Problem 8.3. In the class, there are 29 students: some are excellent students and some are troublemakers. Excellent students always tell the truth, while troublemakers always lie.

All students in this class sat around a round table.

- Several students said: “There is exactly one troublemaker next to me.”
- All other students said: “There are exactly two troublemakers next to me.”

What is the minimum number of troublemakers that can be in the class?
Answer: 10.

Solution. If there were three straight-A students in a row, the middle one would definitely lie. Therefore, among any three consecutive people, there must be at least one troublemaker.

Choose an arbitrary troublemaker. Assign this person the number 29, and number all the people following them clockwise from 1 to 28. Since in each of the non-overlapping groups $(1,2,3),(4,5,6), \ldots,(25,26,27),(29)$ there is at least one troublemaker, there must be at least $\frac{27}{3}+1=10$ troublemakers in total.

Note also that there could have been exactly 10 troublemakers. Again, numbering the people clockwise from 1 to 29, let the students with numbers $3,6,9,12,15,18,21,24$, 27, 29 be the troublemakers, and the students with all other numbers be straight-A students. In this case, all straight-A students except the one with number 28 said the first phrase, while the student with number 28 and all the troublemakers said the second phrase. It is easy to see that all the conditions of the problem are satisfied.