Four teams competed in a football tournament where each played once against each of the others. When a match ended in a draw, each team received one point; otherwise, the winner received three points and the loser, zero. The table shows the final score of the tournament. How many were the draws?

| Team | Points |
| :--- | :---: |
| Cruzínthians | 5 |
| Flameiras | 3 |
| Nauritiba | 3 |
| Greminense | 2 |

A) 2
B) 3
C) 4
D) 5
E) 6
Solution

## ALTERNATIVE D

$1^{st}$ solution: Each team played three times. With 5 points, Cruzinthians could only have won one match and drawn two, because if they had won two matches, they would have at least 6 points, and if they hadn't won any, they would have a maximum of 3 points. Greminense did not win any matches, as they only scored 2 points; thus, they drew two matches and lost one. Flameiras, in second place with 3 points, did not win any matches, because if this had happened, they would have had to lose two; since Greminense did not win any and Cruzinthians only one, they would have lost to Nauritiba. But the same reasoning shows that then Nauritiba, having won the match against Flameiras, should have lost to Flameiras! Since this cannot happen, we conclude that Flameiras and Nauritiba drew all three of their matches. Therefore, the number of draws was $3+3-1=5; the -1$ appears in this expression because the draw between Flameiras and Nauritiba should be counted only once. The table below shows the championship scoring.

|  | Cruzinthians | Flameiras | Nauritiba | Greminense |
| :---: | :---: | :---: | :---: | :---: |
| Points scored by Cruzinthians |  | 1 | 1 | 3 |
| Points scored by Flameiras | 1 |  | 1 | 1 |
| Points scored by Nauritiba | 1 | 1 |  | 1 |
| Points scored by Greminense | 0 | 1 | 1 |  |

$2^{nd}$ solution: Another solution is to note that in each match played, 2 points are distributed in the case of a draw or 3 points if there is no draw. Since each of the four teams played once against each of their three opponents, a total of six matches were played, in which $5+3+3+2=13$ points were distributed. The only way to partition 13 into six parts of 2 or 3 is $13=3+2+2+2+2+2$; thus, five of the six matches ended in a draw.