In a chess tournament, two 7th-grade students and a certain number of 8th-grade students participated. The two 7th-graders scored 8 points, and each of the 8th-graders scored the same number of points. How many 8th-graders participated in the tournament? (Each participant plays one game with each of the others. A win earns 1 point, a draw earns 1/2 point, and a loss earns 0 points.)

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Let $x$ be the number of eighth graders, $y$ be the number of points scored by each eighth grader. By calculating the total points scored by all participants in the tournament in two ways, we arrive at the equation $x y + 8 = \frac{(x+2)(x+1)}{2}$, which simplifies to $2 y = \frac{(x+2)(x+1)-16}{x} = x + 3 - \frac{14}{x}$.

Therefore, $x$ can take one of the values $1, 2, 7, 14$. The values 1 and 2 are eliminated, as in these cases $y < 0$.

When $x = 7$, we get $y = 4$. Such a tournament is possible: for example, all matches ended in a draw; then each participant scored 4 points.

When $x = 14$, we get $y = 8$. Such a tournament is also possible: for example, one of the seventh graders lost all their matches, and all other matches ended in a draw; then each participant, except the last one, scored 8 points.

## Answer

7 or 14.