GK 2019 Q23

Let the largest number of teams that can enter the tournament be \(n\). Hence there would be \(3n\) players in total. Each player in a team will play a game against every player from all other teams and hence the total number of games played is \(\tfrac{3n \times (3n - 3)}{2} = \tfrac{9n(n - 1)}{2}\). Since no more than 250 games can be played, we have \(\tfrac{9n(n - 1)}{2} \leqslant 250\). Therefore \(n(n - 1) \leqslant \tfrac{500}{9} = 55\tfrac{5}{9}\). Since \(8 \times 7 = 56 > 55\tfrac{5}{9}\), we have \(n < 8\). Also, since \(7 \times 6 = 42 < 55\tfrac{5}{9}\), \(n = 7\) is a possible solution. Therefore the largest number of teams that can enter the tournament is 7.