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

Question

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?

Answer 1

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$(1,20)=20$

Answer 2

$33^{100} \equiv (33^2)^{50} \equiv 1^{50} \equiv 1 \bmod 4$, 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 $100$.

Answer 3

By the binomial theorem applied to $(30+3)^{100}$, they are the same as the digits of $3^{100}$.

Then, $\bmod 100$,

$$3^2=9,3^4=81,3^5=43,3^{10}=49,3^{20}=01\implies 3^{20k}=01.$$

Answer 4

33^4 = 21 mod 100 ;33^20 = 01 mod 100 ;33^100 = 01 mod 100 ;Yes the last 02 digits are 01