Irish Mathematical Olympiad

A number starts with the digit $1$ followed by $r$ zeros if and only if it can be written in the form $10^{r+s} + A$, with $0 \le A < 10^s$. If $k > r \ge 0$ we have $10^{r+s} + 10^k A + A < 10^{r+s} + 10^{k+s} + 10^s < 10^{k+s+1}$. Hence $(10^k + 1)(10^{r+s} + A) = 10^{k+r+s} + 10^{r+s} + 10^k A + A$ is a number which has at least $r-1$ zeros after its leading digit $1$. By induction, we obtain for $k > r \ge 0$ and any $1 \le m < r$ that $(10^k + 1)^m (10^{r+s} + A)$ has at least $r-m$ zeros after the leading digit $1$. With $A = 1$, $s = 1$, $r = k-1$ we obtain that the number $(10^k + 1)^{2011} = (10^k + 1)^{2010} (10^k + 1)$ has at least $k-1-2010$ zeros after the leading digit $1$, provided that $k > 2011$. Hence, any number $10^k + 1$ with $k \ge 4022$ gives a solution.