4. In a chess tournament, two girls and several boys participated. Each participant played exactly one game with each other. The two girls together scored 8 points, and all the boys scored an equal number of points. How many boys could have participated in the tournament? (1 point is awarded for a win, $\frac{1}{2}$ point for a draw, and 0 points for a loss.)
Answer: 7 or 14.

Solution. Let $n$ be the number of boys participating in the tournament, and each of them scored $k$ points. Then all the players together scored $8 + kn$ points. On the other hand, there were $n + 2$ participants in the tournament, and they played $\frac{(n+2)(n+1)}{2}$ matches. In each match, exactly one point was at stake. Therefore, the total number of points scored is equal to the number of matches played:

$$
8 + kn = \frac{(n+2)(n+1)}{2}
$$

This equation can be rewritten as

$$
n(3 + n - 2k) = 14
$$

From this, it follows that $n$ is a divisor of the number 14. From the problem statement, it follows that $n > 2$. Therefore, $n = 7$ or $n = 14$. We will show that both cases are possible.

For $n = 7$, we have $k = 4$, and the desired tournament can be achieved, for example, by assuming that all matches between the 9 participants ended in a draw.

For $n = 14$, we have $k = 8$, and the desired tournament can be achieved, for example, by assuming that one of the girls lost to all other participants in the tournament, and all matches between the remaining 15 participants ended in a draw.

Evaluation. 14 points for a correct solution. If the desired values of $n$ are found, but it is not shown that the corresponding tournaments actually exist, 7 points.

## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

## Final Stage

2017-2018 academic year

## Problems, Answers, and Evaluation Criteria

## 8th Grade Variant 1

 physics5. A small load hangs in the air on a spring. When this load is fully submerged in water on the same spring, the magnitude of the spring's deformation remains the same. Determine the density of the material of the load. The density of water $\rho = 1000$ kg $/ \mathrm{m}^{3}$. (15 points)

Answer: $500 \mathrm{kg} / \mathrm{m}^{3}$

Solution. In the air, the spring is stretched: $kx = mg$ (4 points). In the water, the spring is compressed: $mg + kx = F_{A} = \rho_{\text{liquid}} g V$ (4 points). Since the mass of the load is: $m = \rho_{2} V$ (3 points), we get: $\rho_{2} = \frac{\rho_{\text{liquid}}}{2} = \frac{1000}{2} = 500 \mathrm{kg} / \mathrm{m}^{3}$ (4 points).