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:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$

Another approach:
$$\text{Euler's formula:  }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$$\phi(100)=\phi(2^2)\phi(5^2)=2\cdot 20=40$$
$$\gcd(100,33)=1$$
$$33^{40}\equiv 1 \pmod {100}$$
$$33^{100}\equiv (33^{40})^{2}3^{20}\equiv 3^{20}\pmod {100}$$
$$\equiv (3^5)^4\equiv43^4\equiv 49^2\equiv 01\pmod {100}$$
