2. A three-digit number $X$ was written with three different digits $A B C$. Four schoolchildren made the following statements. Petya: “The largest digit in the number $X$ is $B$”. Vasya: “$C=8$”. Tolya: “The largest digit is $C$”. Dima: “$C$ is the arithmetic mean of $A$ and $B$”. Find the number $X$, given that exactly one of the children is wrong.
Answer: 798. Solution. If Tolya told the truth and the largest number is $-C$, then immediately two people (Petya and Dima) would be wrong, which cannot be the case, so Tolya must be wrong, and the others told the truth. Thus, the number $C=8 ; B$ is the largest number, i.e., $B=9 ; C$ is the average of numbers $A$ and $B$, i.e., $A=2 C-B=7$

Grading criteria: Full solution - 7 points. Proven that only Tolya could be wrong - $\mathbf{4}$ points. Only answer - $\mathbf{2}$ points.