dhruv solanki wrote:

Bunuel wrote:

Official Solution:

How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.

Answer: D

hi,

the question ask for prime factors so for 7 it will be 1 for 8 it will be 3 as in (2^3) and 9 it will be (3^2).

i am not able to comprehend properly as the question asks number of prime factors and not the prime numbers it is constituted of..please explain..

7 has only one prime factor, which is 7.

8 has only one prime factor, which is 2.

9 has only one prime factor, which is 3.

But for example, 36 has two primes: 2, and 3.

but doesnt 8 constitute of 3 2's if we consider prime factor it would be 3 right??