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 $14^{5532}$?

Consider the following class of exam questions:

If $n$ is even find the last two digits of $n^k$.

There is a quick way of cranking out the answers by giving the multiplicative semigroup,

contained in $\mathbb {Z} / \text{100} \mathbb {Z}$, a central/absorbing role.
We know that $76 \times 76 \equiv 76 \pmod{100}$ and more generally that for any $a$ in the semigroup
$\quad 76 \times a \equiv a \pmod{100}$
(we actually have a group with the identity being $76$)
We know that there is an easy calculation algorithm that can be used to multiply any two of these residue numbers.
Example: For $56 \times 96 \pmod{100}$ the tens digit is equal to the residue $5 + 9 + 3 \pmod{10}$:
$\quad 56 \times 96 \equiv 76 \pmod{100}$
We know that given an even integer $n$ such that $n \not\equiv 0 \pmod{10}$, at least one of the residue classes
$\quad [n], [n^2], [n^4]  \quad \pmod{100}$
belongs to our semigroup.
We know that given an even integer $n$ such that $n \not\equiv 0 \pmod{10}$,
$\quad n^{20} = 76 \pmod{100}$
Now with these preliminaries we can mentally turn the crank (recall that $14^2 = 196$) ,
$\; 14^{5532} \equiv 76 \times 14^{12} \equiv  76 \times (14^2)^6 \equiv 76 \times (96)^2  \times (96)^2 \times (96)^2 \equiv$
$\quad \quad  \quad \quad 76 \times 16  \times 16 \times 16 \equiv 96 \pmod{100}$
