Show that for every integer $a$ such that $\gcd(a,100)=1$, $a^{100}= 1 \mod 100$ and use this to compute the last two digits of $(11)^{102}$. Can you say something more precise about the order of an element in $(\mathbf{Z}/100 \mathbf{Z} )^*$?
So, I worked out that the if $S=(\mathbf{Z}/100 \mathbf{Z} )^*$ and $a$ is in $S$, then $aS=S$ because they're all coprimes. Thus for any element $x$ in $S$, $x^{40}=1 \mod 100$. But I can't understand why it works for $20$ when $40$ is the number of elements coprime to $100$.
Using the fact that $11^{20}=1 \mod 100$ its easy to see that $11^{102}= (11^{20})^5\cdot (11^2)=121 \mod 100=21$, but I'm still not sure why $20$ works.
Also, I don't know what I can say about elements of $(\mathbf{Z}/100 \mathbf{Z} )^*$ except that their order must be less than $20$?