2. A 200-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 a factor of 5. Find all numbers for which this is possible.
Answer: $125 a \cdot 10^{97}$ for $a=1,2,3$.

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,

$$
m+10^{k} a+10^{k+1} n=5\left(m+10^{k+1} n\right) \Longleftrightarrow 4 m=10^{k}(a-40 n)
$$

Note that $n=0$, otherwise $m<0$. Therefore, $a$ will be the leading digit of the original number, from which $k=199$. Then

$$
4 m=10^{k} a=10^{199} a \Longleftrightarrow m=25 a \cdot 10^{197},
$$

and the original number is $125 a \cdot 10^{197}$. Since $m<10^{199}$, the digit $a$ takes the values $1,2,3$.