2. A 100-digit natural number had two adjacent digits erased. As a result, the number decreased by 87 times. Find all numbers for which this is possible.
Answer: $435 \cdot 10^{97}, 1305 \cdot 10^{96}, 2175 \cdot 10^{96}, 3045 \cdot 10^{96}$.

Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+2} n$, where $a, k, m, n$ are non-negative integers, and $m<10^{k}$ and $a<100$. By erasing the digits in the $k$-th and $(k+1)$-th positions, we get the number $m+10^{k} n$. According to the condition,

$$
m+10^{k} a+10^{k+2} n=87\left(m+10^{k} n\right) \Longleftrightarrow 86 m=10^{k}(a+13 n)
$$

The number $a+13 n$ is positive and divisible by 43. Moreover, $a+13 n<86$, otherwise $m \geqslant 10^{k}$. Therefore, $a+13 n=43$, and the pair $(n, a)$ takes the values $(0,43),(1,30),(2,17),(3,4)$. In the first case, the number of digits is $(k+2)$, from which $k=98$. Then $m=5 \cdot 10^{97}$, and the original number is

$$
m+10^{k} a=5 \cdot 10^{97}+43 \cdot 10^{98}=435 \cdot 10^{97}
$$

In the other cases, the number of digits is $(k+3)$, from which $k=97$. Then $m=5 \cdot 10^{96}$, and the original numbers are

$$
10^{99}+30 \cdot 10^{97}+5 \cdot 10^{96}, \quad 2 \cdot 10^{99}+17 \cdot 10^{97}+5 \cdot 10^{96}, \quad 3 \cdot 10^{99}+4 \cdot 10^{97}+5 \cdot 10^{96}
$$

which gives the answer.