HMMT-Nov 2019 guts, Problem 23

For a positive integer $n$, let, $\tau(n)$ be the number of positive integer divisors of $n$. How many integers $1 \leq n \leq 50$ are there such that $\tau(\tau(n))$ is odd?

Solution: Note that $\tau(n)$ is odd if and only if $n$ is a perfect square. Thus, it suffices to find the number of integers $n$ in the given range such that $\tau(n)=k^{2}$ for some positive integer $k$. If $k=1$, then we obtain $n=1$ as our only solution. If $k=2$, we see that $n$ is either in the form $p q$ or $p^{3}$, where $p$ and $q$ are distinct primes. The first subcase gives $8+4+1=13$ solutions, while the second subcase gives 2 solutions. $k=3$ implies that $n$ is a perfect square, and it is easy to see that only $6^{2}=36$ works. Finally, $k \geq 4$ implies that $n$ is greater than 50 , so we've exhausted all possible cases. Our final answer is $1+13+2+1=17$. $\fbox{17}$.

Answer: 17