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This is a similar puzzle to

Enigma 239(there are only 3 teams in this version). But because of the smaller solution space an even simpler approach suffices (and is slightly faster with a slightly shorter program).This Python 3 program runs in 66ms

Solution:The scores in the matches are: A v B = 2-2; A v C = 4-2; B v C = 3-2.Analytically:

C cannot win or draw either of their games (A v C, B v C). Their 4 points can only come from goals scored.

As it is not possible to gain 20 or 21 points from a single game with no more than six goals scored it means that A must win A v C and B must win B v C and the other match (A v B) must be played. Neither side can win the A v B match and only have 20 or 21 points, so the match must be drawn.

So A and B have 15 points each from match outcomes, and C has 0 points. Their remaining points come from goals scored: A scores 6, B scores 5, C scores 4.

The score in the A v B match is one of: 0-0, 1-1, 2-2, 3-3.

Considering A v B = 0-0:

A’s goals must all be scored in the A v C match, so A must win that match 6-0. So the score in the B v C match must be 5-4, but this is not possible as no match has more than 6 goals scored in it. So A v B = 0-0 is not possible.

We can similarly eliminate A v B = 1-1 and A v B = 3-3.

This leaves A v B = 2-2, and the only possible scores for the other matches are A v C = 4-2, B v C = 3-2.