 0 and all the other players score 1 + 14/2 = 8. That also satisfies the conditions. 8th CanMO 1976 © John Scholes jscholes@kalva.demon.co.uk 10 June 2002