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 $14^{5532}$?

The OP quickly realizes that we can't write $14^k \equiv 1 \pmod{100}$ with $k \gt 0$, but there are still relations to be found in (multiplicative) semigroups.
If the last digit of integers $a$ and $b$ end in $6$, then the last digit of the product ends in $6$. This motivates us to write
$\quad 14^2 \equiv 96 \equiv -4 \pmod{100}$
and
$\quad 96^2 \equiv 16 \pmod{100}$
$\quad 96^3 \equiv 36 \pmod{100}$
$\quad 96^4 \equiv 56 \pmod{100}$
$\quad 96^5 \equiv 76 \pmod{100}$
$\quad 96^6 \equiv 96 \pmod{100}$
and
$\quad 96^{36} \equiv 96 \pmod{100}$
$\quad 96^{216} \equiv 96 \pmod{100}$
$\quad 96^{1296} \equiv 96 \pmod{100}$
So
$\; 14^{5532} = (14^2)^{2766} \equiv 96^{2766} \equiv (96^{1296})^2 (96^{174}) \equiv 96^{176} \equiv (96^{36})^4 (96^{32}) \equiv 96^{36} \equiv 96 \pmod{100}$
