In a soccer tournament each team plays exactly one game with all others. The winner gets 3 points, the loser zero and each team gets 1 point in case of a draw.
It is known that teams () took part in a tournament and the final classification is given by an arithmetical progression of points, the last team having only 1 point.
a) Prove that this is not possible in the Championship of the Republic of Moldova (with ).
b) Find all values of and all configurations when this is possible.
, 2010
Solution
a) The total number of matches is . Let be the number of games ended with a victory and the number of games ended in a draw ( due to the last team). Thus, . If is the step of the arithmetical progression, we have that the total number of points in the final classification is
or
The case is obviously impossible (each team should have only 1 point). For we get , a contradiction (for one has 3 games, , all games ended in a draw, but the last team has only 1 point; the case implies ). If , then in contradiction with the fact that the total number of matches is .
Thus, the only possible value is . In this case and the number of points of each team in decreasing order is the sequence .
Denote by and by the number of victories and draws of the -th classified team (in decreasing order). Note that . Considering the number of points obtained by the first three teams, that is , we get , that is . Analogously, and .
For we obtain that , a contradiction with the fact that the total number of victories is . It implies that and it is impossible to have .
b) It is clear that such a configuration does not exist for ().
The case is possible, where the points in the final classification are realized by the following results of the matches of teams: T1-T2 (a draw), T1-T3 (T1 won), T1-T4 (T1 won), T2-T3 (T2 won), T2-T4 (a draw), T3-T4 (T3 won).
In this case we have to write 7 points of the third as a sum of at most five numbers of 1's, which is impossible.
Therefore, the only possibility is , the configuration being described above. ☐