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PS: Divisors

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PS: Divisors [#permalink] New post 25 Feb 2009, 15:19
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What is the sum of all common divisors of 48 and 36?
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Re: PS: Divisors [#permalink] New post 25 Feb 2009, 18:48
48 = 2^4 * 3
36 = 2^2 * 3^2
HCF= 2^2 * 3

The Sum of all the factors of N=a^p * b*q ( where a and b are prime numbers) is expressed as

(a ^p+1 - 1)/a-1 * (b^q+1 -1)/b-1

sum = (2^3 -1)/2-1 * (3^2 -1)/3-1 = 7 * (8/2) = 7*4 = 28
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Re: PS: Divisors [#permalink] New post 25 Feb 2009, 20:01
tkarthi4u wrote:
48 = 2^4 * 3
36 = 2^2 * 3^2
HCF= 2^2 * 3

The Sum of all the factors of N=a^p * b*q ( where a and b are prime numbers) is expressed as

(a ^p+1 - 1)/a-1 * (b^q+1 -1)/b-1

sum = (2^3 -1)/2-1 * (3^2 -1)/3-1 = 7 * (8/2) = 7*4 = 28



Thansk for the formula

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Re: PS: Divisors [#permalink] New post 25 Feb 2009, 23:01
i did it manually
1,2,3,4,6,12 are the common divisors of 36,48
hence sum is 28

for bigger sum, formula is v useful.thx

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Re: PS: Divisors [#permalink] New post 25 Feb 2009, 23:30
Hey thanks a lot for the formula but there is a typo it shld be b^q
tkarthi4u wrote:
48 = 2^4 * 3
36 = 2^2 * 3^2
HCF= 2^2 * 3

The Sum of all the factors of N=a^p * b*q ( where a and b are prime numbers) is expressed as

(a ^p+1 - 1)/a-1 * (b^q+1 -1)/b-1

sum = (2^3 -1)/2-1 * (3^2 -1)/3-1 = 7 * (8/2) = 7*4 = 28
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Re: PS: Divisors [#permalink] New post 26 Feb 2009, 02:59
Typo error.U are correct.
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Re: PS: Divisors [#permalink] New post 26 Feb 2009, 06:36
That formula looks really complex. :?

Is it different than this?

If N = a^p * b^q, where a and b are prime numbers,

Then number of factors = (p+1)(q+1)?

So is my formula used to calculate of the NUMBER of FACTORS, and the one you expressed is used to calculate the SUM of FACTORS?
Re: PS: Divisors   [#permalink] 26 Feb 2009, 06:36
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