Kanada 2011

We claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring.

Let the decimal expansion of $a$ be $a_k{a}_{k-1}\dots a_0$. For $i=0,\dots ,2011$, let $b_i$ be the number with decimal expansion $a_i{a}_{i-1}\dots a_0$. Then by the pigeonhole principle, $b_i\equiv b_j(\mathrm{mod}2011)$ for some $i<j\le 2011$.

It follows that $2011$ divides $b_j-b_i=c\cdot 10^i$. Here $c$ is the substring $a_j\dots a_{i+1}$. Since $2011$ and $10$ are relatively prime, it follows that $2011$ divides $c$.

$\square$