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:

stmt 1: says that n^2 is an integer. So n could be 4, which n^16 or n could be 5.477 and n^2 would be 30. This gives us two different options so insuff

stmt 2: says that sqrt n is an integer. that means n is the square of an (or integer x integer) which will always give you an integer. suff

Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

Show Tags

20 Jun 2013, 14:07

Stament 1 : n^2=integer n=+/- sqrt(integer) => Sqrt of integer may or may not be integer depending on whether chose integer is perfect square or not. insufficient

Statement 2: sqrt (n) = integer => Squaring both sides => n= integer^2 = some other integer (as intg*intg=intg) = sufficient

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?
_________________

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?

How can square of non-integer be an integer? 1.73 is a non- integer and so is its square

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?

How can square of non-integer be an integer? 1.73 is a non- integer and so is its square

The square root of 3 is a number which when squared gives 3.
_________________

Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

Show Tags

21 Jul 2015, 20:29

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: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

Show Tags

12 Sep 2015, 07:28

Statement 1 says \(n^2\) is an integer. According to the solution, \((\sqrt{2})^2\) is an integer. But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx. And when we square it we would get 1.9999999 approx. We do not get an integer on squaring an irrational number but rather a value close to an integer. So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\) For this reason, i marked D. Can someone please explain why is my thinking wrong and why are we taking approximate values ?

Statement 1 says \(n^2\) is an integer. According to the solution, \((\sqrt{2})^2\) is an integer. But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx. And when we square it we would get 1.9999999 approx. We do not get an integer on squaring an irrational number but rather a value close to an integer. So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\) For this reason, i marked D. Can someone please explain why is my thinking wrong and why are we taking approximate values ?

The square root of 2 is a number (whatever it is) which when squared gives 2.
_________________

Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

Show Tags

04 Apr 2016, 13:00

1

This post received KUDOS

Bunuel wrote:

vwjetty wrote:

Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer

Please explain. Thanks.

Is n an integer?

(1) n^2 is an integer --> not sufficient, as if \(n^2=4\) answer is YES, but if \(n^2=3\) answer is NO. <<<< n will be square root of the value. So it can be an integer or non-integer, depending on the value whether it is a perfect square or not. So, not sufficient.

(2) n^(1/2) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2=integer\). Sufficient.

Answer: B.

I have added few more explanation for the 1st option, as i got this question wrong because of that problem. Hope you guys will not repeat the same mistake. #kudos
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...