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 :)
 A: Among the first 2014 numbers of this kind, two must have the same remainder mod $2013$. 
A: 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 $2,5\not|2013$)
A: note $3$ has to divide the number. So what can you say about the sum of the digits?
$11 has to divide the number. So what can you say about the sum of digits in the odd location and even location.
Once you have fixed these two, you want $61$ divide the number.
I am glad that 2013 is not a prime number :)

In response to the comment by OP:
Suppose the number starts as $12$. It is not clear if the length has to be even, but I will assume that. Then every time you add another $12$ you are muliplying  by 100 and adding 12.
So
$$
P_{n+1} = 100 P_n + 12
$$
Solving we have
$$
P_n = 4 \times \frac{100^n-1}{33}
$$
Since $61$ divides $P_n$, by Fermat's theorem
$$
n = 60 k
$$
Number is clearly divisible by 3.
Working modulo $11$ we have
$$
P_{n+1} = P_n + 1
$$
So 
$$
P_1 = 1\\
P_2 = 2 \\
P_3 = 3 \\
P_4 = 4 
$$
Hence $n$ has to be a multiple of 11.
The smallest $n$ is 660.
So the number is $12$ repeated 660 times.
