There exists an integer with alternating digits $1$ and $2$ which is divisible by $2013$

Question

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 :)

Answer 1

Among the first 2014 numbers of this kind, two must have the same remainder mod $2013$.

Answer 2

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 $2,5\not|2013$)

Answer 3

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 :)

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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.