If $n=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$ 
If $N=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$

My solution:

$$n^2=2^{20}\times 3^{10}$$
Factors of $n^2=\left(20+1\right)(10+1)=21\times 11=231$ 
Factors which are greater than or equal to $n$ can be found as we have choices for power of 2 $\in [10,11,\dots 20]$ and that of 3 $\in[5,6,\dots10]$ 
We get the factors of $n^2$ which are greater than or equal to $n$ as $11\times 6=66$

Factors of $n$ which are less than $n=(10+1)(5+1)-1=65$ ($-1$ as $n$ has been counted in the above condition)

$\therefore$ factors which satisfy the condition are $231-66-65=100$

But the answer is $50$. What is the correct way to solve it and what is the error in my solution?

$n = 248832.$
You want all numbers of the form $2^a \times 3^b$ where

*

*$a \geq 11$ or $b \geq 6$.


*$a \leq 20$ and $b \leq 10$.


*$2^a \times 3^b < n.$
Consider the following two charts:
\begin{array}{| r | r | r |}
  \hline
2^a & b & \text{number of solutions} \\
\hline
2^{11} & b \leq 4 & 5 \\
\hline
2^{12} & b \leq 3 & 4 \\
\hline
2^{13} & b \leq 3 & 4 \\
\hline
2^{14} & b \leq 2 & 3 \\
\hline
2^{15} & b \leq 1 & 2 \\
\hline
2^{16} & b \leq 1 & 2 \\
\hline
2^{17} & b = 0 & 1 \\
\hline
\end{array}
\begin{array}{| r | r | r |}
  \hline
3^b & a & \text{number of solutions} \\
\hline
3^{6} & a \leq 8 & 9 \\
\hline
3^{7} & a \leq 6 & 7 \\
\hline
3^{8} & a \leq 5 & 6 \\
\hline
3^{9} & a \leq 3 & 4 \\
\hline
3^{10} & a \leq 2 & 3 \\
\hline
\end{array}
$50$ solutions in the two charts combined.
Edit 
Thanks to Calvin Lin for pointing out that $3^{11}$ was not a valid solution, since it does not divide $n^2.$
