$120^{1492}\pmod{100}$ How can I solve it?
I can solve it using the fact: if $b\mid a$ then $b^{n}\mid a^{n}$  (I have proved it using induction)
$\frac{120}{10}=12$ ,then $\frac{120^{2}}{100}$ 
this mean $120^{2}\equiv0\pmod{10}$
so, $120^{1492}\equiv (120^{2})^{746}\equiv 0^{746}\equiv 0\pmod{100}$
I'm asking for another way.
I tried this: $120=2^{3}\cdot3\cdot5$,so:$$[2^{4467}\pmod{100}\cdot3^{1492}\pmod{100}\cdot5^{1492}\pmod{100}]\pmod{100}$$
$3^{1492}\equiv41\pmod{100}$ using Euler's theorem
but I stuck with $$[2^{4467}\pmod{100}\cdot41\cdot5^{1492}\pmod{100]\pmod{100}}$$

Use this teorem: If $a \equiv b \pmod{m} \implies a^n \equiv b^n \pmod{m}$, then: $$120 \equiv 20 \pmod{100} \implies 120^{1492} \equiv 20^{1492} \pmod{100}$$
If you look the values of $20^n$ you gonna see this:
$$\begin{align}
20^1 \equiv 20 \pmod {100} \\
20^2 \equiv 0 \pmod {100} \\
20^3 \equiv 0 \pmod{100} \\
... \\
20^{1492} \equiv 0 \pmod {100}
\end{align}$$
Then, for transitive property $120^{1492} \equiv 0 \pmod{100}$
