<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] N= 11^2 * 13^4 * 17^6
No of Factors of N = 3*5*7 = 105
There will be one factor which is N, So remaining = 105-1 = 104
No of factors of N^2= 5*9*13 = 585, of these N will be one of the factors.
Out of the remaining 584 factors, half will be less than N and half will be grater than N , So no. of factors of N^2 which are less than N= 584/2= 292
Hence the number of factors of N^2 are less than N but not a factor of N = 292-104= 188 
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