3. In a football tournament, seven teams played: each team played once with each other. Teams that scored thirteen or more points advance to the next round. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round?
Solution. Evaluation. Note that the total number of games will be $\frac{7 \cdot 6}{2}=21$. Then the total number of points does not exceed $21 \cdot 3=63$. This means that no more than $\left[\frac{63}{13}\right]=4$ teams can advance to the next round.

Example. We will show that 4 teams can advance to the next round. For this, we will provide an example of a suitable tournament table (the symbol "x" denotes insignificant results):

| $№$ | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{1}$ | - | 1 | 3 | 3 | 3 | 3 | 0 |
| $\mathbf{2}$ | 1 | - | 0 | 3 | 3 | 3 | 3 |
| $\mathbf{3}$ | 0 | 3 | - | 3 | 3 | 3 | 1 |
| $\mathbf{4}$ | 0 | 0 | 0 | - | $\mathrm{x}$ | $\mathrm{x}$ | 0 |
| $\mathbf{5}$ | 0 | 0 | 0 | $\mathrm{x}$ | - | $\mathrm{x}$ | 0 |
| $\mathbf{6}$ | 0 | 0 | 0 | $\mathrm{x}$ | $\mathrm{x}$ | - | 0 |
| $\mathbf{7}$ | 3 | 0 | 1 | 3 | 3 | 3 | - |

Answer: 4.

Comment. The correct evaluation is obtained - no more than 3 points.