3. In a tournament with six teams, each team plays once against every other team. If a team wins a match, it receives 3 points and the loser gets 0 points. If a match ends in a draw, both teams receive 1 point.

Can the final scores of the teams be exactly six consecutive numbers $a, a+1, \ldots, a+5$? If so, determine all values of $a$ for which this is possible.
3. In total, 15 matches are played. In each match, a total of 2 or 3 points are earned. The sum of the six scores therefore lies between $15 \cdot 2=30$ (all matches end in a draw) and $15 \cdot 3=45$ (no match ends in a draw).

The sum of the six scores is also equal to $a+(a+1)+\cdots+(a+5)=15+6a$. We see that $30 \leqslant 15+6a \leqslant 45$, so $3 \leqslant a \leqslant 5$. We will prove that only $a=4$ is possible.

First, consider the case $a=5$. The total number of points is then $15+30=45$, so no match ends in a draw. In each match, a team earns either 0 or 3 points, so all scores are divisible by 3. The scores cannot therefore be six consecutive numbers.

Now consider the case $a=3$. The sum of the scores is $3+4+5+6+7+8=33$. The teams with 6 and 7 points have each won at least one of their five matches.

The team with 8 points has even won at least twice, because $3+1+1+1+1=7<8$. Therefore, there are at least 4 matches that did not end in a draw, so the sum of the scores is at least $4 \cdot 3+11 \cdot 2=34$. But we already saw that this sum is equal to 33. Therefore, the case $a=3$ cannot occur.

Finally, we show that $a=4$ is possible: in the table below, you see one of the possible game outcomes where teams $A$ through $F$ have scores from 4 to 9. The last column shows the scores of the six teams.

|  | $A$ | $B$ | $C$ | $D$ | $E$ | $F$ |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | - | 3 | 1 | 0 | 0 | 0 | 4 |
| $B$ | 0 | - | 1 | 0 | 3 | 1 | 5 |
| $C$ | 1 | 1 | - | 3 | 0 | 1 | 6 |
| $D$ | 3 | 3 | 0 | - | 1 | 0 | 7 |
| $E$ | 3 | 0 | 3 | 1 | - | 1 | 8 |
| $F$ | 3 | 1 | 1 | 3 | 1 | - | 9 |