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

Question

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?

Answer 1

Hints:

1. How many factors of $n$ are there?
2. How many factors of $n^2$ are there? OP stated 231.
3. How many factors of $n^2$ are there, which are $\leq n$?

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

4. 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?)

Answer 2

Let's start with all numbers fitting the desired criteria that have no factors of 2, and increment the power on 2 as we go, and record how many values we find:

$3^6,...,3^{10}$ (5)

$2\times3^6,...,2\times3^{10}$ (5)

$2^2\times3^6,...,2^2\times3^{10}$ (5)

$2^3\times3^6,...,2^3\times3^{9}$ (4)

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Notice that higher powers on one base may limit the highest power on the other. List out all possibilities this way and you will get the correct answer.

Answer 3

$n = 248832.$

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 $n^2.$