HMMT November, 2024

Solution:

In general, $\operatorname{Top}(r, s)=\frac{r}{\operatorname{gcd}(r, s)}$. We characterize all possible $(a, z)$ as follows.

Claim 1. For any positive integers $a$ and $z$, we have $\operatorname{Top}(a, z)=\operatorname{Top}(z, a-1)$ if and only if there exists positive integers $d$ and $e$ such that $e \mid d^{2}-1$, $a=d^{2}$, and $z=d e$.

Proof. $(\Leftarrow)$ From $e \mid d^{2}-1$, we deduce that $\operatorname{gcd}(d, e)=1$. Thus, $\operatorname{Top}(a, z)=\frac{d^{2}}{\operatorname{gcd}\left(d^{2}, d e\right)}=\frac{d^{2}}{d}=d$. We also have $\operatorname{gcd}(z, a-1)=\operatorname{gcd}\left(d e, d^{2}-1\right)=e$, so $\operatorname{Top}(z, a-1)=\frac{d e}{e}=d$ as well.

$(\Rightarrow)$ Let $d=\operatorname{gcd}(a, z)$ and $e=\operatorname{gcd}(z, a-1)$. We have that $\operatorname{gcd}(d, e)=1$ because it divides both $a$ and $a-1$. The equation implies that $\frac{a}{d}=\frac{z}{e}$, or $\frac{a}{z}=\frac{d}{e}$. The left side has simplified form $\frac{a / d}{z / d}$, and the right side is already simplified. Thus, $a=d^{2}$ and $z=d e$. Finally, $e=\operatorname{gcd}(z, a-1) \mid a-1=d^{2}-1$.

The condition that $200 \leq a \leq 300$ implies $d \in\{15,16,17\}$. Once we select $d$, each divisor $e$ of $d^{2}-1$ yields a solution. Thus, the answer is
$$\begin{aligned} \tau\left(15^{2}-1\right)+\tau\left(16^{2}-1\right)+\tau\left(17^{2}-1\right) & =\tau(14 \cdot 16)+\tau(15 \cdot 17)+\tau(16 \cdot 18) \\ & =12+8+18=38 \end{aligned}$$
where $\tau(n)$ is the number of divisors of $n$.