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:

Re: If x is a positive integer, is root(x) an integer? [#permalink]

Show Tags

05 Jun 2013, 06:58

[quote="alexBLR"]If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer.

(1) since \(\sqrt{4x}\) is an integer, it is the same as 2*\(\sqrt{x}\) that means \(\sqrt{x}\) also must be ans integer - sufficient

(2) \(\sqrt{3x}\) is not an integer, in this case \(\sqrt{x}\) could be or couldn't be an integer. Case 1) x=4, where \(\sqrt{x}\) is an integer, then \(\sqrt{3*4}\)=2\(\sqrt{3}\) still not an integer Case 2) x=5, where \(\sqrt{x}\) is NOT an integer, then \(\sqrt{3*5}\)=\(\sqrt{15}\) still not an integer. This statement is not sufficient. Answer is A
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

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 \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...).

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice.

From First statement we get 2 sqrt (x) is an integer Therefore sqrt (x) must be an integer

Just to elaborate a bit further:

The thing is that sqrt (x) can't be an integer/2, cause the square root of any positive integer (as stated in the givens) is always >=1. Therefore, sqrt (x) will have to be an integer.

Suff

From second setatement we get the sqrt (3x) is not an integer Therefore, 3x is not a perfect square or x is not perfect square that is multiple of 3 But it could very well be a perfect square Matter is we still don't know

Just to clarify, sqrt (x) could or could not be an integer. If sqrt root (x) is an integer then sure 1.7 (integer) not an integer. But, sqrt (x) could also be say x=2, therefore sqrt (2)=1.4. Then 1.4*1.7 is clearly not an integer either. Answer A stands

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

Show Tags

09 Feb 2014, 08:55

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, 00:36

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

04 Dec 2014, 23:00

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, \(\sqrt{2}*\sqrt{2}=2\).
_________________

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

Show Tags

10 Dec 2014, 12:21

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?

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

Show Tags

11 Dec 2014, 02:35

my approach

I try to be careful in such types of questions because of a trap where a fraction multiplied by an integer can lead to an integer.

Statement 1 sqrt(4x)= +-2 * sqrt(x) (-2 not for gmat) so sqrt(4x)= 2*sqrt(x) here sqrt(x) has to be an integer or an irrational no. no matter what form x is in (complex nos out of scope so ignore) as per the statement 1 it will be an integer. so sufficient.

now lets check the trap if sqrt(x) is of the form integer/2 then still statement 1 will hold true but since x is an integer then at best x will of the form 4*perfect square. so again this statement will also give us a positive answer.

Statement 2 this basically gives us two opposite answers we need to compare x when it is a 1. perfect square (e.g. 25, sqrt(x)=5 and statement 2 is true) 2. not a perfect square (e.g 10 sqrt(10)=3.16 and statement 2 is still true) can't deduce anything
_________________

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

Show Tags

27 Sep 2015, 07:36

Bunuel wrote:

ezinis wrote:

If x is a positive integer, is \sqrt{x} an integer? (1) \(\sqrt{4x}\) is an integer4. (2) \(\sqrt{3x}\) is an integer.

I am not satisfied with the official explanation. Please give yours, thanks.

I think (2) should be \(\sqrt{3x}\) is NOT 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.

Answer: A.

i think it does not matter whether statement 2 is integer or not, as in both the cases we are getting different solution. and answer will be one in both the cases.

Re: If x is a positive integer, is root(x) an integer? [#permalink]

Show Tags

24 Dec 2015, 03:30

sqrt (4x) is an integer implies that 2 * sqrt(x) is also an integer. That implies x is an integer. Statement 1 is sufficient

sqrt (3x) is not an integer. For x = 2 it is possible. For a number such as 4 also it is. But sqrt () is not an integer while sqrt (4) is. Insufficient

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

Show Tags

12 Feb 2017, 13:41

I am new to gmatclub so I apologize if I am using this forum incorrectly... I instinctively want to plug in numbers: The way i solved was: Sqrt(4*1)= 2= integer but sqrt(1) is not an integer Sqrt(4*4)= 4= integer and sqrt(4)=2=integer

so I would say statement 1 is not sufficient. Clearly my logic is flawed since statement 1 is sufficient, but can someone tell me why making up numbers for X does not work for this problem?

I am new to gmatclub so I apologize if I am using this forum incorrectly... I instinctively want to plug in numbers: The way i solved was: Sqrt(4*1)= 2= integer but sqrt(1) is not an integer Sqrt(4*4)= 4= integer and sqrt(4)=2=integer

so I would say statement 1 is not sufficient. Clearly my logic is flawed since statement 1 is sufficient, but can someone tell me why making up numbers for X does not work for this problem?

Highlighted part is not correct. \(\sqrt 1 = 1 = integer\)
_________________

Re: If x is a positive integer, is root(x) an integer? [#permalink]

Show Tags

31 May 2017, 10:38

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.
_________________