10.1. (12 points) The surface of a round table is divided into $n$ identical sectors, in which numbers from 1 to $n (n \geqslant 4)$ are written sequentially clockwise. Around the table sit $n$ players with numbers $1,2, \ldots, n$, going clockwise. The table can rotate around its axis in both directions, while the players remain in place. The players are seated at equal distances from each other, so when the table stops rotating, exactly one player is opposite each sector and receives the number of coins written on that sector. After $m$ rotations of the table, player №1 received 74 fewer coins than player №4, and player №2 received 50 fewer coins than player №3. Find $m$, given that player №4 received 3 coins twice as often as 2 coins, but half as often as 1 coin.
Answer: 69.

Solution. Let player No. 3 get one coin exactly $k$ times, then we get the equation $(m-k)-(n-1) k=50$. Exactly in $7 k$ cases, one coin was received by someone from players $2,3,4$, then we get the equation $3(m-7 k)-7 k(n-3)=74$. We expand the brackets, combine like terms, and get a system of two equations $m-n k=50,3 m-7 n k=74$. We multiply the first equation by 7, subtract the second from it, and get $m=69$.