<s>[INST] 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?
 [/INST] Hints:

*

*How many factors of $n$ are there?

*How many factors of $n^2$ are there? OP stated 231.

*How many factors of $n^2$ are there, which are $\leq n$?


 $\frac{231+1}{2} = 116$. (But, why?)



*Show that the factors of $n^2$ that are $\leq n$, but not factors of $n$, is the difference between 2 of the above values.


 $116 -66 = 50.$ (But, why?)

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