Scores of six soccer matches

Question

In the first round of the city soccer tournament, the teams in group A finished as follows:

Team        Goals For           Goals Against           Points
Naranja     4                   2                       7
Bleu        5                   2                       6
Midori      1                   3                       3
Gelb        1                   4                       1

A victory earns three points, a tie one point, and a loss no points. Each team played the other three once. What were the scores of all the matches?

What I'm doing is making different combinations and doing guess and check. Is there a simpler way to do this (mathematical way)?

Answer 1

First, figure out how many games each team won, lost, and tied, based only on their total points (the third column). For example, Naranja has 7 points. How can you make 7 points by adding up 0's, 1's, and 3's?

Then use that, plus the number of goals scored, to deduce the goals for / goals against for each team's three games.

Then pair up the teams that were in matches with the same scores.

Answer 2

This can be done pretty systematically. Since N has 7 points, they must have won twice and tied once. Since B has 6 points, they must have won twice and lost once. Since G has one point, they must have tied once and lost twice. As for M with three points, they could have either tied thrice or won once and lost twice. Since B never tied any game, M cannot have tied thrice. So, we know:

• N: WWT
• B: WWL
• M: WLL
• G: TLL

Since only one tie occurred, N and G must have tied, so N beat B and M. This accounts for B's single loss, so B beat M and G. This accounts for M's two losses, so M beat G.

As for scores, we begin by noting that M only has one goal for, so it must have been during their winning game against G: M 1-0 G. Since M has three goals against, their other two games were 0-1 and 0-2, but we don't yet know which against whom. G's tie with N was either 0-0 or 1-1, and G's loss against B was either 1-3 or 0-2. Summarizing, we have four cases:

1. M 0-1 N and M 0-2 B and G 0-0 N and G 1-3 B
2. M 0-2 N and M 0-1 B and G 0-0 N and G 1-3 B
3. M 0-1 N and M 0-2 B and G 1-1 N and G 0-2 B
4. M 0-2 N and M 0-1 B and G 1-1 N and G 0-2 B

Case 1.
B scored 5 goals against G and M, hence no goals on N. But N ceded 2 goals from B because it shut out M and G. Contradiction!

Case 2.
Similarly, G and M scored 0 against N, so N suffered 2 goals from B. But B scored 4 against G and M, so it scored 1 on N. Contradiction!

Case 3.
B scored 4 goals against G and M, hence 1 goal on N. B allowed no goals from G and M, so it allowed 2 from N. Conversely, N scored 2 goals against G and M, hence 2 goals on B. And N lost one goal to G, so it lost the other to B. Consistent!

Case 4.
B scored 3 goals against G and M, hence 2 goals on N. But N let in a goal against G and M, so it must have let only 1 in against B. Contradiction!

Thus, there is one solution:

• N 2-1 B
• N 1-0 M
• N 1-1 G
• B 2-0 M
• B 2-0 G
• M 1-0 G.

And indeed, we see that N has 4 for, 2 against; B has 5 for, 2 against; M has 1 for, 3 against, and G has 1 for, 4 against.