Q: Algebra and Number theory Consider $N=20132013\cdots 2013$, where N consists of the number $2013$ concatenated (repeated) $2013$ times. What is the remainder of $N$ when divided by $1001$?

A: Notice that $1000 \equiv -1\pmod{1001}$. So powers of $10$ repeat as follows
$$1,\ 10,\ 100,\ -1,\ -10,\ -100,\ 1,\ \cdots$$
which has period $6$ so this allows us to remove $\operatorname{lcm}(6,\ 4) = 12$ digits, or three concatenated terms from the modulus and replace it with 
$$201320132013 \equiv 0 \pmod {1001}$$
As $2013$ is divisible by $3$, this means that all terms will be eliminated, giving
$$\underbrace{2013}_{2013\ \text{times}}\equiv0\pmod{1001}$$
