Two precocious students from Level 3 participated in a university chess tournament. Each participant plays against all the others exactly once. A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0 points. The sum of the scores of the two Level 3 students is 6.5. All university students scored the same. How many university students participated in the competition?
Solution

Let $x$ be the number of college students and $p$ the common score of all of them. Since exactly 1 point is contested in each game, it follows that the total score of the tournament, which is $6.5 + p x$, coincides with the number of games, which is $\frac{(x+2)(x+1)}{2}$. In addition, the score of each participant is an integer multiple of 0.5, so we can write $p = k / 2$, for some positive integer $k$. Therefore:

$$
\begin{aligned}
\frac{(x+2)(x+1)}{2} & = 6.5 + p x \\
(x+2)(x+1) & = 13 + k x \\
x^2 + 3x + 2 & = 13 + k x \\
x(x + 3 - k) & = 11
\end{aligned}
$$

Since $x$ and $x + 3 - k$ are integers, we can conclude that $x$ is a positive divisor of 11, that is, $x = 1$ or $x = 11$. We cannot have $x = 1$, because in that case the tournament would have only 3 participants, and it would not be possible for two students to score 6.5 points. To show that $x = 11$ is a solution, consider the tournament formed by the college students $U_1, U_2, \ldots, U_{11}$ and the Level 3 students $E_1$ and $E_2$ with the following results:

I) All games between two college students ended in a draw.

II) $E_1$ lost to $U_1, U_2, \ldots, U_{11}$ and $E_2$.

III) $E_2$ drew with $U_1, U_2, \ldots, U_{11}$.

The tournament with these results satisfies the statement.