GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Mar 2019, 20:41

### 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

# If p is a positive integer, is p^2 divisible by 96?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53738
If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

05 Jun 2015, 06:20
00:00

Difficulty:

45% (medium)

Question Stats:

68% (01:53) correct 32% (01:54) wrong based on 296 sessions

### HideShow timer Statistics

If p is a positive integer, is p^2 divisible by 96?

(1) p is a multiple of 8.
(2) p^2 is a multiple of 12.

_________________
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2844
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

05 Jun 2015, 10:09
3
2
Bunuel wrote:
If p is a positive integer, is p^2 divisible by 96?

(1) p is a multiple of 8.
(2) p^2 is a multiple of 12.

Questions: If p is a positive integer, is p^2 divisible by 96?

96 = $$2^5 * 3$$

i.e. for $$P^2$$ to be a multiple of 96, it must be a multiple of $$(2^5 * 3)$$

Since P is an Integer therefore $$P^2$$ must be a perfect square

i.e. $$P^2$$ MUST be a multiple of$$(2^6 * 3^2)$$

i.e P MUST be a multiple of $$(2^3 * 3)$$

Question Stem Modified: Is P a multiple of $$(2^3 * 3)$$?

Statement 1: p is a multiple of 8

This confirms that p is a multiple of $$2^3$$ but since it's still unknown whether p is a multiple of 3 or not therefore

NOT SUFFICIENT

Statement 2: p^2 is a multiple of 12

This confirms that p is a multiple of 3 and 2 but since it's still unknown whether p is a multiple of $$2^3$$ or not therefore

NOT SUFFICIENT

Combining the two statements

Statement 1 confirms that p is a multiple of $$2^3$$

Statement 2 confirms that p is a multiple of 3

i.e. p is a multiple of $$2^3 * 3$$

SUFFICIENT

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

##### General Discussion
Senior Manager
Joined: 28 Feb 2014
Posts: 294
Location: United States
Concentration: Strategy, General Management
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

05 Jun 2015, 07:30
1
If p is a positive integer, is p^2 divisible by 96?

(1) p is a multiple of 8.
If p^2=96^2 yes
if p^2=16 no
Insufficient

(2) p^2 is a multiple of 12.
if p^2=144 no
if p^2= 144^2 yes
insufficient

Combined, p^2 has a minimum of five 2s and one 3
sufficient

Manager
Joined: 18 Nov 2013
Posts: 78
Concentration: General Management, Technology
GMAT 1: 690 Q49 V34
If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

05 Jun 2015, 19:01
peachfuzz you need one correction in your solution for stmt #1

peachfuzz wrote:
(1) p is a multiple of 8.
If p^2=96^2 yes
if p^2=16 no -- > as p is a multiple of 8, so $$p^2\geq{8*8}$$ ; or $$p^2\neq{16}$$

_________________

_______
- Cheers

+1 kudos if you like

Manager
Joined: 18 Nov 2013
Posts: 78
Concentration: General Management, Technology
GMAT 1: 690 Q49 V34
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

05 Jun 2015, 19:21
1
If p is a positive integer, is p^2 divisible by 96?

Stmt (1) p is a multiple of 8.
lets pick numbers, take {p=8, p=96}
p = 8 ; --------------> $$p^2/96^2 = 64/96^2 = No$$
p = 96 ; --------------> $$p^2/96^2 = 96^2/96^2 = Yes$$
not sufficient

Stmt (2) p^2 is a multiple of 12.
lets take {p=6, p=96 } ; Note we picked p=6 because p is a positive integer, we can't say p^2 = 12; need a perfect square value for p^2;
p = 6 ; --------------> $$p^2/96^2 = 36/96^2 = No$$
p = 96 ; --------------> $$p^2/96^2 = 96^2/96^2 = Yes$$
not sufficient

together we know p is a multiple of 8 (p^2 must have 8 * 8 ), and p^2 is a multiple of 12 (so it must have 3 * 3 )
minimum value p^2 = 8 * 8 * 3 * 3 = 576
$$p^2/96^2 = yes$$
sufficient

Ans : C
_________________

_______
- Cheers

+1 kudos if you like

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13765
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

06 Jun 2015, 11:31
2
Hi All,

This question can be solved with a mix of prime-factorization, Number Properties and TESTing VALUES.

