HMMT February

As $100=2^{2} \cdot 5^{2}$, there are $3 \cdot 3=9$ divisors of 100, so there are 900 possible pairs of $d$ and $z$ that can be chosen. If $d$ is chosen, then there are $\frac{100}{d}$ possible values of $z$ such that $d$ divides $z$, so the total number of valid pairs of $d$ and $z$ is $\sum_{d \mid 100} \frac{100}{d}=\sum_{d \mid 100} d=(1+2+2^{2})(1+5+5^{2})=7 \cdot 31=217$. The answer is therefore $\frac{217}{900}$.