5. In a football tournament where each team played against each other once, teams A, B, C, D, and E participated. For a win, a team received 3 points, for a draw 1 point, and for a loss 0 points. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\mathrm{E}$ could have?
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. There were a total of 15 matches, as is easily verified. Therefore, all teams together scored no more than $2 \cdot 3 + 3 \cdot 12 = 42$ points. Of these, teams A, B, C, D, and E scored 35 points. Therefore, team F scored no more than $42 - 35 = 7$ points. How it could have scored exactly 7 points is shown in the table on the right.

Comment: A realization is proposed - 3 points, an estimate is made - 4 points, points are summed. In the absence of a solution, 2-3 points for potentially useful ideas and approaches. Answer without justification - 0 points.