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: If p is a positive integer and p^2 is divisible by 12 [#permalink]

Show Tags

29 Dec 2014, 18:56

1

This post received KUDOS

4

This post was BOOKMARKED

Since p is an integer so p cannot have a 2 and sqrt 3 (because squaring this will give us a 2^2 and 3 (making the product as 12, and making p^2 as a multiple of 12)) p^2 is divisible by 12 (12 = 2*2*3), So, p should have at least one 2 and one 3 so that p^2 has a 2^2 and two 3

So, p will have a 2 and a 3. Or p will be a multiple of 6 So, largest possible integer than should divide p^3 is 6^3 So, Answer will be D

Hope it helps!

sunita123 wrote:

If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p ^3 is

The point of all of this is that each "P" has to be made up of the same "pieces" - in this example, P is made up of one "2" and one "5."

Here, we're told that P is a positive integer and P^2 is DIVISIBLE by 12. This causes a bit of extra work for us because 12 = (2)(2)(3). P CANNOT = 12 though because there aren't enough "pieces" to make both Ps the same. We can put a "2" into each "P" but we can't split up the 3....

P^2 = (P)(P) = (2 x something)(2 x something) ....but the 3 has to go in there somewhere....

If we had ANOTHER 3 though, then we'd have P^2 = (2)(2)(3)(3) and then we can put one "2" and one "3" into each P...

P^2 = (2x3)(2x3) = 36 P = 6

Having that extra 3 is the simplest/smallest value for P that fits this question. From here, P^3 = 6^3.

In questions like this that give and ask information about divisibility of higher powers of an integer, it's a good idea to start with the prime factorized form of the integer itself.

That is, as the first step, express \(P =\)\(P1^{a}*P2^{b}*P3^{c}*P4^{d}\) . . . where P1, P2, P3, P4 etc. are prime numbers and a, b, c, d etc. are non-negative integers.

We are given that \(P^2\) is divisible by \(12=\) \(2^{2}*3^{1}\)

This means, P1 = 2 and 2a >= 2 That is, a > = 1. So, minimum possible value of a = 1

Also, P2 = 3 and 2b > = 1 That is, b > = 0.5 But b must be an integer. So, minimum possible value of b = 1

So, we see that P = \((2^{1}*3^{1})(something)\) . . .

We don't have any idea about what the value of this 'something' is, but we can say for sure that P must be divisible by \((2^{1}*3^{1})\) and so, \(P^3\) must be divisible by \((2^{3}*3^{3})\)

Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

Show Tags

06 May 2015, 09:41

Hello! Apologies if I am puting an absurd question. What if there was 6^4 and also 6^3 in the answer options. Right now we selected 6^3... but what if with 6^3 there is also 6^4 ? will we choose 6^4? Thanks Celestial

Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

Show Tags

06 May 2015, 09:58

Celestial09 wrote:

Hello! Apologies if I am puting an absurd question. What if there was 6^4 and also 6^3 in the answer options. Right now we selected 6^3... but what if with 6^3 there is also 6^4 ? will we choose 6^4? Thanks Celestial

All we know is that the prime box of p has at least One 3 and One 2. So p^3 will have at least Three 3s' and Three 2s' (3^3 x 2^3) hence the highest factor possible is 6^3. We do not have any evidence to conclude that p^3 can have more than three 3s' and 2s'. Hence 6^4 can't be a factor of p.

Please note it is possible that p has more than three 3s' and 2s' but we cant conclude that from the statement.

Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

Show Tags

19 Aug 2016, 10:34

P^2 has atleast two 2's and two 3's (squares has even powers of integers)

P must thus contain atleast one 2 and one 3. P^3=2*2*2*3*3*3*x P^3=6^3*x P^3 is thus divisible by atleast at most 6^3 given the information known to us.

Posted from my mobile device _________________

I love being wrong. An incorrect answer offers an extraordinary opportunity to improve.

Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

Show Tags

21 Sep 2016, 05:46

sunita123 wrote:

If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p ^3 is

(A) 2^3 (B) 2^6 (C) 3^3 (D) 6^3 (E) 12^2

the wording is kind of confusing...just because of re-reading everything i lost ~20 sec.

since p^2 is divisible by 12, and 12 contains 2^2 and one 3, it must be true that p^2 must contain at least one more 3. since we are given p^3, then 6^3 would work, as 6 has both 2 and 3, and everything is raised to the power of 3. we know for sure that in this case p^3 is divisible by 6^3.

12^2 on the other hand...contains 2^4 * 3^2 - we are not sure whether we have another factor of 3... so D works.

Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

Show Tags

26 Nov 2017, 02:48

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.
_________________