Last visit was: 18 Dec 2024, 10:04 It is currently 18 Dec 2024, 10:04
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

[m]N = m^4n^2p^3[/m]

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
KarunMendiratta
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3259 [4]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
[m]N = m^4n^2p^3[/m] [#permalink] New post   Updated on: 21 Mar 2021, 22:50
4
Bookmarks
00:00

Question Stats:

67% (00:57) correct 32% (01:35) wrong based on 43 sessions

Hide Show timer Statistics

N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


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
C

Originally posted by KarunMendiratta on 10 Mar 2021, 20:20.
Last edited by Carcass on 21 Mar 2021, 22:50, edited 1 time in total.
Formatted
avatar
B
eldorado21
Manager
Manager
Joined: 26 Nov 2020
Posts: 110
Own Kudos [?]: 102 [0]
Given Kudos: 31
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   11 Mar 2021, 16:46
Some more details in the question are needed,

is m = n = p ? Is m,n,p greater than 1?
User avatar
D
KarunMendiratta
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3259 [1]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   11 Mar 2021, 23:35
1
eldorado21 wrote:
Some more details in the question are needed,

is m = n = p ? Is m,n,p greater than 1?


Thanks, I have made the changes.
User avatar
D
KarunMendiratta
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3259 [2]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   14 Mar 2021, 01:03
1
1
Bookmarks
KarunMendiratta wrote:
N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


OA Explanation

If \(N = (a^x)(b^y)(c^z)....\)
Then, Total number of factors = \((x + 1)(y + 1)(z + 1)....\)

Col. A: Number of factors = \((4 + 1)(2 + 1)(3 + 1) = (5)(3)(4) = 60\)
Col. B: \(60\)

Hence, option C
avatar
S
FCOCGALVAN
Manager
Manager
Joined: 01 Dec 2018
Posts: 87
Own Kudos [?]: 35 [1]
Given Kudos: 38
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   21 Mar 2021, 16:46
1
Carcass

KarunMendiratta


Help here!

Ans must be D,

m n and p can be compund numbers, what if m is 11, n is 26 and p is 77?

Kind regards!
User avatar
D
KarunMendiratta
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3259 [1]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   21 Mar 2021, 21:54
1
FCOCGALVAN

The questions says Factor Form!

\((11)(27)(77)\) would be written as \((3^3)(7)(11^2)\)
avatar
S
COolguy101
Senior Manager
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Own Kudos [?]: 172 [0]
Given Kudos: 81
Concentration: , International Business
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   28 May 2021, 19:55
KarunMendiratta wrote:
KarunMendiratta wrote:
N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


OA Explanation

If \(N = (a^x)(b^y)(c^z)....\)
Then, Total number of factors = \((x + 1)(y + 1)(z + 1)....\)

Col. A: Number of factors = \((4 + 1)(2 + 1)(3 + 1) = (5)(3)(4) = 60\)
Col. B: \(60\)

Hence, option C


Your calculated factor is total no of factor sir, but Quantity A ask for no of positive factor which must be 60/2 =30. Hence, quantity B must be greater.
User avatar
D
KarunMendiratta
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3259 [2]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   29 May 2021, 01:29
1
1
Bookmarks
kumarneupane4344 wrote:
KarunMendiratta wrote:
KarunMendiratta wrote:
N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


OA Explanation

If \(N = (a^x)(b^y)(c^z)....\)
Then, Total number of factors = \((x + 1)(y + 1)(z + 1)....\)

Col. A: Number of factors = \((4 + 1)(2 + 1)(3 + 1) = (5)(3)(4) = 60\)
Col. B: \(60\)

Hence, option C


Your calculated factor is total no of factor sir, but Quantity A ask for no of positive factor which must be 60/2 =30. Hence, quantity B must be greater.


kumarneupane4344
Dear Sir, I have only calculated the positive factors of N here, there are 60 more factors of N which are negative.

Example:
Let N = 8 = \(2^3\)

Total number of positive factors = 3 + 1 = 4 (1, 2, 4, and 8)
Number of negative factors = 4 (-1, -2, -4, and -8)

Hence, Total number of factors of N = 4 + 4 = 8
avatar
S
COolguy101
Senior Manager
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Own Kudos [?]: 172 [2]
Given Kudos: 81
Concentration: , International Business
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   29 May 2021, 04:30
2
Thank you KarunMendiratta sir, I have a misconception about this formula that it gives total no of factors. NOw I am clear sir. Many thanks, Stay safe!!
avatar
B
AppK
Intern
Intern
Joined: 25 Oct 2021
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 19
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   19 Jul 2022, 11:08
Carcass Can factor form be assumed as "Prime Factor"? I am still unsure why D doesn't hold true

Posted from my mobile device
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [0]
Given Kudos: 26080
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   19 Jul 2022, 11:14
Expert Reply
I am not sure what you meant

However, when you perform the factorization they are all primes

usually we do have composite numbers but also just prime numbers that can be divided only by 1 and themselves
avatar
B
AppK
Intern
Intern
Joined: 25 Oct 2021
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 19
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   19 Jul 2022, 11:18
Carcass

I meant if a factored form is given in the question stem does it always have to be Prime raised to exponents.

From your reply, it seems it always is

Posted from my mobile device
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [0]
Given Kudos: 26080
Send PM
[m]N = m^4n^2p^3[/m] [#permalink] New post   19 Jul 2022, 11:25
Expert Reply
AppK wrote:
Carcass

I meant if a factored form is given in the question stem does it always have to be Prime raised to exponents.

From your reply, it seems it always is

Posted from my mobile device


\(8^2 = 64\) and neither 8 nor 64 are primes

But at its core \(8^2\) is just \(2^3 * 2^3 = 2^6\) and 2 is prime
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [1]
Given Kudos: 26080
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   19 Jul 2022, 11:27
1
Expert Reply
Carcass wrote:
AppK wrote:
Carcass

I meant if a factored form is given in the question stem does it always have to be Prime raised to exponents.

From your reply, it seems it always is

Posted from my mobile device


\(8^2 = 64\) and neither 8 nor 64 are primes

But at its core \(8^2\) is just \(2^3 * 2^3 = 2^6\) and 2 is prime


In fact, when you see question like this they are always factorization problems and they are always problems prime-related
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5090
Own Kudos [?]: 76 [0]
Given Kudos: 0
Send PM
Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   07 Mar 2024, 04:41
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
gmatclubot
Re: [m]N = m^4n^2p^3[/m] [#permalink]
07 Mar 2024, 04:41
Moderators:
adewale223
GRE Instructor
88 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1115 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