2. A 100-digit natural number had one of its digits erased (not the leading one). As a result, the number decreased by 13 times. Find all numbers for which this is possible.
Answer: $1625 \cdot 10^{96}, 195 \cdot 10^{97}, 2925 \cdot 10^{96}$, and $13 b \cdot 10^{98}$ for $b=1,2,3$.

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

$$
m+10^{k} a+10^{k+1} n=13\left(m+10^{k} n\right) \Longleftrightarrow 12 m=10^{k}(a-3 n)
$$

The digit $a$ must be divisible by 3, so $a=3 b$ for $b \in\{1,2,3\}$. Then $4 m=10^{k}(b-n)$. The digit $a$ is not the leading digit of the original number, so $n>0$. Note also that $n \leqslant 3$, otherwise $m<0$. Therefore, $n$ will be the leading digit of the original number, from which $k=98$. Let's consider three cases.

1) $b=n$. Then $m=0$, and the original number is $13 b \cdot 10^{98}$.
2) $b=n+1$. Then $m=25 \cdot 10^{96}$, the pair $(b, n)$ is $(2,1)$ or $(3,2)$, and the original number is $1625 \cdot 10^{96}$ or $2925 \cdot 10^{96}$, respectively.
3) $b=n+2$. Then $m=5 \cdot 10^{97},(b, n)=(3,1)$, and the original number is $195 \cdot 10^{97}$.