How do I find the last two digits $2012^{2013}$? How do I find the last two digits 20122013
My teacher said this was simple arithmetics(I still don't see how this is simple).
I thought of using Congruence equation as 2012 is congruent to 2 mod 10...but i can't get 22013
Can anyone help please?

We work modulo $100$, so we work modulo $2^2=4$ and $5^2=25$. The given number  is of course zero modulo four, so we only need it modulo $25$. The Euler indicator of $25$ is $\frac 45\cdot 25=20$, so $12^{20}=1$ modulo $25$. So
$$
2012^{2013}=12^{2013}=12^{20\cdot 100+13}=(12^{20})^{100}\cdot 12^{13}
=12^{13}=22
$$
modulo $25$. Among $22$, $22+25$, $22+50$, $22+75$ the one divisible by four is $22+50=72$. So this is the answer.
Computer check, here sage:
sage: R = Zmod(100)
sage: R(2012)^2013
72

