Task B-2.5. At a chess tournament, three first-grade students and several second-grade students participated. The three first-grade students won a total of 7 points, and each second-grade student won the same number of points. How many second-grade students were at the tournament if each win brings one point, a loss 0 points, and an unresolved result half a point? The tournament is played so that each player plays one game with each of the remaining players.
## Solution.

Let $x$ be the number of second-grade students participating in the tournament, and let each of them have scored $y$ points. Then the total number of points collected by the students of both the first and second grades is $x y + 7$. Since there are $x + 3$ students participating in the tournament, the total number of points scored is $\frac{(x+3)(x+2)}{2}$. Therefore, we have

$$
\begin{aligned}
& x y + 7 = \frac{(x+3)(x+2)}{2} \\
& 2 x y + 14 = (x+3)(x+2) \\
& 2 y = x + 5 - \frac{8}{x}
\end{aligned}
$$

Since $x, 2 y \in \mathbb{N}$, it follows that $x \in \{2, 4, 8\}$. Therefore, there were 2, 4, or 8 second-grade students at the tournament.

Let's show that the obtained solutions are possible. Denote the first-grade students as $A, B$, and $C$, and the second-grade students as $1, 2, \ldots$ For each solution, the outcomes of the matches are given in the corresponding part of the following table (for $x=2$, we consider the upper left $5 \times 5$ table, for $x=4$, we consider the upper left $7 \times 7$ table, and for $x=8$, we consider the entire $11 \times 11$ table). The number in the row labeled $r$ and the column labeled $s$ represents the points that $r$ scored in the match against $s$.

|  | $A$ | $B$ | $C$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| $B$ | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| $C$ | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | $1 / 2$ | 1 | 1 | 1 | 1 | 0 | $1 / 2$ |
| 2 | 1 | 0 | 0 | $1 / 2$ | 0 | 1 | 1 | 1 | 0 | $1 / 2$ | 1 |
| 3 | 1 | 1 | 1 | 0 | 0 | 0 | $1 / 2$ | 0 | $1 / 2$ | 1 | 1 |
| 4 | 1 | 1 | 1 | 0 | 0 | $1 / 2$ | 0 | $1 / 2$ | 1 | 1 | 0 |
| 5 | 1 | 1 | 1 | 0 | 0 | 1 | $1 / 2$ | 0 | 1 | 0 | $1 / 2$ |
| 6 | 1 | 1 | 1 | 0 | 1 | $1 / 2$ | 0 | 0 | 0 | $1 / 2$ | 1 |
| 7 | 1 | 1 | 1 | 1 | $1 / 2$ | 0 | 0 | 1 | $1 / 2$ | 0 | 0 |
| 8 | 1 | 1 | 1 | $1 / 2$ | 0 | 0 | 1 | $1 / 2$ | 0 | 1 | 0 |

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - $B$ variant  Dubrovnik, April 3, 2013.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE AND GRADE THAT APPROACH APPROPRIATELY.