5. In a football tournament, 10 teams participated, each playing against every other team exactly once. For a win, a team gets 3 points, for a loss 0 points, and in the case of a draw, each team gets 1 point. At the end, the total number of points for each team was calculated. Let $M$ be the highest of these numbers, and $m$ be the lowest of these numbers.

a) If the total number of points for all teams is 98, what is the maximum value that $M$ can have?

b) If the total number of points for all teams is 98 and $M=19$, what can $m$ be?

## Ministry of Science and Education of the Republic of Croatia

Agency for Education and Upbringing

Croatian Mathematical Society

## NATIONAL COMPETITION IN MATHEMATICS  Poreč, April 24 - 26, 2023.  6th grade - elementary school

Each task is worth 10 points. In addition to the final result, the procedure is also scored. To earn all points, it is necessary to find all solutions and confirm that there are no others, write down the procedure, and justify your conclusions.
5. In a football tournament, 10 teams participated, each playing against each other exactly once. For a win, a team gets 3 points, for a loss 0 points, and in the case of a draw, each team gets 1 point. At the end, the total number of points for each team was calculated. Let $M$ be the highest of these numbers, and $m$ the lowest of these numbers.

a) If the total number of points of all teams is 98, what is the maximum value that $M$ can take?

b) If the total number of points of all teams is 98 and $M=19$, what can $m$ be?

## First Solution.

If a match ends in a draw, the total number of points earned in that match is 2, and if it ends in a win for one team, it is 3.

If $x$ is the number of matches that ended in a draw, and $y$ is the number of matches that ended in a win for one team, then $2x + 3y = 98$.

The total number of matches is $9 + 8 + 7 + \cdots + 2 + 1 = 45$. Thus, $x + y = 45$, i.e., $x = 45 - y$.

| $y$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | $\mathbf{8}$ | 9 | 10 | $\ldots$ |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $x$ | 45 | 44 | 43 | 42 | 41 | 40 | 39 | 38 | 37 | 36 | 35 | $\ldots$ |  |  |
| $2x + 3y$ | 90 | 91 | 92 | 93 | 94 | 95 | 96 | 97 | $\mathbf{9 8}$ | 99 | 100 |  |  |  |

Increasing $y$ by 1 increases the total number of points $2x + 3y$ by 1.

Therefore, the only solution is $y = 8, x = 37$.

The total number of matches played by each team is 9.

If only 8 matches ended in a win for one team, then the team that achieved the maximum number of points could have at most 8 wins and 1 draw, i.e., $M = 8 \cdot 3 + 1 = 25$.

This can be achieved if that team won against all teams except one, and all other matches ended in a draw. Therefore, the answer to part a) is 25.

If $M = 19$, this means that the team with the most points, let's call it Team A, could have achieved:

I. 6 wins, 1 draw, and 2 losses

II. 5 wins and 4 draws

In the first case, the remaining 9 teams must have played all their matches against each other in a draw, as all matches that did not end in a draw were those involving Team A. This means that two of the remaining teams have 8 draws and 1 win, and six of the remaining teams have 8 draws and 1 loss, so $m = 8$.

In the second case, the remaining 9 teams have 3 wins and losses and 33 draws in their matches against each other. We consider the following scenarios.

If all three losses are by the same team, then that team has a total of three or four losses and five or six draws, depending on the result against A, so $m = 5$ or $m = 6$.

If two of the three losses are by the same team, then that team has a total of two or three losses and six or seven draws, depending on the result against A. Since the team with one of the three losses cannot have more than one additional loss against A, then $m = 6$ or $m = 7$.

If the three losses are by three different teams, then they have one or two losses (depending on the result against A), and since the other teams cannot have more than one loss (possibly against A), then $m = 7$ or $m = 8$.

Therefore, $m$ can be 5, 6, 7, or 8.

## NATIONAL MATHEMATICS COMPETITION

Poreč, April 24 - 26, 2023.