Find the last two digits of $33^{100}$ 
Find the last two digits of $33^{100}$

By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of  @Lucian:
$33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$
By using the binomial expansion, we have:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$

You can use fast exponentiation: modulo $100$
$$33^2\equiv -11,\quad 33^4\equiv 21,\quad 33^8\equiv 441\equiv 41,\quad 33^{16}\equiv1681\equiv -19$$
whence  $\,33^{20}\equiv -19\cdot 21 =-(20-1)(20+1)\equiv 1$.
