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:

Thank you for solving. However I am confused why not x is divisible by16.

if x is 16 then x^2 is 256 which is divisible by 32. Appreciate your help.

The largest positive integer that must divide \(x\), means for lowest value of \(x\) which satisfies the given statement in the stem.

Given: \(32k=x^2\), where \(k\) is an integer \(\geq1\) (as \(x\) is positive).

\(32k=x^2\) --> \(x=4\sqrt{2k}\), as \(x\) is an integer \(\sqrt{2k}\), also must be an integer. The lowest value of \(k\), for which \(\sqrt{2k}\) is an integer is when \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(x=4\sqrt{2k}=4*2=8\)

Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

Show Tags

29 Oct 2013, 17:09

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 and x^2 is divisible by 32, then [#permalink]

Show Tags

05 Jul 2014, 07:15

Expert's post

1

This post was BOOKMARKED

parul1591 wrote:

Hello Bunuel

"The largest positive integer that must divide x, means for lowest value of x which satisfies the given statement in the stem."

Why are we looking for the smallest value of x ?

Because we need the largest positive integer that MUST divide x. So, we should find the least value of x for which x^2 is divisible by 32, and if that x is divisible by some number then so will be every other x's (MUST condition will be satisfied). The least positive x for which x^2 is divisible by 32 is 8 (8^2 = 64, which is divisible by 32). So, even if x = 8 it is divisible by 8 but not divisible by 6, 12 or 16.

Check similar problems in my post above. _________________

Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

Show Tags

26 Oct 2015, 11:43

Bunuel wrote:

prasadrg wrote:

Thank you for solving. However I am confused why not x is divisible by16.

if x is 16 then x^2 is 256 which is divisible by 32. Appreciate your help.

The largest positive integer that must divide \(x\), means for lowest value of \(x\) which satisfies the given statement in the stem.

Given: \(32k=x^2\), where \(k\) is an integer \(\geq1\) (as \(x\) is positive).

\(32k=x^2\) --> \(x=4\sqrt{2k}\), as \(x\) is an integer \(\sqrt{2k}\), also must be an integer. The lowest value of \(k\), for which \(\sqrt{2k}\) is an integer is when \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(x=4\sqrt{2k}=4*2=8\)

Hope it's helps.

so if 18 was on the list would that also satisfy the statement?

Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

Show Tags

14 Mar 2016, 01:40

Excellent Question... here the same old rule applies that x and x^n have the same number of prime factors.. hence 2 is the prime factor of x. now x^2/32=integer so x=8 is the least value hence 8 can be the largest integer that divides 8 hence C

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...