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:

If y is a positive integer, is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer. (2) \(\sqrt{5y}\) is an integer.

If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) can not be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

Re: If y is a positive integer, is square_root of y [#permalink]

Show Tags

30 Jun 2012, 04:08

Bunuel wrote:

metallicafan wrote:

If y is a positive integer, is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer. (2) \(\sqrt{5y}\) is an integer.

If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) can not be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

Answer: D.

Hope it helps.

Thank you Bunuel. I arrived to the same conclusion. However, I have a doubt: Ok, we know that \(x\sqrt{5}\). But how can we be so sure that x is not for example a a huge number like 10*10^10000000.... to make \(x\sqrt{5}\) a integer? Maybe, I am forgeting a concept. Please, your help.
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

If y is a positive integer, is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer. (2) \(\sqrt{5y}\) is an integer.

If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) can not be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

Answer: D.

Hope it helps.

Thank you Bunuel. I arrived to the same conclusion. However, I have a doubt: Ok, we know that \(x\sqrt{5}\). But how can we be so sure that x is not for example a a huge number like 10*10^10000000.... to make \(x\sqrt{5}\) a integer? Maybe, I am forgeting a concept. Please, your help.

The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).

Re: If y is a positive integer, is square_root of y [#permalink]

Show Tags

30 Jun 2012, 18:22

Bunuel wrote:

The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).

Hope it's clear.

Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational? Thanks!
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).

Hope it's clear.

Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational? Thanks!

Since no prime can be a perfect square (or perfect cube, ...), then \(\sqrt{prime}=irrational\).
_________________

Re: If y is a positive integer, is square_root of y [#permalink]

Show Tags

12 Aug 2015, 19:52

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.
_________________

No, your calculations are fine and are consistent with the question. If you have y = 125, \(\sqrt{125}\) \(\neq\) Integer and thus get an unambiguous "no" making statement 2 sufficient. The original question is

"\(\sqrt{y}\) an integer?" So getting an unambiguous yes or no will be sufficient. Per Bunuel's solution, it is shown that y can not be a perfect square leading to \(\sqrt{y}\neq Integer\)

Re: If y is a positive integer, is square_root of y [#permalink]

Show Tags

06 Nov 2016, 08:27

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.
_________________

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...