2. A 200-digit natural number had its leading digit and the digit standing one place after it erased. As a result, the number decreased by 44 times. Find all numbers for which this is possible.
Answer: $132 c \cdot 10^{197}$ for $c=1,2,3$.

Solution. Represent the original number in the form $m+10^{k} \cdot \overline{c b a}$, where $a, b, c$ are decimal digits, $k, m$ are non-negative integers, with $c>0$ and $m<10^{k}$. By erasing the digits $a$ and $c$, we obtain the number $m+10^{k} b$. According to the condition,

$$
m+10^{k} a+10^{k+1} b+10^{k+2} c=44\left(m+10^{k} b\right) \Longleftrightarrow 43 m=10^{k}(a-34 b)+10^{k+2} c
$$

The number $m$ is divisible by $10^{k}$ and is less than $10^{k}$, so it is equal to zero. Then, after canceling out $10^{k}$, the equation will take the form $34 b=a+100 c$. Since $0=(100 c+a) \bmod 34=(a-2 c) \bmod 34$, we get $a=2 c$, from which $b=3 c$. Since $b<10$, the digit $c$ takes the values $1,2,3$. It remains to note that the original number is $(k+3)$-digit, so $k=197$.