It is currently 15 Dec 2017, 02:27

# Decision(s) Day!:

CHAT Rooms | Wharton R1 | Stanford R1 | Tuck R1 | Ross R1 | Haas R1 | UCLA R1

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If p is a positive integer and p^2 is divisible by 12

Author Message
TAGS:

### Hide Tags

Manager
Joined: 13 Oct 2013
Posts: 133

Kudos [?]: 60 [4], given: 129

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

### Show Tags

29 Dec 2014, 15:51
4
KUDOS
2
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

69% (01:08) correct 31% (01:20) wrong based on 187 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

---------------------------------------------------------------------------------------------
Kindly press +1 Kudos if my post helped you in any way

Kudos [?]: 60 [4], given: 129

Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 613

Kudos [?]: 810 [1], given: 59

Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

### Show Tags

29 Dec 2014, 18:56
1
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

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

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

_________________

Ankit

Check my Tutoring Site -> Brush My Quant

GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief

How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

Kudos [?]: 810 [1], given: 59

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1849

Kudos [?]: 2790 [1], given: 193

Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

### Show Tags

30 Dec 2014, 20:09
1
KUDOS

Least value of p for which p^2 divisible by 12 is 6

$$\frac{6^2}{12} = 3$$

Largest value which can divide 6^3 is 6^3
_________________

Kindly press "+1 Kudos" to appreciate

Kudos [?]: 2790 [1], given: 193

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10401

Kudos [?]: 3693 [1], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]

### Show Tags

30 Dec 2014, 20:28
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
Hi sunita123,

This question is essentially about prime factorization and how exponents "work."

When a question uses an exponent, you have to remember what the exponent implies:

Here, we have P^2, which means (P)(P)

For example, if P is a positive integer and P^2 = 100, then P = 10

This can be written in a variety of ways:

P^2 = 100
P^2 = 10^2
P^2 = (10)(10)
P^2 = [2x5][2x5]

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.

[Reveal] Spoiler:
D

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3693 [1], given: 173

Manager
Joined: 08 Oct 2013
Posts: 51

Kudos [?]: 17 [0], given: 16

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

### Show Tags

05 May 2015, 18:17
If p^2 is divisible by 12 then we can conclude that the prime box of p has at-least 3 & 2 in it. (note that we are not sure if there is a 4)

So p^3 will be divisible by (3x2)^3, that is 6^3

Kudos [?]: 17 [0], given: 16

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 755

Kudos [?]: 2242 [2], given: 123

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

### Show Tags

06 May 2015, 02:56
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.

This means

$$P^2$$ =$$P1^{2a}*P2^{2b}*P3^{2c}*P4^{2d}$$ . . .

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 $$(2^{3}*3^{3})$$

Hope this was useful!

Japinder
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Kudos [?]: 2242 [2], given: 123

Retired Moderator
Status: I Declare War!!!
Joined: 02 Apr 2014
Posts: 257

Kudos [?]: 97 [0], given: 546

Location: United States
Concentration: Finance, Economics
GMAT Date: 03-18-2015
WE: Asset Management (Investment Banking)
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

Kudos [?]: 97 [0], given: 546

Manager
Joined: 08 Oct 2013
Posts: 51

Kudos [?]: 17 [0], given: 16

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.

Hope it is clear.

Regards,
Aj

Kudos [?]: 17 [0], given: 16

Intern
Joined: 20 Oct 2014
Posts: 5

Kudos [?]: 11 [0], given: 1

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

### Show Tags

19 Aug 2016, 07:59
I took P=12 for a sample, and ended up picking 12^2

Kudos [?]: 11 [0], given: 1

Manager
Joined: 15 Mar 2015
Posts: 113

Kudos [?]: 29 [0], given: 7

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.

Kudos [?]: 29 [0], given: 7

Board of Directors
Joined: 17 Jul 2014
Posts: 2698

Kudos [?]: 451 [0], given: 208

Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
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.

Kudos [?]: 451 [0], given: 208

Non-Human User
Joined: 09 Sep 2013
Posts: 14830

Kudos [?]: 287 [0], given: 0

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.
_________________

Kudos [?]: 287 [0], given: 0

Re: If p is a positive integer and p^2 is divisible by 12   [#permalink] 26 Nov 2017, 02:48
Display posts from previous: Sort by