Problem 9.6. A card game is played by five persons. In a group of 25 persons all like to play that game. Find the maximum possible number of games which can be played if no two players are allowed to play simultaneously more than once.

Ivailo Kortezov
9.6. The number of all pairs of players is $\frac{25.24}{2}=300$ and after each game 10 of them become impossible. Therefore at most $300: 10=30$ games are possible.

We shall prove that 30 games are possible. We denote the pairs of players by ( $m, n$ ), where $1 \leq m, n \leq 5$ are integers (in other words, we put them in a table $5 \times 5$ ).

In the game $i, 1 \leq i \leq 5$, we put the five pairs with $m=i$ (i.e. those from the $i$-th row of the table). In the game $6+5 k+i, 0 \leq i \leq 4,0 \leq k \leq 4$, we set
the pair ( $m, n$ ) such that $m k+n$ is congruent to $i$ modulo 5 . It is clear that for any fixed values of $k, i, m$ there exists a unique $n$ such that $m k+n \equiv i$ $(\bmod 5)$. Thus we have one pair in each row, i.e. the pairs are five and they have not played in the first 5 games.

For every two pairs $(m, n)$ and $\left(m^{\prime}, n^{\prime}\right), m^{\prime} \neq m$, the numbers $k\left(m-m^{\prime}\right)$, $k=0,1,2,3,4$, give different remainders modulo 5 . Hence there exists a unique $k$ such that $k\left(m-m^{\prime}\right) \equiv n-n^{\prime}(\bmod 5)$. Equivalently, $k m-n$ and $k m^{\prime}-n^{\prime}$ have the same remainder $i$ modulo 5 and the numbers $k$ and $i$ determine the unique game in which the pairs $(m, n)$ and $\left(m^{\prime}, n^{\prime}\right)$ participate.