Let $n = 2^{31}*3^{19}$. Find the number of positive divisors $d$ of $n^2$ such that $1\leq d\leq n$ and $d \nmid n$

Question

Let $n=2^{31}*3^{19}$. Find the number of positive divisors $d$ of $n^2$ such that $1\leq d\leq n$ and $d$ does not divide $n$.

My attempt

$n^2 = 2^{62} * 3^{38}$

Total divisors $= 1 + 62 + 38 + 62*38 = 2457$

Answer 1

As you pointed out, the total number of divisors of $n^2$ is $2457$. Since if $d | n$, $\frac{n^2}{d} | n$, we can pair each divisor of $n^2$ less than $n$ with one greater than $n$, and then the only divisor remaining is $n$. Thus, there are $\frac{2457-1}{2} = 1228$ divisors of $n^2$ less than $n$, one equal to $n$, and $1228$ greater than $n$. Thus, we have $1228+1 = 1229$ divisors $d$ that satisfy the property $1 \leq d \leq n$.

Since all divisors of $n$ are in this set of $1229$ numbers, we can just subtract the number of divisors of $n$ from $1229$ to get the required answer. Since there are $(31+1)(19+1) = 32\cdot20 = 640$ divisors of $n$, our answer is $1229-640 = 589$.

Answer 2

The hard part is the $\leq n$ part, avoiding $d\mid n$ is just requiring the exponent of either $2$ or $3$ to be larger than $31$ respectively $19$.

Right now I can't think of a better way, than going through the possible exponents of $3$ (as there a fewer of those than of $2$), and calculating the number of potential powers of $2$.