5. In a football tournament where each team played against each other once, teams $A$, B, V, G, D, and E participated. Teams received 3 points for a win, 1 point for a draw, and 0 points for a loss. In the end, it turned out that teams $A$, B, V, G, and D each scored 7 points. What is the maximum number of points that team E could have scored?
Answer: 7 points. Solution: In a match where one of the teams won, the teams together score 3 points, in a match that ended in a draw - 2 points. Since 7 is not divisible by 3, the team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament. In total, as is easy to check, 15 matches were played. Therefore, all teams together scored no more than $2 \times 3 + 3 \times 12 = 42$ points. Of these, 35 points were scored by teams A, B, V, G, and D. Therefore, team $\mathrm{E}$ scored no more than 42-35 = 7 points. How it could-

|  | $A$ | $B$ | $B$ | $\Gamma$ | $D$ | $E$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | $\mathrm{x}$ | 3 | 3 | 1 | 0 | 0 |
| $B$ | 0 | $\mathrm{x}$ | 3 | 3 | 1 | 0 |
| $B$ | 0 | 0 | $\mathrm{x}$ | 3 | 3 | 1 |
| $\Gamma$ | 1 | 0 | 0 | $\mathrm{x}$ | 3 | 3 |
| $D$ | 3 | 1 | 0 | 0 | $\mathrm{x}$ | 3 |
| $E$ | 3 | 3 | 1 | 0 | 0 | $\mathrm{x}$ |

have scored exactly 7 points is shown in the table on the right.