Exercise 8. In a tournament organized between 6 teams, each team plays against each other team exactly once. When a team wins, it gets 3 points, and the losing team receives 0 points. If the game is a draw, both teams receive 1 point. Determine the values of \( a \) for which it is possible that the final scores of the teams can be the six consecutive numbers \( a, a+1, \ldots, a+5 \)?
Solution to Exercise 8: In each match, between two and three points are awarded in total. Given that there are 15 matches in total, there must have been between 30 and 45 points distributed among the teams. Therefore, we must have:

$$
\begin{aligned}
& 30 \leqslant a+(a+1)+(a+2)+\cdots+(a+5) \leqslant 45 \\
& 30 \leqslant 6 a+15 \leqslant 45
\end{aligned}
$$

This means that \(a\) can only be 3, 4, or 5.

If \(a=5\), no match could have ended in a draw, so all scores are multiples of 3. It is therefore not possible to have 6 consecutive numbers for the scores, and this case is not possible.

If \(a=3\), the scores are 3, 4, 5, 6, 7, 8. The teams with 6 and 7 points have each won at least one match, and the team with 8 points has won at least 2 matches, so at least 4 matches did not end in a draw. Thus, the total number of points distributed among the teams is at least 34, but \(3+4+5+6+7+8=33\), so this case is not possible.

The case \(a=4\) is possible. The scores are 4, 5, 6, 7, 8, 9. Here is a table that works:

| $\mathrm{X}$ | 3 | 1 | 0 | 0 | 0 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | $\mathrm{X}$ | 1 | 0 | 3 | 1 |
| 1 | 1 | $\mathrm{X}$ | 3 | 0 | 1 |
| 3 | 3 | 0 | $\mathrm{X}$ | 1 | 0 |
| 3 | 0 | 3 | 1 | $\mathrm{X}$ | 1 |
| 3 | 1 | 1 | 3 | 1 | $\mathrm{X}$ |

The number written in the j-th column of the i-th row is the score obtained by team \(i\) in its match against team \(j\).

Comment from the graders: The exercise was very well done by the students who attempted it.