Number of divisors

Let the two prime divisors of $N$ be $p$ and $q$ and $N = p^mq^n$ , where $m$ and $n$ are positive integers. Then $N^2=p^{2m}q^{2n}$ and it is given that $N^2$ has $77$ divisors. That means its number of divisors is given by $(2m+1)(2n+1) = 77$ , $\Rightarrow m = 3$ and $n = 5$ or reverse. Therefore, the number of divisors of $N$ is $(m+1)(n+1) = 4 \times 6 = \boxed{24}$ .