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: Division & Factor [#permalink]
24 Dec 2009, 06:21
5
This post received KUDOS
Expert's post
7
This post was BOOKMARKED
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 \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(x=4\sqrt{2k}=4*2=8\)
Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]
29 Oct 2013, 16:09
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]
05 Jul 2014, 06:15
Expert's post
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]
26 Oct 2015, 10:43
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?
gmatclubot
Re: If x is a positive integer and x^2 is divisible by 32, then
[#permalink]
26 Oct 2015, 10:43
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...