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:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
 A: 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  $\,33^{20}\equiv -19\cdot 21 =-(20-1)(20+1)\equiv 1$.
A: Hint: $33^2\equiv-11\bmod100$.
A: Another approach:
$$\text{Euler's formula:  }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$$\phi(100)=\phi(2^2)\phi(5^2)=2\cdot 20=40$$
$$\gcd(100,33)=1$$
$$33^{40}\equiv 1 \pmod {100}$$
$$33^{100}\equiv (33^{40})^{2}3^{20}\equiv 3^{20}\pmod {100}$$
$$\equiv (3^5)^4\equiv43^4\equiv 49^2\equiv 01\pmod {100}$$
A: $$(33)^{100}\equiv(33)^{5\cdot5\cdot4}\equiv(93)^{5\cdot4}\equiv(93)^4\equiv 01\pmod{100}$$
A: $$\begin{align}
33^{100}&=9^{50}\cdot11^{100}\\
&=(1-10)^{50}(1+10)^{100}\\
&=(1-50\cdot10+\cdots)(1+100\cdot10+\cdots)\\
&\equiv1\mod100
\end{align}$$
A: It can be proved that for any odd number $n$ not ending with 5,
${{n}^{20}}\equiv 1\,(\bmod \ 100)$
By Euler's theorem, $\phi (25)=25\left( 1-\frac{1}{5} \right)=20$ and $n$ is coprime to 25, so
${{n}^{20}}\equiv 1\,(\bmod \ 25)$        ...... (1)
As square of any natural number is either of the type $4k$ or $4k+1$, hence
${{n}^{20}}\equiv 1(\bmod \ 4)$           ....... (2)
Combining the results of (1) and (2), we get,
${{n}^{20}}\equiv 1\,(\bmod \ 100)$
So last two digits of numbers like ${{33}^{20}},\ {{17}^{20}},\ {{19}^{20}},\,{{47}^{20}}$ etc are $01$.
