10.5. In a chess tournament, 10th graders and 9th graders participated. There were 9 times more 10th graders, and they scored 4 times more points than the 9th graders. Which class did the winner of the tournament belong to, and how many points did he score? In chess, 1 point is awarded for a win, 0.5 points for a draw, and 0 points for a loss.
Answer: 9th grader, 9 points.

Solution. Let there be $x$ 9th graders and $9x$ 10th graders in the tournament. Estimate the maximum number of points that 9th graders could score and the minimum number of points that 10th graders could score. This would happen if all 9th graders won their matches against 10th graders. Then the 9th graders would collectively score $\frac{x(x-1)}{2}$ points in matches among themselves and $x \cdot 9x$ in matches against 10th graders. The 10th graders would collectively score $\frac{9x(9x-1)}{2}$.

From the condition, it follows that $\left(9x^2 + \frac{x(x-1)}{2}\right) \cdot 4 \geq \frac{9x(9x-1)}{2} \Leftrightarrow x \leq 1$.

Therefore, the only possible situation is when one 9th grader participated in the tournament and won all matches against 10th graders, that is, scored 9 points.

Comment. Only an example of one tournament - 1 point.

10.6-1. A polynomial $P(x)$ with integer coefficients has an integer root. Can the product $P(0) \cdot P(1)$ end in 4321?

Answer: No.

Solution. Let $x_0$ be an integer root of $P(x)$, that is, $P(x_0) = 0$. Since the polynomial $P(x)$ has integer coefficients, $P(x_0 + 2k)$ is an even number (this follows from the fact that $(x_0 + 2k)^m - x_0^m$ is divisible by $(x_0 + 2k) - x_0 = 2k$). Therefore, at least one of the numbers $P(0), P(1)$ is even, and thus the product $P(0) \cdot P(1)$ is even and cannot end in 4321.

Comment. Correct answer without justification - 0 points.

10.6-2. A polynomial $P(x)$ with integer coefficients has an integer root. Can the product $P(1) \cdot P(2)$ end in 2021?

Answer: No.

Solution. Let $x_0$ be an integer root of $P(x)$, that is, $P(x_0) = 0$. Since the polynomial $P(x)$ has integer coefficients, $P(x_0 + 2k)$ is an even number (this follows from the fact that $(x_0 + 2k)^m - x_0^m$ is divisible by $(x_0 + 2k) - x_0 = 2k$). Therefore, at least one of the numbers $P(1), P(2)$ is even, and thus the product $P(1) \cdot P(2)$ is even and cannot end in 4321.

Comment. Correct answer without justification - 0 points.