9.1. At the school for slackers, a competition on cheating and giving hints was organized. It is known that $75 \%$ of the students did not show up for the competition at all, and all the rest participated in at least one of the competitions. When the results were announced, it turned out that $10 \%$ of all those who showed up participated in both competitions, and that those who participated in the hinting competition were one and a half times more than those who participated in the cheating competition. Find the smallest possible number of students in the school for slackers.
Answer: 200.

Solution. Let the number of students in our school be $n$ people. $\frac{n}{4}$ people attended the competitions. Taking into account $\frac{n}{40}$ people who participated in both competitions, the number of participants in the competition with hints was $\frac{3}{5}\left(\frac{n}{4}+\frac{n}{40}\right)=\frac{33 n}{200}$. For this number to be an integer, given the mutual simplicity of the numbers 33 and 200, $n$ must be divisible by 200. In the competition involving copying, $\frac{2}{5}\left(\frac{n}{4}+\frac{n}{40}\right)=\frac{11 n}{100}$ people participated, and it is sufficient for $n$ to be divisible by 100. Considering everything, it is necessary and sufficient for $n$ to be divisible by 200, in particular, to be no less than 200.

Example: 200 students: 50 people participated in the competitions, 33 people in the competition with hints, and 22 people in the copying competition, with 5 people participating in both.

Criterion. Answer with verification: 2 points.

Correct formulas: 3 points.

Correct conclusions about the divisibility of $n$ by 200: 2 points.

Lack of an example with 200: minus 1 point.