<s>[INST] Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$?
$Approach$:
$N$=$11^2$.$13^4$.$17^6$
$N^2$=$11^4$.$13^8$.$17^{12}$
This means $N$ has $(2+1) * (4+1) * (6+1)  = 105$ factors and $N^2$ has $ (4+1) * (8+1) * (12+1) = 585 $ factors.
Therefore, there are 480 numbers that are not a factor of N. They are any combination of :
$11^3$,$11^4$,$13^5$,$13^6$,$13^7$,$13^8$,$17^7$,$17^8$,$17^9$,$17^{10}$,$17^{11}$,$17^{12}$. 
But how many of these combinations are less than N?  Not really sure how to do that in a easy way.
Please guide me how to do so.
Any help will be appreciated. Thanks.
 [/INST] If a divisor $d$ of $N^2$ is smaller than $N$ then $\frac{N^2}{d}$ is a divisor of $N^2$ larger than $N$. 

Let's say that there are $n$ divisors of $N^2$ that are smaller than $N$. Then there are $n$ divisors of $N^2$ that are larger than $N$.
In total there are 585 divisors of $N^2$ and each one is either smaller, larger or equal to $N$. 
From these you should be able to find $n$.
Now $480$ divisors of $N^2$ do not divide $N$ and $n$ of them are larger than $N$ (some explanation here) therefore $480-n$ of them are smaller than $N$.
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