Get the last two digits of $16^{100}$ and $17^{100}$ I try to get the last two digits of $16^{100}$ and $17^{100}$
I started with:
$6^{1} = 6$
$6^{2} = 36$
$6^{3} = 216$ means last digit for $6$ is always $6$
That is why the last digit for $16^{100}$ is $6$
And for $17^{100}$ i calculated:
$7^{1} = 7$
$7^{2} = 49$
$7^{3} = 343$
$7^{4} = 2401$
$7^{5} = 16807$
And because $4*25 = 100$ or better said $100mod4 = 0$
The last digit for  $17^{100}$ is $1$
But how do i get the penultimate digits? Thanks
 A: As $(17,100)=1,$ and using Carmichael function $\displaystyle\lambda(100)=20$
$\displaystyle\implies17^{20}\equiv1\pmod{100}$

$16^{100}=(2^4)^{100}=2^{400}$
As $(2^{400},100)=4=2^2\ne1$  let us find $2^{400-2}\pmod{25}$
As $\displaystyle\lambda(25)=\phi(25)=20, 2^{20}\equiv1\pmod{25}$
and $\displaystyle400-2\equiv18\pmod{20}\implies 2^{400-2}\equiv2^{18}\pmod{25}$
Now, $\displaystyle2^9=512\equiv12\pmod{25}\implies 2^{18}=(2^9)^2\equiv12^2=144\equiv19\pmod{25}$
$\displaystyle\implies 2^{400-2}\equiv19\pmod{25}\  \ \ \ (1)$
Now as $\displaystyle a\equiv b\pmod m\implies a\cdot n\equiv b\cdot n\pmod{m\cdot n}$ for any integer $n,$
multiply either sides of $(1)$ by $4$ 
A: Another way for $16$
We have already found $\displaystyle16^{100}=2^{400}$
$\displaystyle2^{10}=1024\equiv24\pmod{100}\implies 2^{20}\equiv24^2=576\equiv76$
Now, $\displaystyle2^{20a+b}-2^b=2^b(2^{20a}-1)\equiv0\pmod{100}$
if  $4$ divides $2^b\iff$ integer $b\ge2$ 
and $2^{20a}-1$ is divisible by $2^{20}-1$ for any positive integer $a$ and  $2^{20}\equiv1\pmod{25}$ by the other answer
So, $\displaystyle2^{20a+b}\equiv2^b\pmod{100}$ if integer $b\ge2,a\ge1$
Setting $\displaystyle b=20\implies2^{20(a+1)}\equiv2^{20}\pmod{100}\equiv76$
Now, set $a+1=20$
