1. (10 points) Chess clubs from Moscow, Saint Petersburg, and Kazan agreed to hold a tournament. Each Muscovite played exactly 9 Saint Petersburg residents and $n$ Kazan residents. Each Saint Petersburg resident played exactly 6 Muscovites and 2 Kazan residents. Each Kazan resident played exactly 8 Muscovites and 6 Saint Petersburg residents. What is $n$?
# Answer: 4.

Solution. Let the team from Moscow consist of $m$ participants, the team from Saint Petersburg - of $p$ participants, and the team from Kazan - of $k$ participants.

According to the problem, each Muscovite, i.e., each of the $m$ people, played exactly 9 games with the Saint Petersburg residents; and each Saint Petersburg resident, i.e., each of the $p$ people, played exactly 6 games with the Muscovites. Clearly, the games played by Muscovites with Saint Petersburg residents are the same games played by Saint Petersburg residents with Muscovites, i.e., $m \cdot 9 = p \cdot 6$.

Similarly, $m \cdot n = k \cdot 8$ and $p \cdot 2 = k \cdot 6$. We have

$$
\left\{\begin{array} { l } 
{ 9 m = 6 p , } \\
{ n m = 8 k , } \\
{ 2 p = 6 k ; }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 3 m = 2 p , } \\
{ n m = 8 k , } \\
{ p = 3 k ; }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 3 m = 2 \cdot 3 k , } \\
{ n m = 8 k , } \\
{ p = 3 k ; }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ m = 2 k , } \\
{ n m = 8 k , } \\
{ p = 3 k ; }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
m=2 k \\
n \cdot 2 k=8 k \\
p=3 k
\end{array}\right.\right.\right.\right.\right.
$$

From this, $n=4$.