XV Junior Macedonian Mathematical Olympiad

The numbers $n$ and $S(n)$ have the same remainder when divided by $3$ and $9$. Namely, if $n=\overline{a_m{a}_{m-1}\dots a_1{a}_0}$ then $$\begin{array}{rl}n& =\overline{a_m{a}_{m-1}\dots a_1{a}_0}=10^m{a}_m+10^{m-1}{a}_{m-1}+\cdots +10a_1+a_0\\ & =[(10^m-1)a_m+(10^{m-1}-1)a_{m-1}+\cdots +(10-1)a_1]+[a_m+a_{m-1}+\cdots +a_1+a_0]\end{array}$$ Each expression in the brackets is divisible by $9$, therefore $n$ and $S(n)$ have the same remainder when divided by $3$ and $9$. The numbers $n$, $S(n)$, $S(S(n))$ have the same remainder when divided by $3$. Therefore the sum $n+S(n)+S(S(n))$ is always divisible by $3$ while $2011$ is not. Therefore the equation has no solution.