rahul.s wrote:How many distinct integers are factors of 90?

A) 6
B) 8
C) 9
D) 10
E) 12

OA: E

Solution:

Approach 1: List down the factors. 1, 2, 3, 6, 5, 9, 10, 15, 30, 45, 90 = 12
Approach 2: I just learnt about this method. So I thought I'd share it with you'll.

Prime factorization of 90 = 2^1 * 3^2 * 5^1
Now, we add 1 to each exponent and multiply them.
(1+1) * (2+1) * (1+1) = 12

It works with any number. I tried it. Let's take the example of 50. It's factors are 1, 2, 5, 10, 25, 50 = 6
or 2^1 * 5^2 = (1+1) * (2+1) = 6

I'm quite weak with math, so I thought I'd share it with others like me. Hope it helped

Mom4MBA wrote:In general, for any number N, which can be written as N=m^a x n^b x p^c..

where m,n p ... are the prime factors of N and a,b,c .... are positive integers

the number of factors is equal to (a+1)(b+1)(c+1).....