Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.
My try:
So I have to compute $33^{100}\mod 100$
Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$
So we have $33^{40}\equiv 1 \pmod{100}$
Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$
Since $100=2\cdot40+20$ so we have $33^{100}=1\pmod{100}$
So last two digits are $01$
Is it right?

$33^{100} \equiv (33^2)^{50} \equiv 1^{50} \equiv 1 \bmod 4$, because $\varphi(4)=2$.
And $33^{100} \equiv (33^{20})^5 \equiv 1^5\equiv 1 \bmod 25$, because $\varphi(25)=20$.
So $33^{100} \equiv 1 \bmod 100$.
Similarly, $a^{100} \equiv 1 \bmod 100$ for all $a$ coprime to $100$.
