Find the number of positive divisors of $N^2$ such that the positive divisors are less than N and do not divide N for $N=2^{17}*3^9*5^3$ Find the number of positive divisors of $N^2$ such that the positive divisors are less than N and do not divide N for $N=2^{17}*3^9*5^3$
My doubts
What I did firstly:-
$N^2=2^{34}*3^{18}*5^6$
Total number of factors will be $(34+1)(9+1)(3+1)=4655$
As given in the solution
There will be $(4655-1)/2=2327$ pairs of factors.
Why are we subtracting $1$ and why are we dividing it by $2$?
How these $2337$ factors of $N^2$ will be less than N?
Total number of factors will be $N=18*10*4=720$
So $720-1=719$ factors of N which are less than N and all of these will be a factors of $$N^2. 
Why subtracting $1$ again from $720$?
I will be so glad if anyone of you can please clarify my doubts.
 A: The reasonning step by step:


*

*You correctly found the total number of positive divisors of $N^2$: $4655$.

*One of these is $N$. Let's discard it. The other ones come in pairs whose products is $N^2$; there are $(4655-1)/2=2327$ such pairs.

*In each of these pairs one factor is above $N$ and one factor is below $N$. Hence the number is positive divisors of $N^2$ that are inferior to $N$ equals the number of pairs: $2327$
edit: for clarification about these two steps, let's see what happens with $M=6$ instead of $N$. The positive divisors of $M^2=36$ are $1,2,3,4,6,9,12,18,36$. $M=6$ stands alone, and all the other ones can be regrouped in four pairs whose product is $M^2=36$: $1*36=2*18=3*12=4*9=36$. That's how we find the number of divisors of $M^2$ strictly inferior to $M$: one per pair, hence $(9-1)/2=4$.


*

*Now we want to know how many of these do not divide $N$. The easiest way is to find the complement, which is exactly the number of positive divisors of $N$, since a divisor of $N$ is necessarily a divisor of $N^2$.
There are $720$ divisors of $N$, one of them is $N$ itself so there are $719$ divisors of $N$ inferior to $N$.

*Finally, the solution is $2327-719=1608$
There are 1608 positive divisors of $N^2$ which are (strictly) smaller than $N$ and do not divise $N$.
