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Prime factors

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Manager
Joined: 23 Jul 2008
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18 Nov 2008, 09:46
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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
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18 Nov 2008, 11:56
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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.

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Manager
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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.

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Re: Prime factors   [#permalink] 18 Nov 2008, 12:33
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