Hello David!

We have that $53\cdot17 = 901$, this means that $53\cdot 17 \equiv 1 \pmod{100}$ (the remainder of $901$ when dividing by $100$ is $1$), so $53$ is the multiplicative inverse of $17$ modulo $100$. Now, $17^{1999} \equiv 1\cdot 17^{1999} \equiv 53\cdot 17\cdot 17^{1999} \equiv 53\cdot 17^{20000}\pmod{100}$.

Note that this does not mean that $17^{1999}=17^{2000}\cdot53$, but that they have the same last two digits.