Question Link here: gregmat(dot)com/problems/problem/if-m-and-n-are-integers-for-how/

The answer stated here is E. 2(m+1)(n+1).

Sadly, Greg’s explanation is useless here because in case either (m+1) & (n+1) is a prime factor like (4+1)=5 or (6+1)=7 which isn’t divisible by either 3 or 2 then the solution falls apart.

We tried back-solving every answer and the problem with the prime factors other than 2 or 3.

I think the trick here is that the information “m and n are greater than 3” is not actually as important. Basically, we need to find the highest possible powers in the denominator such that the result remains an integer. Hence we need to find out the number of factors of the numerator, which is (n+1)(m+1).

We then multiply this result by 2 to consider the possibilities of negative integers. Since for every positive integer factor, we have a negative integer factor, we consider twice the above result.

First make prime factorization of an integer n=a^p∗b^q∗c^n where a, b, \text{ and } c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1)
So, you’re finding total number of factors and not just a factor(Which you’re thinking)!

for eg:

8 = 2^3 \\
\text{total number of factors of 8} = 3+1 \text{ or } 4 \\
\text{Those 4 factors are} = 1,2,4,8