There exists an integer with alternating digits $1$ and $2$ which is divisible by $2013$ Could someone give me hints in how to solve the following (rather interesting) problem? 

Prove that there exists an integer consisting of an alternance of $1$s
  and $2$s with as many $1$s as $2$s (as in $12$, $1212$, $1212121212$, etc.), and
  which is divisible by $2013$.

Source: Problem 4 in this document. 
I just had to ask it before 2013 ends :)

Hint:
Consider the numbers: the sequence of integers $12,1212,121212,12121212,.......$ 
Two of them must be equal modulo $2013$.Hence, the difference of these two must be divisible by $2013$. Now use this difference to obtain the desired number (note that $2,5\not|2013$)
