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counting factors

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Senior Manager
Joined: 09 Oct 2007
Posts: 466
counting factors [#permalink]

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30 May 2008, 11:46
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Hi, I remember there was a formula to count factors, but I can't find the notebook where I wrote it it.

If I recall correctly, you had to find all the prime factors, then turn them into base-exponents format, then drop the base, then add 1 to each exponents and then multiply them. Is this correct?

IE: Factors of 200?

200 =
(2)(2)(2)(5)(5), therefore: (2^3)(5^2),
(3+1)(2+1) = (4)(3) = 12

200 has 12 factors total.

Can someone confirm?
Manager
Joined: 28 May 2008
Posts: 94
Re: counting factors [#permalink]

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30 May 2008, 20:48
asdert wrote:
Hi, I remember there was a formula to count factors, but I can't find the notebook where I wrote it it.

If I recall correctly, you had to find all the prime factors, then turn them into base-exponents format, then drop the base, then add 1 to each exponents and then multiply them. Is this correct?

IE: Factors of 200?

200 =
(2)(2)(2)(5)(5), therefore: (2^3)(5^2),
(3+1)(2+1) = (4)(3) = 12

200 has 12 factors total.

Can someone confirm?

Yes that is correct !
Here s the formula
The number of divisors of a given number N(including 1 and the number itself)
where N= (a^m)(b^n)(c^m), where a,b,c are prime numbers ,are
(1+m)(1+n)(1+p)

for eg 70=(2^1)(5^1)(7^1)
Therefore, No of divisors= (1+1)(1+1)(1+1)=8
those would be 1,2,5,7,10,14,35,70

Hope thats clear !

Cheers !
Re: counting factors   [#permalink] 30 May 2008, 20:48
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counting factors

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