1008. Does there exist a natural number $n$ such that

$$
n+S(n)=125
$$

where $S(n)$ is the sum of the digits of $n ?$
$\triangle$ The number $n$ can only be a three-digit number, and its first digit is equal to 1. Let $n=\overline{1 a b}$. We get:

$$
\overline{1 a b}+1+a+b=125, \quad 100+10 a+b+1+a+b=125, \quad 11 a+2 b=24
$$

The last equation has a unique solution in digits: $a=2, b=1$.

Answer: It exists and is unique - 121.