Determine the last two digits of $3^{3^{100}}$

Question

Determine the last two digits of $3^{3^{100}}$

This is one of the problems in the past exam my modern algebra course. I think I need to use euler-fermat theorem but can't figure out how to use it for this problem. Can anyone help me out?

Answer 1

We want to compute $3^{3^{100}}\bmod\; 100$. Since $\phi(100)=\phi(2^2)\phi(5^2)=(2^2-2)(5^2-5)=40$ the value of the exponent only matters $\bmod\; 40$. Since $\phi(40)=16$ and $100\equiv 4\bmod 16$ we have $3^{100}\equiv 3^4\equiv 1\bmod\; 40$. Thus $$3^{3^{100}}\equiv 3^1\equiv 3\bmod \;100$$ so the last two digits are $03$.

Answer 2

To find the last two digits of a simple base like 3, the knowledge of power cyclicity can be very useful.

The cyclicity of powers of 3 is 20, i.e, after every 20 power iterations the last two digits repeat themselves.

$3^1$ = 03;
$3^2$ = 09;
$3^3$ = 27;
$3^4$ = 81;
$3^5$ = 43;
.....
$3^{10}$ = 49;
.....
.....
$3^{20}$ = 01;
$3^{21}$ = 03; .... and so on...

So, last two digits of the main base (3 here) are the last two digits in $3^{100}$, i.e, 01.

So, the last two digits in $3^{3^{100}}$ are nothing but the last two digits in $3^{01}$ = 03.

Answer 3

$$3^{3^{100}}\equiv3^{3^{100}\pmod{\lambda(100)}}\pmod{100}$$ where $\lambda$ is the Carmichael function

Now $\displaystyle\lambda(100)=20,$

$\displaystyle3^4=81\equiv1\pmod{20}$ or $\displaystyle\lambda(20)=4\implies3^4\equiv1\pmod{20}$

$\displaystyle\implies3^{100}=(3^4)^{25}\equiv1^{25}\equiv1\pmod{20}$