3-ча 1. In a chess tournament, students from grades IX and X participated. There were 10 times more students from grade X than from grade IX, and they scored 4.5 times more points in total than all the students from grade IX. How many points did the students from grade IX score? Find all solutions.
Solution 1. Let $x$ be the number of ninth graders in the tournament. Then there were a total of $11x$ participants, and they scored $\frac{11x(11x-1)}{2}$ points. According to the problem, the ratio of the number of points scored by the ninth graders to the number of points scored by the tenth graders is $1:4.5$. Therefore, the ninth graders scored $x(11x-1)$ points, which means each ninth grader won all $11x-1$ games they played. However, if there were two ninth graders among the participants, they would have had to both win their game against each other, which is impossible. Therefore, only one ninth grader participated in the tournament; he scored 10 points.

Part 2. Given the sequence of numbers: $0,1,1,2,3,5,8,13,21,34,55,89, \ldots$, where each number, starting from the third, is the sum of the two preceding numbers. Will there be a number among the first hundred million ($10^8 + 1$) terms of this sequence that ends with four zeros?

Solution 2. Answer: Yes, there will be. Replace each of the given numbers with its remainder when divided by 1000. Let $a_{1}=0, a_{2}, \ldots$ be the resulting numbers. If we know the numbers $a_{k}$ and $a_{k+1}$, then we also know $a_{k-1}$, since in the original sequence the $(k-1)$-th term is the difference between the $(k+1)$-th and $k$-th terms. Therefore, if for some $k$ and $n$ the equalities $a_{k}=a_{k+n}$ and $a_{k+1}=a_{k+n+1}$ hold, then $a_{k-1}=a_{k+n-1}, a_{k-2}=a_{k+n-2}, \ldots, a_{1}=a_{n+1}$. But $a_{1}=0$, so $a_{n+1}=0$, i.e., in the original sequence, the number at the $(n+1)$-th position ends with four zeros.

It remains to prove that among the pairs $\left(a_{1}, a_{2}\right),\left(a_{2}, a_{3}\right), \ldots,\left(a_{10^{8}}, a_{10^{8}+1}\right),\left(a_{10^{8}+1}, a_{10^{8}+2}\right)$, there will be two identical pairs. But from the numbers $0,1,2, \ldots, 9999$, it is impossible to form more than $10^8$ different pairs, while we are considering $10^8 + 1$ pairs.