Harvard-MIT Mathematics Tournament

Solution:
Suppose $n$ is sparkly; then its smallest divisor other than $1$ is some prime $p$. Hence, $n$ has $p$ divisors. However, if the full prime factorization of $n$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{r}^{e_{r}}$, the number of divisors is $\left(e_{1}+1\right)\left(e_{2}+1\right) \cdots\left(e_{r}+1\right)$. For this to equal $p$, only one factor can be greater than $1$, so $n$ has only one prime divisor - namely $p$ - and we get $e_{1}=p-1 \Rightarrow n=p^{p-1}$. Conversely, any number of the form $p^{p-1}$ is sparkly. There are just three such numbers in the desired range $\left(2^{1}, 3^{2}, 5^{4}\right)$, so the answer is $3$.