11.6. In a chess tournament, 11th graders and 10th graders participated. There were 10 times more 11th graders, and they scored 4.5 times more points than the 10th graders. Which class did the winner of the tournament belong to, and how many points did they score? In chess, 1 point is awarded for a win, 0.5 points for a draw, and 0 points for a loss.
Answer: 10th-grader, 10 points.

Solution: Let $x$ be the number of 10th-graders and $10x$ be the number of 11th-graders in the tournament. We will estimate the maximum number of points that 10th-graders could have scored and the minimum number of points that 11th-graders could have scored. This will happen if all 10th-graders won their matches against 11th-graders. Then, 10th-graders would collectively score $\frac{x(x-1)}{2}$ points in matches among themselves and $x \cdot 10x$ points in matches against 11th-graders. Meanwhile, 11th-graders would collectively score $\frac{10x(10x-1)}{2}$ points.

From the condition, it follows that $\left(10x^2 + \frac{x(x-1)}{2}\right) \cdot 4.5 \geq \frac{10x(10x-1)}{2} \Leftrightarrow x \leq 1$.

Therefore, the only possible scenario is that one 10th-grader participated in the tournament and won all matches against 11th-graders, thus scoring 10 points.

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

11.7-1. A polynomial $P(x)$ with integer coefficients has an integer root. Can the product $P(0) \cdot P(1) \cdot P(2) \cdot P(3)$ end in 1010?

Answer: No.

Solution: Let $x_0$ be an integer root of $P(x)$, i.e., $P(x_0) = 0$. 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 two of the numbers $P(0), P(1), P(2), P(3)$ are even, which means the product $P(0) \cdot P(1) \cdot P(2) \cdot P(3)$ is divisible by 4 and cannot end in 1010.

Comment: Correct answer without justification - 0 points.

11.7-2. A polynomial $P(x)$ with integer coefficients has an integer root. Can the product $P(2) \cdot P(3) \cdot P(4) \cdot P(5)$ end in 2030?

Answer: No.

Solution: Let $x_0$ be an integer root of $P(x)$, i.e., $P(x_0) = 0$. 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 two of the numbers $P(2), P(3), P(4), P(5)$ are even, which means the product $P(2) \cdot P(3) \cdot P(4) \cdot P(5)$ is divisible by 4 and cannot end in 2030.

Comment: Correct answer without justification - 0 points.