2. Let $s(n)$ denote the sum of the digits (in decimal notation) of a natural number $n$. Find all natural $n$ for which $n+s(n)=2011$.
Answer: 1991. Hint. Since $n<2011$, then $s(n) \leq 2+9+9+9=29$. Therefore, $n=2011-$ $s(n) \geq 1982$. Since the numbers $n=2011$ and $n=2010$ obviously do not fit, the first three digits of the number $n$ can be one of three possibilities: 198, or 199, or 200. Let the last (fourth) digit of the number $n$ be $x$. Then in each of the cases mentioned, we get the equation: either 1) $1980+x+18+x=2011$, or 2) $1990+x+19+x=2011$, or 3) $2000+x+2+x=2011$. Only in the second case does the equation have an integer solution, namely, $x=1$.