Quick way to determine total # of positive factors?

by **ddohnggo** Thu Jul 24, 2008 12:36 am

i can't remember for the formula for it, but does anyone happen to know a quick way to determine the total # of positive factors?

for example, trying to find them for the number 441. i would first go about this by breaking 441 into prime numbers, yielding 7, 7, 3, and 3. there are a total of 9 different factors of 441, which results from these primes. does anyone know a fast way to find this total number of 9 without having to go through all of the different combinations?

this could become quite tedious with bigger numbers.

thanks!

by **tontonio** Mon Jul 28, 2008 4:24 pm

Hi. After you write the prime factorization base form you drop the bases and add one to the exponent. Then multiply for the total number of factors.

IE:

441 = (3)(3)(7)(7) = (3^2)(7^2)
Then drop the bases:
(2)(2), add one: (3)(3)=9

by **Guest** Tue Jul 29, 2008 5:43 pm

HI,

How would you do 96 with this formula?

Thanks

by **tontonio** Thu Jul 31, 2008 12:50 pm

96 = (2)(6)(8) = (2)(2)(2)(2)(2)(3) = (2^5)(3^1)

Drop the bases and add one to the exponents:
(5+1)(1+1) = 12

by RA Sun Aug 10, 2008 8:32 pm

Thanks tontonio for a good tip

by **RonPurewal** Wed Sep 17, 2008 4:53 am

yep - take the EXPONENTS of all the primes in the prime factorization, ADD ONE to each of the exponents, and then MULTIPLY the resulting numbers.

good stuff!

incidentally, this is going to appear in the next edition of the MGMAT number properties guide, which will probably hit the printers within the next couple of months.