We're told that P is a POSITIVE INTEGER. We're asked if P^2 is divisible by 96. This is a YES/NO question.

Before jumping to the two Facts, we can dissect the prompt a bit. 96 = (2^5)(3), so for a number to be divisible by 96, that number MUST have at least (2^5)(3) in its prime-factorization. Since P is an INTEGER and it's P^2 that's divisible by 96, we have to make sure that we're following ALL of the rules in this prompt before we TEST VALUES...

Fact 1: P is a multiple of 8.

8 = (2^3), so P^2 will include (2^6) at the minimum. To be divisible by 96 though, we need to have a '3' in the prime-factorization...

IF....
P = 8, P^2 = (2^6) and the answer to the question is NO

IF....
P = 24, P^2 = (2^6)(3^2) and the answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: P^2 is a multiple of 12.

Here, we have an interesting 'wrinkle'....P MUST be an integer, but there is no integer (when squared) that equals 12...The smallest multiple of 12 that is a perfect square is 36...

36 = 6^2 = (2^2)(3^2) --> this has the necessary '3', but not enough 2's...

IF....
P = 6, P^2 = (2^2)(3^2) and the answer to the question is NO

IF....
P = 24, P^2 = (2^6)(3^2) and the answer to the question is YES
Fact 2 is INSUFFICIENT

Combined, we know....
P is a multiple of 8
P is a multiple of 6

From this, we know that P COULD be 24, 48, 72, etc. ALL of those, when squared, will be divisible by 96. The answer to the question is ALWAYS YES.
Combined, SUFFICIENT

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!***** SVP Joined: 26 Mar 2013 Posts: 2098 Re: If p is a positive integer, is p^2 divisible by 96? [#permalink] ### Show Tags 07 Jun 2015, 03:59 EMPOWERgmatRichC wrote: Hi All, This question can be solved with a mix of prime-factorization, Number Properties and TESTing VALUES. We're told that P is a POSITIVE INTEGER. We're asked if P^2 is divisible by 96. This is a YES/NO question. Before jumping to the two Facts, we can dissect the prompt a bit. 96 = (2^5)(3), so for a number to be divisible by 96, that number MUST have at least (2^5)(3) in its prime-factorization. Since P is an INTEGER and it's P^2 that's divisible by 96, we have to make sure that we're following ALL of the rules in this prompt before we TEST VALUES... Fact 1: P is a multiple of 8. 8 = (2^3), so P^2 will include (2^6) at the minimum. To be divisible by 96 though, we need to have a '3' in the prime-factorization... IF.... P = 8, P^2 = (2^6) and the answer to the question is NO IF.... P = 24, P^2 = (2^6)(3^2) and the answer to the question is YES Fact 1 is INSUFFICIENT Fact 2: P^2 is a multiple of 12. Here, we have an interesting 'wrinkle'....P MUST be an integer, but there is no integer (when squared) that equals 12...The smallest multiple of 12 that is a perfect square is 36... 36 = 6^2 = (2^2)(3^2) --> this has the necessary '3', but not enough 2's... IF.... P = 6, P^2 = (2^2)(3^2) and the answer to the question is NO IF.... P = 24, P^2 = (2^6)(3^2) and the answer to the question is YES Fact 2 is INSUFFICIENT Combined, we know.... P is a multiple of 8 P is a multiple of 6 From this, we know that P COULD be 24, 48, 72, etc. ALL of those, when squared, will be divisible by 96. The answer to the question is ALWAYS YES. Combined, SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich Hi Rich, In order to test fact number, what is the best way to 36. is it trial and error? Is it wrong to use 144 to test this fact? Thanks EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13765 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If p is a positive integer, is p^2 divisible by 96? [#permalink] ### Show Tags 07 Jun 2015, 15:10 1 Hi Mo2men, For Fact 2, the number 144 is a fine number to TEST (and it's relatively easy since you know that 12^2 is a multiple of 12). When it comes to TESTing VALUES, I'm always looking for the easiest numbers to TEST (and that often means looking for the 'smallest'). As far as how I ended up choosing 36, we know that P is an integer and that P^2 is a multiple of 12. So I was looking for a PERFECT SQUARE that was a multiple of 12. If you know your perfect squares, then choosing 36 isn't that big of a 'leap' in logic. 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!*****

