last two digits of $14^{5532}$? This is a exam question, something related to network security, I have no clue how to solve this!
Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?

Finding the last two digits necessarily implies $\pmod{100}$
As $(14^n,100)=4$ for $n\ge2$
Let use start with $14^{5532-2}\pmod{100/4}$ i.e., $14^{5530}\pmod{25}$
As $14^2\equiv-2^2\pmod{25}$
Now $2^5\equiv7,2^{10}\equiv7^2\equiv-1\pmod{25}$
$\implies14^{10}=(14^2)^5\equiv(-2^2)^5=-2^{10}\equiv-1(-1)\equiv1$
As $5530\equiv0\pmod{10},14^{5530}\equiv14^0\pmod{25}\equiv1$
Now use $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod {m\cdot c} $
$\displaystyle14^{5530}\cdot14^2\equiv1\cdot14^2\pmod{25\cdot14^2}$
As $100|25\cdot14^2,$
$\displaystyle14^{5530+2}\equiv14^2\pmod{100}\equiv?$
