The encryption games involved 168 players in 50 teams, which had two to five members. The most were four-member teams, there were 20 three-member teams, and the games were attended by at least one five-member team. How many two-member, four-member, and five-member teams were there? (M. Mach)
There were 20 three-member teams, which represents 60 players. The remaining 108 players from the total number need to be divided into 30 teams of two, four, and five players. Four-member teams were the most, i.e., at least 21, which represents at least 84 players. The remaining 24 players need to be divided into 9 teams of two, four, and five players.

There was at least one five-member team. Additionally, from 24 players, a maximum of four five-member teams can be formed. Now let's consider all the cases that may arise:

- If there was exactly 1 five-member team, then 19 players remain to be divided into 8 teams of two and four players. This is not possible because 19 is an odd number.
- If there were 2 five-member teams, then 14 players remain to be divided into 7 teams of two and four players. This can only be realized in one way - all these players will be in two-member teams, and no additional four-member team will be formed.
- If there were 3 five-member teams, then 9 players remain to be divided into 6 teams of two and four players. This is not possible because at least 12 players are needed for 6 teams.
- If there were 4 five-member teams, then 4 players remain to be divided into 5 teams of two and four players. This is not possible because at least 10 players are needed for 5 teams.

From the above discussion, the only possibility is that the game was attended by 7 two-member, 20 three-member, 21 four-member teams, and 2 five-member teams.

Evaluation. 2 points for considering that it is sufficient to divide 24 players into 9 teams of 2, 4, and 5 players; 3 points for justifying that there could only be two five-member teams; 1 point for the correct result.