SVP
Joined: 26 Mar 2013
Posts: 2098
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

07 Jun 2015, 15:27
EMPOWERgmatRichC wrote:
Hi Mo2men,

For Fact 2, the number 144 is a fine number to TEST (and it's relatively easy since you know that 12^2 is a multiple of 12). When it comes to TESTing VALUES, I'm always looking for the easiest numbers to TEST (and that often means looking for the 'smallest').

As far as how I ended up choosing 36, we know that P is an integer and that P^2 is a multiple of 12. So I was looking for a PERFECT SQUARE that was a multiple of 12. If you know your perfect squares, then choosing 36 isn't that big of a 'leap' in logic.

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

Hi Rich,
Thanks for your prompt and easy logic.
Math Expert
Joined: 02 Sep 2009
Posts: 53738
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

08 Jun 2015, 04:09
2
Bunuel wrote:
If p is a positive integer, is p^2 divisible by 96?

(1) p is a multiple of 8.
(2) p^2 is a multiple of 12.

MANHATTAN GMAT OFFICIAL SOLUTION:

The prime factorization of 96 is (2)(2)(2)(2)(2)(3) = (2^5)(3^1). In order for p^2 to be divisible by 96, p^2 would have to have the prime factors 2^5*3^1 in its prime box. The rephrased question is therefore “Does p^2 have at least five 2's and one 3 in its prime box?”

(1) INSUFFICIENT: If p is a multiple of 8 = (2)(2)(2), p has 2^3 in its prime box. Therefore, p^2 has (2^3)^2 = 2^6 in its prime box, and therefore has the required five 2's. However, it is uncertain whether p^2 has at least one 3 in its prime box.

Alternatively, we could list numbers:
p = 0, 8, 16, 24, etc.
p^2 = 0, 64, 256, 576, etc.
Divisible by 96? Yes, No, No, Yes

(2) INSUFFICIENT: If p^2 is a multiple of 12 = (2)(2)(3), p^2 has two 2's and one 3 in its prime box. It is uncertain whether p^2 has at least five 2's total, as there may or may not be three more 2's in the prime box.

Alternatively, we could list numbers:
p^2 = 0, 36, 144, 96^2, etc. (perfect squares that are multiples of 12)
Divisible by 96? Yes, No, No, Yes.

(1) AND (2) SUFFICIENT: We know from (1) that p^2 has 2^6 in its prime box, and we know from (2) that p^2 has a 3 in its prime box. Therefore, it is certain that p^2 has at least five 2's and one 3 in its prime box.

Note that the number listing approach would be a little cumbersome for the combined statements.

_________________
Director
Joined: 26 Oct 2016
Posts: 634
Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

12 Mar 2017, 06:14
The prime factorization of 96 is (2)(2)(2)(2)(2)(3) = (25)(31). In order for p2 to be divisible by 96, p^2 would have to have the prime factors 2531 in its prime box. The rephrased question is therefore “Does p^2 have at least five 2's and one 3 in its prime box?”

(1) INSUFFICIENT: If p is a multiple of 8 = (2)(2)(2), p has 23 in its prime box. