$120^{1492}\pmod{100}$

Question

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 $$[2^{4467}\pmod{100}\cdot41\cdot5^{1492}\pmod{100]\pmod{100}}$$

Answer 1

$$120^{1492} = 12^{1492} \cdot 10^{1492}$$

This number will end in $1492$ zeroes. It is divisible by $100$, so it is $0 \pmod{100}$

Answer 2

$120^{1492}=(10\times12)^{1492}=10^{1492}\times 12^{1492}=100\times(10^{1490}\times 12^{1492})\equiv 0 \bmod 100$

Answer 3

Trivial. Get ready to kick yourself.

$120^{1492} = 12^{1492} * 10^{1492} = 12^{1492}*(10^2)^{746} = 12^{1492}*100^{746} = M*100^N \equiv 0 \mod 100$.

Answer 4

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 $120^{1492} \equiv 0 \pmod{100}$