Last visit was: 19 Nov 2025, 16:20 It is currently 19 Nov 2025, 16:20
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT 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 GMAT 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
Back to Forum
Create Topic
605-655 Level|   Number Properties|         
User avatar
amitjash
User avatar
Joined: 17 Mar 2010
Last visit: 22 Feb 2017
Posts: 87
Own Kudos:
674
 [61]
Given Kudos: 9
Posts: 87
Kudos: 674
 [61]
If n = p/q where p and q are nonzero integers, is n an integer ? 21 Sep 2010, 21:43
6
Kudos
Add Kudos
55
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

62% (01:31) correct 38% (01:36) wrong based on 934 sessions

History

Date
Time
Result
Not Attempted Yet
If n = p/q where p and q are nonzero integers, is n an integer ?

(1) n^2 is an integer.
(2) (2n + 4)/2 is an integer.

Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [28]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [28]
If n = p/q where p and q are nonzero integers, is n an integer ? 21 Sep 2010, 22:35
13
Kudos
Add Kudos
15
Bookmarks
Bookmark this Post
If n = p/q where p and q are nonzero integers, is n an integer ?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) cannot be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it cannot be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n + 4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.
General Discussion
User avatar
amitjash
User avatar
Joined: 17 Mar 2010
Last visit: 22 Feb 2017
Posts: 87
Own Kudos:
674
Given Kudos: 9
Posts: 87
Kudos: 674
If n = p/q where p and q are nonzero integers, is n an integer ? 21 Sep 2010, 23:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
OK now i understood the fundamentals about this question. I was thinking what if n=141/100? in that case n^2 will be 2 but n will not be integer.
Thanks bunuel
User avatar
BalakumaranP
User avatar
Joined: 27 Jul 2010
Last visit: 02 Jan 2014
Posts: 12
Own Kudos:
5
Location: Bangalore
Concentration: General
Schools: ISB '15
Posts: 12
Kudos: 5
If n = p/q where p and q are nonzero integers, is n an integer ? 22 Sep 2010, 18:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.
Show more

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [5]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [5]
If n = p/q where p and q are nonzero integers, is n an integer ? 22 Sep 2010, 23:37
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
BalakumaranP
Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or irrational number (for example: \(\sqrt{3}\)), (note that \(n\) cannot be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.
Show more

That's not true. No reduced fraction when squared can equal to an integer.

Or consider this: \(n^2=\frac{p^2}{q^2}=integer\) --> \(n=\frac{p}{q}=\sqrt{integer}\) --> now, square root of an integer is either an integer or an irrational number. But it cannot be an irrational number as we have that \(\frac{p}{q}=\frac{integer}{integer}=\sqrt{integer}\) and we know that an irrational number cannot be expressed as a fraction of two integers, so \(\sqrt{integer}=n=integer\).

Hope it's clear.
User avatar
scheol79
User avatar
Current Student
Joined: 15 Jul 2010
Last visit: 18 Mar 2013
Posts: 118
Own Kudos:
624
Given Kudos: 65
GMAT 1: 750 Q49 V42
GMAT 1: 750 Q49 V42
Posts: 118
Kudos: 624
If n = p/q where p and q are nonzero integers, is n an integer ? 20 Oct 2010, 13:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Originally answered B, but D's clear now.

Thanks Bunuel!
User avatar
Wofford
User avatar
Joined: 10 Oct 2010
Last visit: 03 Dec 2010
Posts: 14
Own Kudos:
4
Given Kudos: 3
Posts: 14
Kudos: 4
If n = p/q where p and q are nonzero integers, is n an integer ? 25 Oct 2010, 13:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Tricky I picked B but makes perfect sense now. Thanks!
avatar
anupamadw
avatar
Joined: 31 Jul 2014
Last visit: 29 Jun 2016
Posts: 106
Own Kudos:
139
Given Kudos: 373
GMAT 1: 630 Q48 V29
GMAT 1: 630 Q48 V29
Posts: 106
Kudos: 139
If n = p/q where p and q are nonzero integers, is n an integer ? 22 Sep 2015, 08:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.
Show more

Hi
I am having difficulty understanding statement1.
If n^2=100 then n=10
If n^2=3 then n=sqrt(3) --> why cant we express it as p/q
sorry It may be a dumbest question, but need to clarify..Thanks
User avatar
AryamaDuttaSaikia
User avatar
Jamboree GMAT Instructor
Joined: 15 Jul 2015
Last visit: 06 Dec 2019
Posts: 252
Own Kudos:
693
Given Kudos: 1
Status:GMAT Expert
Affiliations: Jamboree Education Pvt Ltd
Location: India
Posts: 252
Kudos: 693
If n = p/q where p and q are nonzero integers, is n an integer ? 25 Sep 2015, 21:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We can not represent irrational numbers as p/q as the irrational numbers are non repeating non terminating decimals and there is no way we can write p/q for such expression.

One more thing: To check if a square of a number is irrational or not, we need to check if its prime factors have "even powers"
User avatar
RichaChampion
User avatar
Joined: 28 Sep 2013
Last visit: 02 Jan 2018
Posts: 70
Own Kudos:
88
Given Kudos: 82
GMAT 1: 740 Q51 V39
GMAT 1: 740 Q51 V39
Posts: 70
Kudos: 88
If n = p/q where p and q are nonzero integers, is n an integer ? 13 Jun 2016, 00:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

Show more

You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
If n = p/q where p and q are nonzero integers, is n an integer ? 13 Jun 2016, 00:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
crunchboss
Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.


You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?
Show more

For example, if n = 2/1, or 6/3, or 10/10, ....
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 19 Nov 2025
Posts: 4,844
Own Kudos:
8,945
 [1]
Given Kudos: 225
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,844
Kudos: 8,945
 [1]
If n = p/q where p and q are nonzero integers, is n an integer ? 19 Apr 2020, 07:11
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
User avatar
Harps
User avatar
Joined: 25 Oct 2019
Last visit: 25 Apr 2021
Posts: 14
Own Kudos:
15
 [1]
Given Kudos: 31
Posts: 14
Kudos: 15
 [1]
If n = p/q where p and q are nonzero integers, is n an integer ? 18 Jun 2020, 05:57
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
CrackVerbalGMAT
By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
Show more

Regarding statement 2:
is it possible that n = 1/3 and the equation becomes (2/3+4)/2? in this case too the result will be an integer but n is a fraction
confused here!
User avatar
ShankSouljaBoi
User avatar
Joined: 21 Jun 2017
Last visit: 17 Apr 2024
Posts: 622
Own Kudos:
603
Given Kudos: 4,090
Location: India
Concentration: Finance, Economics
Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)
GMAT 1: 660 Q49 V31
GMAT 2: 620 Q47 V30
GMAT 3: 650 Q48 V31
GPA: 3.1
WE:Corporate Finance (Non-Profit and Government)
Products:
My Reviews
Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)
GMAT 3: 650 Q48 V31
Posts: 622
Kudos: 603
If n = p/q where p and q are nonzero integers, is n an integer ? 07 Nov 2020, 08:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Harps
CrackVerbalGMAT
By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!

Regarding statement 2:
is it possible that n = 1/3 and the equation becomes (2/3+4)/2? in this case too the result will be an integer but n is a fraction
confused here!
Show more
14/6 is not an integer, hence, it is not a valid example :)
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,589
Own Kudos:
1,079
Posts: 38,589
Kudos: 1,079
If n = p/q where p and q are nonzero integers, is n an integer ? 31 Mar 2025, 19:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
Moderators:
Bunuel
Math Expert
105390 posts
kingbucky
496 posts