Last visit was: 22 Nov 2024, 03:53 It is currently 22 Nov 2024, 03:53
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

The sum of all the factors of

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30003
Own Kudos [?]: 36349 [13]
Given Kudos: 25927
Send PM
The sum of all the factors of [#permalink] New post   03 Jun 2020, 05:50
1
Expert Reply
12
Bookmarks
00:00

Question Stats:

56% (01:50) correct 43% (01:13) wrong based on 171 sessions

Hide Show timer Statistics

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.
ShowHide Answer
Official Answer
A
Most Helpful Community Reply
avatar
vndnjn
Retired Moderator
Joined: 16 Oct 2019
Posts: 63
Own Kudos [?]: 175 [15]
Given Kudos: 21
Send PM
Re: The sum of all the factors of [#permalink] New post   03 Jun 2020, 16:13
11
4
Bookmarks
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
General Discussion
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30003
Own Kudos [?]: 36349 [0]
Given Kudos: 25927
Send PM
Re: The sum of all the factors of [#permalink] New post   03 Jun 2020, 05:50
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
avatar
mikearox
Intern
Intern
Joined: 25 May 2020
Posts: 2
Own Kudos [?]: 5 [0]
Given Kudos: 0
Send PM
Re: The sum of all the factors of [#permalink] New post   07 Jun 2020, 15:37
1
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?
avatar
vndnjn
Retired Moderator
Joined: 16 Oct 2019
Posts: 63
Own Kudos [?]: 175 [0]
Given Kudos: 21
Send PM
Re: The sum of all the factors of [#permalink] New post   07 Jun 2020, 15:53
2
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
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [0]
Given Kudos: 136
Send PM

Top Member of the Month
Re: The sum of all the factors of [#permalink] New post   30 Jun 2020, 10:23
1
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
User avatar
Farina
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 349 [0]
Given Kudos: 299
Send PM
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?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30003
Own Kudos [?]: 36349 [0]
Given Kudos: 25927
Send PM
Re: The sum of all the factors of [#permalink] New post   01 Jul 2020, 02:59
1
Expert Reply
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 :)
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [0]
Given Kudos: 136
Send PM

Top Member of the Month
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
User avatar
Farina
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 349 [0]
Given Kudos: 299
Send PM
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
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [0]
Given Kudos: 136
Send PM

Top Member of the Month
Re: The sum of all the factors of [#permalink] New post   02 Jul 2020, 06:58
1
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.
avatar
B
snodoodles234
Intern
Intern
Joined: 11 Aug 2022
Posts: 14
Own Kudos [?]: 5 [0]
Given Kudos: 12
Send PM
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?
Prep Club for GRE Bot
gmatclubot
Re: The sum of all the factors of [#permalink]
28 Apr 2023, 13:05
Moderators:
adewale223
GRE Instructor
84 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1111 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact