Problem 823

Answer: 36, 45 or 72.

Solution. Let $N$ be the two-digit number written on the board. If $N$ were divisible by 30, it would also be divisible by $3, 5, 10, 15$, so Neznaika would have made no more than 3 mistakes, a contradiction. If $N$ were not divisible by 3, it would also not be divisible by $9, 15, 18, 30$, so Neznaika would have made at least 5 mistakes, a contradiction. Therefore, $N$ is divisible by 3 but not by 30. It follows immediately that $N$ is not divisible by 10. Thus, among the statements about divisibility by $4, 5, 9, 15, 18$, exactly three are true. A number $N$ that is divisible by 3 cannot be divisible by both 4 and 5 at the same time (otherwise it would also be divisible by 30). Therefore, among the statements about divisibility by 9, 15, 18, at least two are true. If $N$ were not divisible by 9, it would also not be divisible by 18, a contradiction. Therefore, $N$ is divisible by 9. In addition, $N$ is divisible by 15 or 18. Let's consider two cases.

- Suppose $N$ is divisible by 15. Since it is also divisible by 9, it is either 45 or 90. But $N$ cannot be 90 because $N$ is not divisible by 30. However, $N$ can be 45 (since 45 is divisible by 3, 5, 9, 15 and not divisible by $4, 10, 18, 30$).
- Suppose $N$ is not divisible by 15. Then it is divisible by 18, and it is either 18, 36, 54, 72, or 90. Again, $N$ cannot be 90 because $N$ is not divisible by 30. Also, $N$ cannot be 18 or 54 (since both 18 and 54 are divisible by $3, 9, 18$ and both are not divisible by $4, 5, 10, 15, 30$). However, $N$ can be 36 or 72 (since both 36 and 72 are divisible by $3, 4, 9, 18$ and both are not divisible by $5, 10, 15, 30$).

In total, we have three possible options: 36, 45, and 72.