2. The teacher wrote a positive number $x$ on the board and asked Kolya, Petya, and Vasya to raise this number to the 3rd, 4th, and 12th power, respectively. It turned out that Kolya's number has at least 9 digits before the decimal point, and Petya's number has no more than 11 digits before the decimal point. How many digits are there before the decimal point in Vasya's number?
Answer: 33.

Solution. From the condition, it follows that $x^{3} \geqslant 10^{8}$, and $x^{4}<10^{11}$. Therefore, $10^{32} \leqslant x^{12}<10^{33}$. This means that the integer part of Vasya's number is a 33-digit number.

Evaluation. 13 points for the correct solution.