Distinct Factors

[rahul.s]

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

[thephoenix]

its 12
1,2,3,5,6,9,12,15,18,30,45,90

[bhumika.k.shah]

I get what u've done rahul...but is this some sort of formula ?
could you please explain the main concept behind this?

P.s.: thanks for sharing!

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

[sunil_snath]

thanks Rahul. This really handy....

[Mom4MBA]

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).....

[bhumika.k.shah]

Is there any restriction to keep in mind wiith this formula??

or whenever any question is asked abt the # of factors all we do is apply the formula and get the answer???

for any # ???

be it 10 digits long???

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).....

[Mom4MBA]

This works for any composite no matter how big it is.

[Ian Stewart]

Is there any restriction to keep in mind wiith this formula??

or whenever any question is asked abt the # of factors all we do is apply the formula and get the answer???

for any # ???

be it 10 digits long???

Yes, this will always work, provided you write your prime factorization in the normal way -- that is, using powers whenever a prime is repeated. So if you have a number like 36, you would first need to write the prime factorization as (2^2)(3^2), and then, adding one to each power and multiplying, we find that 36 has 3*3 = 9 positive divisors. If you write the prime factorization of 36 in a different way, say as (2)(3)(2)(3), then the method will not work.