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# Prime factors

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Manager
Joined: 23 Jul 2008
Posts: 189

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18 Nov 2008, 09:46
Hi,
Can you guys tell me how to find the number of divisors a number has if we know its prime factorization? I know there is formula but i don't seem to remember.

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Manager
Joined: 08 Aug 2008
Posts: 228

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18 Nov 2008, 11:56
1
KUDOS
the number of divisors is
= (e1+1)(e2+1)(e3+1) ... (ek+1).
where ek denotes powers of prime factors of the number.

For example, 4200 is $$2^3*3^1*5^2*7^1$$, so it has (3+1)(1+1)(2+1)(1+1) = 48 positive divisors.
Manager
Joined: 23 Jul 2008
Posts: 189

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18 Nov 2008, 12:33
prasun84 wrote:
the number of divisors is
= (e1+1)(e2+1)(e3+1) ... (ek+1).
where ek denotes powers of prime factors of the number.

For example, 4200 is $$2^3*3^1*5^2*7^1$$, so it has (3+1)(1+1)(2+1)(1+1) = 48 positive divisors.

Thanks a lot.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Prime factors   [#permalink] 18 Nov 2008, 12:33
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# Prime factors

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