A number has exactly 32 factors out of which 4 are not compo

The second sentence tells us that the product of 4 prime factors of the original number is 30. The product of 2, 3 and 5 is itself equal to 30, there is no possibility of a fourth factor here. The answer should be E.

Can you please explain why the OA is C.

Let's see data,
i) a number with 32 factors, out of which 4 are not composite (Too vague data)
ii) product of these 4 factors is 30
lets take (ii) data
30 = 2*3*5*1 (1 is natural number i.e not composite nor prime)
so we know 3 factors of number are 2,3,5
now 32 = 2*4*4, 2*2*8
so total numbers possible = 3C1 + 3C1 = 6

There are two types of number when we consider the number of factors they have:-
1) Composite number: More than 2 factors
2) Prime number: Exactly two factors including 1 and itself
number 1 does not fall in any of the two, as it has only one factor itself.


1 is a factor of all numbers, so the 4 factors that are not composite are 1 and 3 prime numbers => \(n=a^xb^yc^z\)
Number of factors = \((x+1)(y+1)(z+1)=32\)

Product of these 4 factors (which are not composite) is 30 : Irrelevant information, although it tells us that the number is \(2^x3^y5^z\).

Now we know that the prime numbers are 3: a, b and c.
cases of 32 as product of three numbers greater than 1.

1) \(32=2*2*8=(1+1)(1+1)(7+1)\)
so x, y and z can be 1, 1 or 7. = \(\frac{3!}{2!}=3\)
\(n=a^1b^1c^7\) or \(n=a^1b^7c^1\) or \(n=a^7b^1c^1\)

2) \(32=2*4*4=(1+1)(3+1)(3+1)\)
so x, y and z can be 1, 3 or 3. = \(\frac{3!}{2!}=3\)
\(n=a^1b^3c^3\) or \(n=a^3b^3c^1\) or \(n=a^3b^1c^3\)

Total = 3+3=6

C