2. A 100-digit natural number had one of its digits replaced by zero (if it was the leading digit, it was simply erased). As a result, the number decreased by 13 times. Find all numbers for which this is possible.
Answer: $325 b \cdot 10^{97}$ for $b=1,2,3$.

Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m<10^{k}$. By replacing the digit $a$ with zero, we obtain the number $m+10^{k+1} n$. According to the condition,

$$
m+10^{k} a+10^{k+1} n=13\left(m+10^{k+1} n\right) \Longleftrightarrow 12 m=10^{k}(a-120 n)
$$

Note that $n=0$, otherwise $m<0$. Therefore, $a$ will be the leading digit of the original number, from which $k=99$. The number $a$ must be divisible by 3, that is, $a=3 b$ for $b \in\{1,2,3\}$. Then the equation becomes $4 m=10^{99} b$. Thus, $m=25 b \cdot 10^{97}$, and the original number is $325 b \cdot 10^{97}$. Since $m<10^{99}$, the number $b$ can take the values $1,2,3$.