NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 16:17 It is currently 25 Apr 2024, 16:17
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

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

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
P
sunita123
Manager
Manager
Joined: 13 Oct 2013
Posts: 117
Own Kudos [?]: 200 [27]
Given Kudos: 530
Concentration: Strategy, Entrepreneurship
Send PM
GMAT ToolKit User
If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   29 Dec 2014, 16:51
6
Kudos
21
Bookmarks
00:00
A
B
C
D
E

Difficulty:

35% (medium)

Question Stats:

67% (01:41) correct 33% (01:52) wrong based on 392 sessions
Hide Show 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
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
L
EMPOWERgmatRichC
GMAT Club Legend
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21846
Own Kudos [?]: 11666 [6]
Given Kudos: 450
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   30 Dec 2014, 21:28
4
Kudos
2
Bookmarks
Expert Reply
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.

Final Answer:
Show Spoiler
D


GMAT assassins aren't born, they're made,
Rich
User avatar
D
BrushMyQuant
Tutor
Joined: 05 Apr 2011
Status:Tutor - BrushMyQuant
Posts: 1777
Own Kudos [?]: 2094 [7]
Given Kudos: 100
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   29 Dec 2014, 19:56
1
Kudos
5
Bookmarks
Expert Reply
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

(A) 2^3
(B) 2^6
(C) 3^3
(D) 6^3
(E) 12^2
General Discussion
User avatar
L
EgmatQuantExpert
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3726
Own Kudos [?]: 16841 [4]
Given Kudos: 165
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   06 May 2015, 03:56
2
Kudos
2
Bookmarks
Expert Reply
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
avatar
PareshGmat
SVP
SVP
Joined: 27 Dec 2012
Status:The Best Or Nothing
Posts: 1562
Own Kudos [?]: 7208 [3]
Given Kudos: 193
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   30 Dec 2014, 21:09
3
Kudos
Answer = (D) 6^3

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
avatar
AjChakravarthy
Intern
Intern
Joined: 08 Oct 2013
Posts: 43
Own Kudos [?]: 39 [1]
Given Kudos: 16
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   05 May 2015, 19:17
1
Bookmarks
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

Answer is D
User avatar
B
MarkusKarl
Manager
Manager
Joined: 15 Mar 2015
Posts: 89
Own Kudos [?]: 63 [1]
Given Kudos: 7
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   19 Aug 2016, 11:34
1
Kudos
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
User avatar
L
GMATBusters
GMAT Tutor
Joined: 27 Oct 2017
Posts: 1905
Own Kudos [?]: 5582 [1]
Given Kudos: 236
WE:General Management (Education)
Send PM
GMAT ToolKit User
Re: If p is a positive integer and p^2 is divisible by 12, then the larges [#permalink] New post   17 Oct 2018, 09:48
1
Kudos
Expert Reply
Since, \(p^2\) is divisible by 12 , \(2^2*3\)
hence the least value of p = 2*3= 6
the largest integer which must divides \(p^3 = 6^3\)
Answer D
User avatar
B
Celestial09
Retired Moderator
Joined: 02 Apr 2014
Status:I Declare War!!!
Posts: 218
Own Kudos [?]: 122 [0]
Given Kudos: 546
Location: United States
Concentration: Finance, Economics
GMAT Date: 03-18-2015
WE:Asset Management (Investment Banking)
Send PM
GMAT ToolKit User
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   06 May 2015, 10:41
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
avatar
AjChakravarthy
Intern
Intern
Joined: 08 Oct 2013
Posts: 43
Own Kudos [?]: 39 [0]
Given Kudos: 16
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   06 May 2015, 10:58
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
avatar
vikram.karthick@gmail.com
Intern
Intern
Joined: 20 Oct 2014
Posts: 4
Own Kudos [?]: 20 [0]
Given Kudos: 1
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   19 Aug 2016, 08:59
I took P=12 for a sample, and ended up picking 12^2 :(
User avatar
P
mvictor
Board of Directors
Joined: 17 Jul 2014
Posts: 2163
Own Kudos [?]: 1180 [0]
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Send PM
GMAT ToolKit User Reviews Badge
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   21 Sep 2016, 06:46
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.
User avatar
D
QuantMadeEasy
Senior Manager
Senior Manager
Joined: 28 Feb 2014
Posts: 471
Own Kudos [?]: 559 [0]
Given Kudos: 74
Location: India
Concentration: General Management, International Business
GMAT 1: 570 Q49 V20
GPA: 3.97
WE:Engineering (Education)
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   02 Jan 2018, 02:55
If p*p is divisible by 12, that means each p must consist 2 and 3; then p has to be 6 only. Now, p^3 becomes 6^3 ie. 2^3 * 3^3. Only D is correct option.
User avatar
B
Mansoor50
Manager
Manager
Joined: 29 May 2017
Posts: 154
Own Kudos [?]: 19 [0]
Given Kudos: 63
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   13 Nov 2018, 04:41
One pitfall i was continuously making in these types of questions was that the erroneous assumption that the numerator contained the EXACT amount of the factors of the divisor.

so p can have several factors of 3 as long as there is at least one 2 present

p^2 would be 2^2 x 3^2

which leads us to p^3 and thus the largest divisor!
User avatar
G
Salsanousi
Senior Manager
Senior Manager
Joined: 19 Oct 2013
Posts: 412
Own Kudos [?]: 306 [0]
Given Kudos: 117
Location: Kuwait
GPA: 3.2
WE:Engineering (Real Estate)
Send PM
GMAT ToolKit User
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   13 Nov 2018, 05:36
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



p^2/12 = integer

that means p^2 must at least be 2^2 and 3 that makes p^2 = 36

216 = 6^3 = 2^3 * 3^3

2^3 is possible but not the largest.
2^6 is not possible.
3^3 is possible but not the largest.
6^3 is the largest.

12^2 = (2^2 * 3)^2 = 2^4 * 3^2 which is not possible.

Answer choice D
avatar
B
kaylaquijas
Intern
Intern
Joined: 05 Mar 2021
Posts: 20
Own Kudos [?]: 1 [0]
Given Kudos: 18
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   16 Oct 2021, 16:50
EMPOWERgmatRichC wrote:
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.

Final Answer:
Show Spoiler
D


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


If I am understanding this correctly, it sounds like you are saying that any factors of p must have the same ratio of twos to ones (1:1). Is that why E does not work in this situation?

This question is similar: https://gmatclub.com/forum/if-m-is-a-po ... fl=similar

I understand that 16 is not an answer because it is missing a factor of 3. The correct answer of 12 has factors of 2x2x3. The ratio of 2s and 3s in the number 12 is the same as the ratio of 2s and 3s in the number 48 (that the ratio of twos to threes is 2:1). Am I understanding this correctly?
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32679
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink] New post   08 Apr 2023, 04:47
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.
GMAT Club Bot
gmatclubot
Re: If p is a positive integer and p^2 is divisible by 12 [#permalink]
08 Apr 2023, 04:47
Moderators:
Bunuel
Math Expert
92915 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions