# of factors

by **Guest** Sun Aug 24, 2008 12:07 am

How many factors does 1284 have ?

Well , I can find the prime numbers and multiply them but IT TAKES forever

The factors of 1284 are:

1 2 3 4 6 12 107 214 321 428 642 1284

The prime factors are: 2 x 2 x 3 x 107

Is there any faster method to find the # factors of a given number ?

Thanks,

Math Curious

by RA Sun Aug 24, 2008 5:05 am

1284 = 2 x 2 x 3 x 107 = 2^2 x 3 x107

To find all the combinations of the factors add 1 to the exponents of all the factors and then multiply them. For the example above that means 3 x 2 x 2 = 12.

by **Guest** Sun Aug 24, 2008 5:31 pm

Sweet!

How do you drive this formula ?

by RA Tue Aug 26, 2008 7:54 pm

This formula is an application of combinatorics.

by **RonPurewal** Sat Sep 20, 2008 12:48 am

Anonymous Wrote:Sweet!

How do you drive this formula ?

it has a stick shift, so i hope you know how to drive manual.

heh heh...

ok seriously:
if you have a prime raised to, say, the 4th power (such as 7^4), then there are 5 options for how many of that prime will occur in a given factor: 0, 1, 2, 3, or 4 of them.
if you have a prime raised to, say, the 6th power (such as 3^6), then there are 7 options for how many of that prime will occur in a given factor: 0, 1, 2, 3, 4, 5, or 6 of them.
you see the pattern: if a prime is raised to the n'th power, then there are (n + 1) different options for the number of times that prime can appear in any factor.

you're making this decision separately for each prime in the factorization, so you multiply all those numbers together.

second example (to make sure you understand)
how many factors does 3500 have?

highlight the area below to read the answer (which is currently written in white on a white background):

3500 = (2^2)(5^3)(7^1), so the exponents are 2, 3, and 1. therefore, (2 + 1)(3 + 1)(1 + 1) = 3 x 4 x 2 = 24 factors.