show there exists an integer k such that $2013^k$ ends with '0001' Prove that there exists an integer k so that $2013^k$ ends with '0001'.
we couldn't figure this out.
i thought we might try to prove that we can find an integer m such that $m*10^4 +1 = 2013^k$, but was unable to get any clues.
both hints and similar solutions are welcome.
Thanks in advance!

You have to prove that for some $k$,
$$ 2013^{k}\equiv 1\pmod{10^4} $$
but since $2013=3\cdot 11\cdot 61$ and $10^4=2^4\cdot 5^4$, we have $\gcd(2013,10^4)=1$, so $2013$ is an element of the group $\mathbb{Z}_{/10^4\mathbb{Z}}^*$, and its order is a divisor of $\varphi(10000)=4000$ by Euler's theorem and Lagrange's theorem.
You may also prove that $k=\color{red}{500}$ works.
