2-0. The number $n$ is such that $8n$ is a 100-digit number, and $81n$ is a 102-digit number. What can the second digit from the beginning of $n$ be?
Answer: 2

Solution. Since $8 n10^{101}$ (equality here is obviously impossible), it means $n>123 \cdot 10^{97}$. Therefore, the second digit from the beginning of the number $n$ is 2.