9.5. In a round-robin chess tournament, two boys and several girls participated. The boys scored a total of 8 points, while all the girls scored an equal number of points. How many girls could have participated in the tournament? (Win - 1 point, draw - 0.5 points, loss - 0 points.)
Answer: 7 or 14.

Solution. Let $n$ be the number of girls participating in the tournament and each of them scored $x$ points, then the total number of points scored is $n x+8$. On the other hand, each of the $n+2$ chess players played $n+1$ games, so the total number of games played in the tournament is $\frac{(n+2)(n+1)}{2}$.

Since one point is awarded in each game, the number of points awarded is equal to the number of games played. Therefore, $n x+8=\frac{(n+2)(n+1)}{2}$.

Transforming this equation, we get: $2 x=n+3-\frac{14}{n}$. Since twice the number of points scored by each girl is an integer, the number of girls must be a natural divisor of the number 14, that is, 1, 2, 7, or 14.

For $n=1$ or $n=2$, the right-hand side of the obtained equation is negative, which is impossible.

If $n=7$, then $x=4$. This situation is possible, for example, if all games in the tournament ended in a draw. Then each participant scored 4 points.

If $n=14$, then $x=8$. This situation is also possible, for example, if one boy scored 0 points (lost all games), and the rest of the games ended in a draw. Then the second boy and each girl scored 8 points.

Grading criteria.

“+” A complete and justified solution is provided

“士” It is proven that the number of girls could be 7 or 14, but examples of their realization are missing

“士” It is proven that the number of girls is a divisor of 14, but only one of the possible answers is indicated and an example of its realization is provided

“Ғ” Both possible answers are provided and it is verified that they satisfy the condition

“-” Only the answer is provided

“-” The problem is not solved or solved incorrectly