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?
 A: Method$\#1:$
$$33^2=(30+3)^2\equiv3^2+2\cdot3\cdot30\equiv190-1\pmod{100}$$
$$\implies33^{100}\equiv(190-1)^{50}$$
$$(190-1)^{50}=(1-190)^{50}\equiv1-\binom{50}1\cdot190\pmod{100}\equiv?$$
Method$\#2:$
Alternatively, Carmichael Function $\lambda(100)=\cdots=20\implies33^{20}\equiv1\pmod{100}$
As $100\equiv0\pmod{20},33^{100}\equiv33^0\pmod{100}$
Method$\#3:$
As $33\equiv1\pmod4$
Again, $33\equiv8\pmod{25}\equiv2^3,2^{10}\equiv-1\pmod{25}$
$\implies33^{10}\equiv(2^3)^{10}\equiv(2^{10})^3\equiv(-1)^3\equiv-1$
$\implies33^{20}\equiv1\pmod{25}$
$\implies$ord$_{100}33=$lcm$(1,20)=20$
A: $33^{100} \equiv (33^2)^{50} \equiv 1^{50} \equiv 1 \bmod 4$, 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 $100$.
A: By the binomial theorem applied to $(30+3)^{100}$, they are the same as the digits of $3^{100}$.
Then, $\bmod 100$,
$$3^2=9,3^4=81,3^5=43,3^{10}=49,3^{20}=01\implies 3^{20k}=01.$$
A: 33^4 = 21 mod 100 ;33^20 = 01 mod 100 ;33^100 = 01 mod 100 ;Yes the
 last 02 digits are 01
