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:

Can somebody post the answer in the systematic explained way?? For the sake of new and weak users? I mean how do you start to think when you see this problem. The systematic approach can be developed for this kind of problems?? And by the way what is POE?

q^2 - 5 = 30k + r now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

That is not true, Eg. 25,55,85,.... are all multiples of 5 of the form 6k+1

Proof that \(q^2-5\) cannot be a multiple of 30

q^2 = 30k + 5 = 6*(5k) + 5 So q^2 leaves remainder 5 when divided by 6.

Consider the various cases for the remainder that q leaves when divided by 6 : q=6k ... q^2 will leave remainder 0 when divided by 6 q=6k+1 ... q^2 will leave remainder 1 when divided by 6 q=6k+2 ... q^2 will leave remainder 4 when divided by 6 q=6k+3 ... q^2 will leave remainder 3 when divided by 6 q=6k+4 ... q^2 will leave remainder 2 when divided by 6 q=6k+5 ... q^2 will leave remainder 1 when divided by 6

So it is not possible to have a perfect square that leaves remainder 5 when divided by 6.

So 30k+5 is not a perfect square for any choice of k. _________________

I am sorry but still i am not sure if i face this problem in exam i will be able to crack this.. How you will target at 30?? I mean you have all other options around...??? Is there any way to crack this type of problems??

q^2 - 5 = 30k + r now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

That is not true, Eg. 25,55,85,.... are all multiples of 5 of the form 6k+1

Proof that \(q^2-5\) cannot be a multiple of 30

q^2 = 30k + 5 = 6*(5k) + 5 So q^2 leaves remainder 5 when divided by 6.

Consider the various cases for the remainder that q leaves when divided by 6 : q=6k ... q^2 will leave remainder 0 when divided by 6 q=6k+1 ... q^2 will leave remainder 1 when divided by 6 q=6k+2 ... q^2 will leave remainder 4 when divided by 6 q=6k+3 ... q^2 will leave remainder 3 when divided by 6 q=6k+4 ... q^2 will leave remainder 2 when divided by 6 q=6k+5 ... q^2 will leave remainder 1 when divided by 6

So it is not possible to have a perfect square that leaves remainder 5 when divided by 6.

So 30k+5 is not a perfect square for any choice of k.

This is a good explanation but how do you apply this on the other answers quickly? 30 is easily divided to 5 and 6. but what about 41 or 29? you have to think about a number which these numbers are multiplied by and than add 5 which is a perfect square. Not trivial at all I think.

Any shortcut suggestions? What's the source of the question? _________________

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3 if q is a multiple of 3, q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...