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The sum of all the factors of

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Carcass
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The sum of all the factors of [#permalink] New post   03 Jun 2020, 05:50
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Quantity A
Quantity B
The sum of all the factors of 220
The sum of all the factors of 285



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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vndnjn
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Re: The sum of all the factors of [#permalink] New post   03 Jun 2020, 16:13
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Factors of 220 = \(2^2*5^1*11^1\)
Sum of factors = \((2^0+2^1+2^2)*(5^0+5^1)*(11^0+11^1)\)
= 7*6*12 = 504

Factors of 285 = \(3^1*5^1*19^1\)
Sum of factors = \((3^0+3^1)*(5^0+5^1)*(19^0+19^1)\)
= 4*6*20 = 480
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Re: The sum of all the factors of [#permalink] New post   03 Jun 2020, 05:50
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Re: The sum of all the factors of [#permalink] New post   07 Jun 2020, 15:37
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vndnjn wrote:
Factors of 220 = \(2^2*5^1*11^1\)
Sum of factors = \((2^0+2^1+2^2)*(5^0+5^1)*(11^0+11^1)\)
= 7*6*12 = 504

Factors of 285 = \(3^1*5^1*19^1\)
Sum of factors = \((3^0+3^1)*(5^0+5^1)*(19^0+19^1)\)
= 4*6*20 = 480




Why do you multiply the factors if the question asks for the sum?
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vndnjn
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Re: The sum of all the factors of [#permalink] New post   07 Jun 2020, 15:53
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mikearox wrote:
vndnjn wrote:
Factors of 220 = \(2^2*5^1*11^1\)
Sum of factors = \((2^0+2^1+2^2)*(5^0+5^1)*(11^0+11^1)\)
= 7*6*12 = 504

Factors of 285 = \(3^1*5^1*19^1\)
Sum of factors = \((3^0+3^1)*(5^0+5^1)*(19^0+19^1)\)
= 4*6*20 = 480




Why do you multiply the factors if the question asks for the sum?


This is the way of the calculating sum of all factors; follow below link of khan academy. You'll get it

https://www.khanacademy.org/math/math-f ... -factors-2
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Re: The sum of all the factors of [#permalink] New post   30 Jun 2020, 10:23
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Carcass wrote:
Quantity A
Quantity B
The sum of all the factors of 220
The sum of all the factors of 285



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


This isn't a very good question.

The test makers specifically tell us that factors can be positive and negative.
See this excerpt from the GRE official guide:
Image

So, for example, the factors of 220 are: 1, -1, 2, -2, 4, -4, 5, -5, ......, 110, -110, 220, -220
So, the sum of all of the factors = 0
The same applies to Quantity B

So, the correct answer here is C.
The authors of this question are assuming that factors (divisors) are only positive when, in fact, factors can be both positive and negative.

As you can see the question above is not an official question.
Rest assured, the GRE will always have some text that restricts the factors (divisors) to POSITIVE values only.

Cheers,
Brent
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Re: The sum of all the factors of [#permalink] New post   30 Jun 2020, 23:29
vndnjn wrote:
Factors of 220 = \(2^2*5^1*11^1\)
Sum of factors = \((2^0+2^1+2^2)*(5^0+5^1)*(11^0+11^1)\)
= 7*6*12 = 504

Factors of 285 = \(3^1*5^1*19^1\)
Sum of factors = \((3^0+3^1)*(5^0+5^1)*(19^0+19^1)\)
= 4*6*20 = 480


Hi, isnt it the formula like power plus one?
if its 220, the factors are
(2^2)*(5^1)*(11^1)

so (power + 1) * (power + 1)
=(2+1) * (1+1) * (1+1)

is this approach correct? What is the difference between your approach and mine? When to use which one can anyone help plz?
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Re: The sum of all the factors of [#permalink] New post   01 Jul 2020, 02:59
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Yes it is correct and the GRE math book says the same thing

https://gre.myprepclub.com/forum/gre-quant ... tml#p51913

@GreenLightTestPrep might confirm :)
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Re: The sum of all the factors of [#permalink] New post   01 Jul 2020, 05:16
Farina wrote:
vndnjn wrote:
Factors of 220 = \(2^2*5^1*11^1\)
Sum of factors = \((2^0+2^1+2^2)*(5^0+5^1)*(11^0+11^1)\)
= 7*6*12 = 504

Factors of 285 = \(3^1*5^1*19^1\)
Sum of factors = \((3^0+3^1)*(5^0+5^1)*(19^0+19^1)\)
= 4*6*20 = 480


Hi, isnt it the formula like power plus one?
if its 220, the factors are
(2^2)*(5^1)*(11^1)

so (power + 1) * (power + 1)
=(2+1) * (1+1) * (1+1)

is this approach correct? What is the difference between your approach and mine? When to use which one can anyone help plz?


Hi Farina,

Yes that approach is correct.
The formula you're using will calculate the total number of POSITIVE divisors.
Some of the discussion above concerned whether divisors can be positive or negative.

Cheers,
Brent
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Farina
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Re: The sum of all the factors of [#permalink] New post   01 Jul 2020, 16:55
GreenlightTestPrep wrote:
Farina wrote:
vndnjn wrote:
Factors of 220 = \(2^2*5^1*11^1\)
Sum of factors = \((2^0+2^1+2^2)*(5^0+5^1)*(11^0+11^1)\)
= 7*6*12 = 504

Factors of 285 = \(3^1*5^1*19^1\)
Sum of factors = \((3^0+3^1)*(5^0+5^1)*(19^0+19^1)\)
= 4*6*20 = 480


Hi, isnt it the formula like power plus one?
if its 220, the factors are
(2^2)*(5^1)*(11^1)

so (power + 1) * (power + 1)
=(2+1) * (1+1) * (1+1)

is this approach correct? What is the difference between your approach and mine? When to use which one can anyone help plz?


Hi Farina,

Yes that approach is correct.
The formula you're using will calculate the total number of POSITIVE divisors.
Some of the discussion above concerned whether divisors can be positive or negative.

Cheers,
Brent


Thank you for confirmation. The formula than vndnjn used i have not read it anywhere yet, i wonder where do we use it
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Re: The sum of all the factors of [#permalink] New post   02 Jul 2020, 06:58
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Farina wrote:
Thank you for confirmation. The formula than vndnjn used i have not read it anywhere yet, i wonder where do we use it


Sorry Farina, I misspoke.
Formula you used earlier helps determine the NUMBER of positive divisors of a number.
The question here asks us to find the SUM of the positive divisors of a number.

The formula used by vndnjn is explained in the following video: https://www.khanacademy.org/math/math-f ... -factors-2

The formula works, but the given values (220 and 285) are small enough to perform the calculations ourselves.

All in all, I consider this a bad question and not representative of official GRE questions.
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Re: The sum of all the factors of [#permalink] New post   28 Apr 2023, 13:05
Based on the factor rules for summing, shouldn't you be dividing by the bases -1 as well?
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