<s>[INST] 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
 [/INST] 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 $4$ 
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