is there a formula to calculate the number of factors? :: Manhattan GMAT Forums

Posted: Sat Mar 08, 2008 12:35 pm

what is the number of factors for 441? It breaks down into 3, 3, 7, 7 ...how do you find the total number of factors? what's the formula?

Posted: Sat Mar 08, 2008 2:39 pm

There is no formula you just take a look at how many different combinations there are of prime factors

So in this case

WE have 3, 7, (7 x 7 = 49), (7 x 3 = 21), (3 x 3 = 9) (9 x 7 = 63), (49 x 3 = 147) , 441, and 1

So for 441 we have total of 9 factors. Just always remember that 1 and the number it self is also a factor.

Posted: Sun Mar 09, 2008 12:25 am

Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.

Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.

So the number of factors for your number would be (a+1)*(b+1)*...(k+1).

For example, if your number is 441 = 3^2 * 7^2 then the number of factors is (2+1)*(2+1)=3*3=9.

One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24.

Posted: Wed Mar 12, 2008 2:59 am

David Pollack wrote:

Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.

Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.

So the number of factors for your number would be (a+1)*(b+1)*...(k+1).

For example, if your number is 441 = 3^2 * 7^2 then the number of factors is (2+1)*(2+1)=3*3=9.

One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24.

this is just awesome. very well done.

i have some unsolicited advice for you:
step 1) drop whatever you're doing
step 2) write a textbook
step 3) resume whatever you're doing

thank you for saving us the time required to answer this ourselves. :)

Posted: Thu Mar 20, 2008 2:08 am

RPurewal wrote:

David Pollack wrote:

Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.

Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.

So the number of factors for your number would be (a+1)*(b+1)*...(k+1).

For example, if your number is 441 = 3^2 * 7^2 then the number of factors is (2+1)*(2+1)=3*3=9.

One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24.

this is just awesome. very well done.

i have some unsolicited advice for you:
step 1) drop whatever you're doing
step 2) write a textbook
step 3) resume whatever you're doing

thank you for saving us the time required to answer this ourselves. :)

That explanation less the examples read more like directions to fix a pinball machine in chinese

Posted: Thu Mar 20, 2008 10:40 pm

Yeah I agree.

Posted: Mon Mar 24, 2008 5:36 am

Hi there ...

what is the quickest way to find the prime factors for a number ?
How would you find the prime factors of 3841, for example ?
Does anyone know of a quick way to do it.

Thanks for the help. Appreciate it.

Posted: Mon Mar 24, 2008 8:49 pm

iil-london wrote:

Hi there ...

what is the quickest way to find the prime factors for a number ?
How would you find the prime factors of 3841, for example ?
Does anyone know of a quick way to do it.

Thanks for the help. Appreciate it.

I guess there is not. If anybody has I appreciate him/her for his/her brilliant idea.

Just factorize it. I know it is not easy and doesnot have factors 2, 3, 5, 7, 11, 13, 17, and 19. but 23 divides it.

3841 = 23 x 167

i guess 167 is also a prime cuz if it were not a prime it has to be divided by 2, 3, 5, 7, or son. none of them divides 167, so its a prime.
this is a tough one.

Posted: Thu Mar 27, 2008 1:29 am

David - I agree, nice process. And for those of you who weren't quite sure how to follow the theory - don't worry about it! Just follow the examples :)

I don't know of a shortcut to find the complete set of prime factors for a number very quickly. I'm guessing there isn't one because one of the weird things about primes is that they pretty much defy our attempts to reduce things to typical patterns or formulas.