## Task 4 - 140524

Students of a 5th grade class held a 14-day chess tournament. Exactly 6 games were played each of the 14 days.

The number of participating boys was greater than that of the participating girls. Each girl played against every other girl and each boy against every other boy exactly twice. None of the girls played against a boy.

Determine the number of girls and the number of boys who participated in this tournament!
According to the problem, because $14 \cdot 6 = 84$, a total of 84 games were played in this tournament. This number is the sum of the number of games played between the girls and the number of games involving the boys.

For $n$ players, where each player plays exactly two games against each of the other $(n - 1)$ players, the total number of games is $n(n-1)$.

Thus, the following table can be constructed:

| Number of Players | Number of Games | Complement to 84 |
| :---: | :---: | :---: |
| 2 | 2 | 82 |
| 3 | 6 | 78 |
| 4 | 12 | 72 |
| 5 | 20 | 64 |
| 6 | 30 | 54 |
| 7 | 42 | 42 |
| 8 | 56 | 28 |
| 9 | 72 | 12 |
| $\geq 10$ | $\geq 90$ | - |

Since a total of 84 games were played according to the problem, the number of participating boys or girls could not have been greater than 9.

The summands can only be the numbers determined in the table, and they must be exactly two numbers of the form $n(n-1)$ whose sum is 84. This is only possible for $42 + 42 = 84$ and $12 + 72 = 84$, as a comparison of the numbers in the second and third columns of the table shows.

Since the number of participating boys was greater than the number of girls, the solution must be:

4 girls and 9 boys participated in the tournament mentioned in the problem.

Solutions of the II. Round 1974 taken from [5]

### 2.16 XV. Olympiad 1975

### 2.16.1 I. Round 1975, Class 5