If n=(p/q) (p and q are nonzero integers), is an integer? : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 Feb 2017, 21:06

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n=(p/q) (p and q are nonzero integers), is an integer?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 16 Mar 2010
Posts: 184
Followers: 3

Kudos [?]: 179 [2] , given: 9

If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

### Show Tags

21 Sep 2010, 20:43
2
KUDOS
6
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

56% (02:02) correct 44% (01:03) wrong based on 277 sessions

### HideShow timer Statistics

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

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

(C) 2008 GMAT Club
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 37102
Followers: 7251

Kudos [?]: 96454 [4] , given: 10751

Re: I cant understand how the OA is??? [#permalink]

### Show Tags

21 Sep 2010, 21:35
4
KUDOS
Expert's post
3
This post was
BOOKMARKED
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.

_________________
Manager
Joined: 16 Mar 2010
Posts: 184
Followers: 3

Kudos [?]: 179 [0], given: 9

Re: I cant understand how the OA is??? [#permalink]

### Show Tags

21 Sep 2010, 22:50
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
Intern
Joined: 27 Jul 2010
Posts: 20
Location: Bangalore
Followers: 0

Kudos [?]: 4 [0], given: 0

Re: I cant understand how the OA is??? [#permalink]

### Show Tags

22 Sep 2010, 17:31
Bunuel wrote:
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.

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.
_________________

Nothing is free.. You 've to earn it!!!

Math Expert
Joined: 02 Sep 2009
Posts: 37102
Followers: 7251

Kudos [?]: 96454 [2] , given: 10751

Re: I cant understand how the OA is??? [#permalink]

### Show Tags

22 Sep 2010, 22:37
2
KUDOS
Expert's post
BalakumaranP wrote:
Bunuel wrote:
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.

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.

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 can not 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.
_________________
Intern
Joined: 29 Sep 2009
Posts: 13
Followers: 0

Kudos [?]: 16 [0], given: 18

### Show Tags

20 Oct 2010, 12:35
Hi,

please, can me somebody explain why is the first the first statement sufficient?
thanks
Jiri

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.
Math Expert
Joined: 02 Sep 2009
Posts: 37102
Followers: 7251

Kudos [?]: 96454 [1] , given: 10751

### Show Tags

20 Oct 2010, 12:40
1
KUDOS
Expert's post
Merging similar topics. Please refer to the solution above.

Hope it's clear.
_________________
Current Student
Joined: 15 Jul 2010
Posts: 257
GMAT 1: 750 Q49 V42
Followers: 10

Kudos [?]: 184 [0], given: 65

Re: I cant understand how the OA is??? [#permalink]

### Show Tags

20 Oct 2010, 12:54
Originally answered B, but D's clear now.

Thanks Bunuel!
_________________

Consider KUDOS if my post was helpful.

My Debrief: http://gmatclub.com/forum/750-q49v42-105591.html#p825487

Intern
Joined: 10 Oct 2010
Posts: 36
Followers: 0

Kudos [?]: 1 [0], given: 3

Re: I cant understand how the OA is??? [#permalink]

### Show Tags

25 Oct 2010, 12:44
Tricky I picked B but makes perfect sense now. Thanks!
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13940
Followers: 590

Kudos [?]: 167 [0], given: 0

Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

### Show Tags

10 Jul 2014, 06:59
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.
_________________
Manager
Joined: 31 Jul 2014
Posts: 152
GMAT 1: 630 Q48 V29
Followers: 0

Kudos [?]: 46 [0], given: 373

Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

### Show Tags

22 Sep 2015, 07:58
Bunuel wrote:
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.

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
Jamboree GMAT Instructor
Status: GMAT Expert
Affiliations: Jamboree Education Pvt Ltd
Joined: 15 Jul 2015
Posts: 294
Location: India
Followers: 67

Kudos [?]: 251 [0], given: 1

Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

### Show Tags

25 Sep 2015, 20:14
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"
_________________

Aryama Dutta Saikia
Jamboree Education Pvt. Ltd.

Manager
Joined: 28 Sep 2013
Posts: 90
GMAT 1: 740 Q51 V39
Followers: 13

Kudos [?]: 13 [0], given: 80

Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

### Show Tags

12 Jun 2016, 23:07
Bunuel wrote:
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?
_________________

Richa Champion | My GMAT Journey - 470 720 740

Target 760+

Not Improving after Multiple attempts. I can guide You.
Contact me richacrunch2@gmail.com

Math Expert
Joined: 02 Sep 2009
Posts: 37102
Followers: 7251

Kudos [?]: 96454 [0], given: 10751

Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

### Show Tags

12 Jun 2016, 23:26
crunchboss wrote:
Bunuel wrote:
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?

For example, if n = 2/1, or 6/3, or 10/10, ....
_________________
Re: If n=(p/q) (p and q are nonzero integers), is an integer?   [#permalink] 12 Jun 2016, 23:26
Similar topics Replies Last post
Similar
Topics:
2 If P and Q are Positive Integers, is P/Q terminating? 2 06 Nov 2016, 08:20
3 If x , p and q are positive integers , then 3 08 Oct 2016, 06:34
2 If p and q are positive integers, p/9 is an integer and q/8 is an inte 1 25 Apr 2016, 02:59
1 If p is an integer, is q an integer? 2 11 Feb 2012, 17:41
if N=p/q, where p and q are nozero integers, is n an 5 22 Jul 2011, 19:19
Display posts from previous: Sort by