Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 563 reviewsTarget Test Prep

gmatclub question

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 318
Joined: Mon Jul 13, 2009 3:55 am
Thanked: 12 times
by nikhilkatira » Wed Jul 28, 2010 10:28 pm
If n = p/q (p and q are nonzero integers), is n an integer?

1. n^2 is an integer.
2. 2n+4/2 is an integer
Best,
Nikhil H. Katira
Top
Join the discussion
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members
Thanked: 215 times

by kvcpk » Wed Jul 28, 2010 11:05 pm
nikhilkatira wrote:If n = p/q (p and q are nonzero integers), is n an integer?

1. n^2 is an integer.
2. 2n+4/2 is an integer
n = p/q

1. n^2 is an integer.
n^2 is an integer and n is not integer occurs when n is square root of intger.
Example when n is sqrt(2).
But, sqrt(2) cannot be expressed in the form p/q, because it is irrational.
Hence n should be an integer.
SUFF

2. 2n+4/2 is an integer

2n+4 = 2k
2n = 2k-4
n=k-2
K is integer. Hence k-2 is also integer.
SUFF

pick D
Top
Join the discussion
User avatar
Legendary Member
Posts: 748
Joined: Sun Jan 31, 2010 7:54 amFollowed by: 3 members
Thanked: 46 times

by outreach » Thu Jul 29, 2010 1:30 am
stmt2

2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff

where am i wrong?



kvcpk wrote:
nikhilkatira wrote:If n = p/q (p and q are nonzero integers), is n an integer?

1. n^2 is an integer.
2. 2n+4/2 is an integer
n = p/q

1. n^2 is an integer.
n^2 is an integer and n is not integer occurs when n is square root of intger.
Example when n is sqrt(2).
But, sqrt(2) cannot be expressed in the form p/q, because it is irrational.
Hence n should be an integer.
SUFF

2. 2n+4/2 is an integer

2n+4 = 2k
2n = 2k-4
n=k-2
K is integer. Hence k-2 is also integer.
SUFF

pick D
-------------------------------------
--------------------------------------
General blog
https://amarnaik.wordpress.com
MBA blog
https://amarrnaik.blocked/
Top
Join the discussion
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members
Thanked: 135 times

by selango » Thu Jul 29, 2010 1:58 am
n=p/q

stmt1,

n^2 is an integer.

p^2/q^2 is an integer.In order to be an integer p^2 and q^2 must have common prime factors so that p^2 is divisble by q^2.

Suff

stmt2,

2n+4/2 is an integer

I think it's (2n+4)/2 is an integer

n+2 is an integer.So n must be surely an integer.

Suff.

Pick D
--Anand--
Top
Join the discussion
User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members
Thanked: 215 times

by kvcpk » Thu Jul 29, 2010 1:59 am
outreach wrote:stmt2

2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff

where am i wrong?
I think it is (2n+4)/2.
You took it as 2n+(4/2).
I am not sure which one the intended statement was.
Nikhil - Can you confirm?
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 318
Joined: Mon Jul 13, 2009 3:55 am
Thanked: 12 times

by nikhilkatira » Thu Jul 29, 2010 2:31 am
kvcpk wrote:
outreach wrote:stmt2

2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff

where am i wrong?
I think it is (2n+4)/2.
You took it as 2n+(4/2).
I am not sure which one the intended statement was.
Nikhil - Can you confirm?
Sorry guys..my mistake
its (2n+4)/2.
Best,
Nikhil H. Katira
Top
Join the discussion
User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members
Thanked: 215 times

by kvcpk » Thu Jul 29, 2010 2:50 am
nikhilkatira wrote: Sorry guys..my mistake
its (2n+4)/2.
No Problem.. is the OA D?
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 318
Joined: Mon Jul 13, 2009 3:55 am
Thanked: 12 times

by nikhilkatira » Thu Jul 29, 2010 4:57 am
kvcpk wrote:
nikhilkatira wrote: Sorry guys..my mistake
its (2n+4)/2.
No Problem.. is the OA D?
OA is D
Best,
Nikhil H. Katira
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 51
Joined: Mon Jul 13, 2009 4:25 pm
Thanked: 1 times

by ymach3 » Thu Nov 18, 2010 12:44 am
hi , got a doubt for stmt 1

n^2=2 then n=sq root(2)=1.414=1414/1000=p/q..

Is'nt this correct???
Top
Join the discussion

GMAT/MBA Expert

GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Thu Nov 18, 2010 4:15 am
ymach3 wrote:...
n^2=2 then n=sq root(2)=1.414=1414/1000=p/q..

Is'nt this correct???
√2 is an irrational number and by definition an irrational number cannot be expressed as the ratio of two integers. We can easily prove that √2 is an irrational number and the proof is an interesting one! :)

Proof: If √2 is rational, then it has the form p/q for integers p, q not both even.
Then, p = (√2)q => p² = 2q², hence p is even
Say, p = 2m. Thus 4m² = 2q² => 2m² = q², hence q is also even, a contradiction.


In fact, √2 = 1.414213562... ≈ 1.414 (When rounded to 3 places after decimal)
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Sat Oct 23, 2010 9:42 am

by kyle8285 » Thu Nov 18, 2010 12:35 pm
Rahul,

Is it then sufficient to say that any number that is not a perfect square (has an integer square root) is irrational and therefore cannot be expressed as the ratio of two integers? For example, sq root(5), sq root(12), etc. Thank you!
Top
Join the discussion

GMAT/MBA Expert

GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Thu Nov 18, 2010 1:09 pm
kyle8285 wrote:Rahul,

Is it then sufficient to say that any number that is not a perfect square (has an integer square root) is irrational and therefore cannot be expressed as the ratio of two integers? For example, sq root(5), sq root(12), etc. Thank you!
Yes. But the generalized proof is different and a bit complicated and not necessary too.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion