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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer

friends please help me in pointing out whee i am going wrong.

if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.

So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.

What's wrong with my explanation??

You forgot that x is a positive integer, so \(\sqrt{x}\) cannot equal to \(\frac{integer}{2}\). Generally \(\sqrt{integer}\) is either an integer or an irrational number.

Complete solution:

If x is a positive integer, is sqrt(x) an integer

If \(x=integer\), is \(\sqrt{x}=integer\)?

(1) \(\sqrt{4x}\) is an integer --> \(2\sqrt{x}=integer\) --> \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient.

(2)\(\sqrt{3x}\) is not an integer --> if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient.

Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!

4/2=2 and is not a fraction (integer/2 means reduced fraction). _________________

Re: if x is a positive integer ,is sqrt x an integer [#permalink]

Show Tags

10 Jan 2010, 08:42

if x is a positive integer ,is sqrt (x) an integer

(1) sqrt(4x) is an integer .

(2) sqrt (3x) is not an integer .

the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer yes. sqrt(x) is an integer.

how i worked it out : sqrt(4x) is an integer ==> 2 * sqrt(x) is an integer

in order that the product of 2 and sqrt(x) be an integer, sqrt(x) must either be 1) an integer 2) exactly half of an integer. i.e a number like 0.5, 1.5, 2.5 etc etc you worked that out as well. sqrt(x) = integer/2

we also know x is an integer. is there any integer whose square root is a half of an integer ? no!

therefore the only other alternative is sqrt(x) is a whole integer.

Re: if x is a positive integer ,is sqrt x an integer [#permalink]

Show Tags

09 Jan 2011, 09:40

I didn't get the explanation:

if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.

So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.

What's wrong with my explanation?? _________________

Re: if x is a positive integer ,is sqrt x an integer [#permalink]

Show Tags

19 Jul 2011, 06:29

Quote:

then sqrt(x) can not be integer/2

I think sqrt(x) can be integer/2 as long as (integer/2) itself is an integer i.e. that integer is multiple of 2. I think that is what bunuel meant. And the answer remains same. _________________

Conquer the Hell and make it Haven. Brain is your hell and Success is your haven!

"Kudos" is significant part of GMAT prep. If you like it, you just click it

Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]

Show Tags

03 Jan 2012, 13:06

Oh nice problem. Took quite some time to answer it but got A. Explanation by Bunuel is more than sufficient to understand the solution. _________________

Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]

Show Tags

03 Jan 2012, 14:31

Both equations and number plugging helps here.

1. sqrt(4x) = integer, that is 2 x sqrt(x) = integer. For this equation to be true, sqrt(x) has to be an integer. SUFFICIENT. If number plugging, use 4x4 and 4x9 combinations 2. sqrt(3x) = frac. Here, sqrt(3) x sqrt(x) = frac, that is frac x sqrt(x) = frac. Difficult to determine if this relationship can infer sqrt(x) as integer. So, let's go with number plugging. sqrt(3x4) = 12 satisfies, and sqrt(4) = integer. sqrt(3x5) = 15 satisfies, but sqrt(5) = frac. So, insufficient.

+1 for A. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: if x is a positive integer ,is sqrt x an integer [#permalink]

Show Tags

15 Dec 2012, 11:21

Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!

The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{x}=\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...).

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

05 Jul 2014, 03:40

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. _________________

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

12 Nov 2014, 01:36

Hello Easiest way i can think of: S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2 - int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But Non-Integer * Integer OR Non-integer = Non- integer So Insufficient (1E) or (AD)

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

05 Dec 2014, 00:00

deeuk wrote:

Hello Easiest way i can think of: S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2 - int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But Non-Integer * Integer OR Non-integer = Non- integer So Insufficient (1E) or (AD)

thoughts?

Bunuel could you please tell me if this approach is correct/incorrect?

Hello Easiest way i can think of: S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2 - int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But Non-Integer * Integer OR Non-integer = Non- integer So Insufficient (1E) or (AD)

thoughts?

Bunuel could you please tell me if this approach is correct/incorrect?

No. That's not correct.

(Non-integer)*(integer) could yield an integer. For example, 1/2*2 = 1.

(Non-integer)*(non-integer) could also yield an integer. For example, \(\sqrt{2}*\sqrt{2}=2\). _________________

If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

10 Dec 2014, 13:21

Thank you for the correction But now i am wondering since \(\sqrt{4}\) is 2, doesn't \(\sqrt{x}\) have to be an integer as well. I mean doesnt x have to be a perfect square?

gmatclubot

If x is a positive integer, is x^1/2 an integer
[#permalink]
10 Dec 2014, 13:21

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Time is a weird concept. It can stretch for seemingly forever (like when you are watching the “Time to destination” clock mid-flight) and it can compress and...