9.2. During breaks, schoolchildren played table tennis. Any two schoolchildren played no more than one game with each other. By the end of the week, it turned out that Petya played half, Kolya - a third, and Vasya - a fifth of the total number of games played during the week. How many games could have been played during the week, given that at least in two games neither Vasya, Petya, nor Kolya participated?
Answer: 30.

Solution: From the condition, it follows that half, a third, and a fifth of the total number of games played are integers. Since the least common multiple of the denominators - the numbers 2, 3, 5 - is 30, the total number of games played is also a multiple of 30. Let this number be $30p$. Then, Pete played $15p$, Kolya $10p$, and Vasya $6p$ games. Let $y$ be the number of games played among Pete, Kolya, and Vasya (this number is 0, 1, 2, or 3), and $z$ be the number of games played without the participation of Pete, Kolya, and Vasya $(z \geqslant 2)$. Then $15p + 10p + 6p - y + z = 30p$, which means $p = y - z$. The only possible positive value of $p$ is 1, and it is achieved when $y = 3, z = 2$.

Remark: The condition is consistent: such a situation could indeed have occurred (i.e., 30 games could have been played according to the condition).

Comment: It is proven that the total number of games is divisible by $30 - 2$ points.

An example with 30 games is provided, but it is not justified that there are no other options -3 points.

It is not necessary to provide an example with 30 games in the solution (it is known from the condition that such games were played).