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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 $1001$?

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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 $$\underbrace{2013}_{2013\ \text{times}}\equiv0\pmod{1001}$$

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    @coffeemath Wolfram - Alpha seems to confirm that it's $0$.2012-12-07
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    @coffeemath: Is $2013$ not $2103$.2012-12-07
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    Pambos: Thanks, I deleted my 2103 result!2012-12-07