3. Consider a prime number $p$. We must design a pachisi tournament subject to the following rules.

- The tournament has $p^{2}$ players.
- In each game, $p$ players play.
- The tournament is divided into rounds. The rounds are divided into games. Each player plays one, or none, game in each round.
- At the end of the tournament, each player has faced each of the other players exactly once.

Determine if it is possible to design such a tournament. If so, find the minimum number of rounds the tournament can have.
Solution. The number of matches each player plays is $\frac{\text { number of players they face }}{\text { number of players they face in each match }}=\frac{p^{2}-1}{p-1}$.

That is, each player plays $p+1$ matches. Therefore, the number of rounds is at least $p+1$, and it is exactly $p+1$ when all players play in all rounds. In that case, $p$ matches are played in each round.

We will try to organize a tournament with $p+1$ rounds. We represent each player as a two-digit number written in base $p$. That is, we write each player in the form

$$
C_{i} C_{d}
$$

where $C_{i}$, the left digit, and $C_{d}$, the right digit, take integer values between 0 and $p-1$.

We make the following schedule.

- Round 0. Group the players by the same $C_{i}$.
- Round 1. Group the players by the same remainder of $C_{i}+C_{d}$ when divided by $p$.
- $\ldots$.
- Round $k$. Group the players by the same remainder of $C_{i}+k \cdot C_{d}$ when divided by $p$.
- $\ldots$
- Round $p-1$. Group the players by the same remainder of $C_{i}+(p-1) \cdot C_{d}$ when divided by $p$.
- Round $p$. Group the players by the same $C_{d}$.

In the arguments that follow, we will say that $D \equiv E$ if both numbers have the same remainder when divided by $p$.

We must observe that the proposed schedule groups the players into sets of $p$ elements. In rounds 0 and $p$, this is clear. Consider a round $k$ with $0<k<p$. For each fixed $C_{i}$, as $C_{d}$ moves, the remainder of $C_{i}+k \cdot C_{d}$ is always different (if not, we would have two players $C_{i} C_{d}, C_{i} C_{d}^{\prime}$ for which $C_{i}+k \cdot C_{d} \equiv C_{i}+k \cdot C_{d}^{\prime}$, and by subtracting we would reach that $k\left(C_{d}^{\prime}-C_{d}\right)$ is a multiple of $p$, without $k$ or $C_{d}^{\prime}-C_{d}$ being so). Therefore, we obtain each possible remainder once, and only once. By varying $C_{i}$, we obtain each remainder $p$ times.

Suppose two players, $C_{i} C_{d}$ and $C_{i}^{\prime} C_{d}^{\prime}$, face each other twice.

If they do so in rounds $0 \leq j<k \leq p-1$, we have that $C_{i}+k \cdot C_{d} \equiv$ $C_{i}^{\prime}+k \cdot C_{d}^{\prime}, C_{i}+j \cdot C_{d} \equiv C_{i}^{\prime}+j \cdot C_{d}^{\prime}$. By subtracting, $(k-j) \cdot C_{d} \equiv(k-j) \cdot C_{d}^{\prime}$. Since $k-j$ is coprime with $p, C_{d} \equiv C_{d}^{\prime}$. Thus, $C_{d}=C_{d}^{\prime}$. Bringing this equality to our hypothesis and subtracting, we easily obtain $C_{i} \equiv C_{i}^{\prime}$. Thus, $C_{i}=C_{i}^{\prime}$.

If the players face each other in rounds $k$ and $p$ with $0 \leq k<p$, we directly obtain that $C_{d}=C_{d}^{\prime}$, and we repeat the previous argument.