Problem 9.8. A group of kids decided to play a computer game. Any two people either play together or against each other; moreover, if player $A$ plays together with $B$, and $B$ plays against $C$, then $A$ also plays against $C$. How many kids are in the group if each player had exactly 15 opponents? List all possible options.
Answer: $16,18,20,30$.

Solution. From the condition of the problem, it follows that all players are divided into several groups such that any two within a group play together, while any two people from different groups play against each other. For each player, the number of opponents is the difference between the total number of people in the company and the number of people in their group. Since this value is the same for all players and equals 15, we conclude that the number of people in each group is the same.

Let $n$ be the number of groups, and $m$ be the number of people in one group. Then the number of opponents for each person is

$$
(n-1) m=15
$$

There are exactly four ways to factorize the number 15 into an ordered product of two natural numbers: $15=1 \cdot 15=3 \cdot 5=5 \cdot 3=15 \cdot 1$. In each of these four cases, we find the pair $(n, m):(2,15),(4,5),(6,3),(16,1)$.

To complete the answer, we need to calculate the product $m n$ for each case and obtain four answers: $30,20,18,16$. All these answers are achieved in cases where the entire company of boys is divided into $n$ groups with $m$ players in each (for the corresponding $m$ and $n$).

## 10th grade