4. In a chess tournament, two girls and several boys participated. Each participant played exactly one game with each other. The two girls together scored 6 points, and all the boys scored an equal number of points. How many boys could have participated in the tournament? (1 point is awarded for a win, $\frac{1}{2}$ point for a draw, and 0 points for a loss.)
Answer: 5 or 10.

Solution. Let $n$ be the number of boys who participated in the tournament, and each of them scored $k$ points. Then all the players together scored $6 + k n$ points. On the other hand, there were $n + 2$ participants in the tournament, and they played $\frac{(n+2)(n+1)}{2}$ matches. In each match, exactly one point was at stake. Therefore, the total number of points scored is equal to the number of matches played:

$$
6 + k n = \frac{(n+2)(n+1)}{2}
$$

This equation can be rewritten as

$$
n(3 + n - 2k) = 10
$$

From this, it follows that $n$ is a divisor of the number 10. From the problem statement, it follows that $n > 2$. Therefore, $n = 5$ or $n = 10$. We will show that both cases are possible.

For $n = 5$, we have $k = 3$, and the desired tournament can be achieved, for example, if all matches between the 7 participants ended in a draw.

For $n = 10$, we have $k = 6$, and the desired tournament can be achieved, for example, if one of the girls lost to all other participants in the tournament, and all matches between the remaining 11 participants ended in a draw.

Evaluation. 14 points for a correct solution. If the desired values of $n$ are found, but it is not shown that the corresponding tournaments actually exist, 7 points.

## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

## Final Stage

2017-2018 academic year

## Problems, Answers, and Evaluation Criteria

## 8th Grade  Variant 2

Physics