Ireland

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.