Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p

Since n = p^3 x q^6 x r^7, the factors which are squares of an integer other than n have to be of the form (p^2) x (q^2) x (r^2)
Taking P = p^2, Q = (q^2), R = (r^2), max possible value of N which is a square is = P x Q^3 X R^3

So number of possible factors = (1+1) x (3+1) x (3+1) - 1 (because question mentions squares greater than 1)
= 32 - 1 = 31

let x be a factor of n which is a perfect square. x = p^(0/2) * q^(0/2/4/6) * r^(0/2/4/6). So there are 2*4*4=32 combinations which result in x being a perfect square. One of these combinations is p^0*q^0^r^0 which is 1. x must be >1 so the number of factors greater than 1 is 32-1 = 31

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Powers that p can take = 0, 2

Powers that q can take = 0, 2, 4, 6

Powers that r can take = 0, 2, 4, 6

Total possibilities = 2 * 4 * 4 = 32

However this also includes = p^0 * q^0 * r^ 0 = 1

32 - 1= 31

Option D

Inorder to make a number square of any other number that it has to has even exponent. This is the underlying concept here.
Since n is \(p^3*q^6*r^7\)
We can have only 0,2 as exponent of p. Similarly 0,2,4,6 as exponent of q and 0,2,4,6 as exponent of r.
We know total combinations selecting A and B = Number of ways to select A * Number of ways to select B
Using this we have total combinations of p,q,r is \(2*4*4=32\).
But it also involve case when exponent of p,q,r will all have 0. But we need squares of number greater than 1. Thus answer would be 32-1=31.

If n=p^3∗q^6∗r^7, where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

p contributes - p^0 and p^2 - 2
q contributes - q^0, q^2, q^4, q^6 - 4
r contributes - r^0, r^2, r^4, r^6 - 4
So in all we have 2*4*4 = 32 factors
but, we need to eliminate p^0*q^0*r^0 = 1, as we are required to find the number of factors on n which are squares of an integer and are greater than 1. So 32 -1 = 31 is the desired answer.
Answer D

Perfect squares = p^2 , q^2, r^2, q^4, r^4 , and any combination of these ( I omit q^6 and r^6 because q^6 is the combination of q^2 and q^4, and r^6 is the combination of r^2 and r^4) .
So with one factor: p^2 , q^2, r^2, q^4, r^4, That is 5
Combination of 2 factors 5C2 = 5! /(3!2!) . That is 10
Combination of 3 factors 5C3 = 5! /(2!3!) . That is 10
Combination of 4 factors 5C4 = 5! /(1!4!) . That is 5
Combination of 5 factors 5C3 = 5! /(0!5!) . That is 1
Total 5+10+10+5+1 = 31
Answer D

Given: n = p^3 * q^6 * r^7 , where p, q, and r are different prime numbers

To find: factors of n which are squares of an integer greater than 1

Now all those factors with even power will meet the above ask
i.e p^2, q^2, q^4, q^6, r^2, r^4 and r^6 and their combinations.

For ease let's assume
p^2 = A
q^2 = B1, q^4 = B2, q^6 = B3
r^2 = D1, r^4 = D2, r^6 = D3

Now total number of factors which are squares of an integer > 1 will be = total number of combinations of the above variables
= total number of single factors + total number of combination of two variable factors + total number of combination of 3 variable factors
= (A, B1, B2, B3, D1, D2, D3) + (ways of choosing combinations of A and B1/2/3 + ways of choosing combinations of B1/2/3 and D1/2/3 + ways of choosing combinations of A and D1/2/3) + (ways of choosing combinations of A, B1/2/3 and D1/2/3)


= 7 + (1C1 * 3C1 + 3C1 * 3C1 + 1C1 * 3C1) + (1C1 * 3C1 * 3C1)
= 7 + 3 + 9 + 3 + 9
= 31

Hence total number of factors which are squares of an integer = 31

Since we need squares, we need to combine
a) the factors of p^3 that have even powers, i.e. p^0 and p^2
b) the factors of q^6 that have even powers, i.e. q^0, q^2, q^4 and q^6
c) the factors of r^7 that have even powers, i.e. r^0, r^2, r^4 and r^6

Thus, the total number of combinations = 2 x 4 x 4 = 32
However, out of these, 1 of them is the number 1 itself formed by combining p^0 x q^0 x r^0
Thus the number of factors greater than 1 is 32 - 1 = 31

Answer D