elementary number theory - 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 $14^{5532}$?

Answer

By the Chinese remainder theorem, it is enough to find the values of $14^{5532}\mod 4$ and $\bmod25$.

Now, clearly $\;14^{5532}\equiv 0\mod 4$.

By Euler's theorem, as $\varphi(25)=20$, and $14$ is prime to$25$, we have:
$$14^{5532}=14^{5532\bmod20}=14^{12}\mod25.$$
Note that $14^2=196\equiv -4\mod25$, so $14^{12}\equiv 2^{12}=1024\cdot 4\equiv -4\mod25$.

Now use the C.R.T.: since $25-6\cdot4=1$, the solutions to $\;\begin{cases}x\equiv 0\mod 4\\x\equiv -4\mod 25\end{cases}\;$ are:
$$x\equiv \color{red}0\cdot25-6\cdot{\color{red}-\color{red}4}\cdot 4= 96\mod 100$$
Thus the remainder last two digits of $14^{5532}$ are $\;96$.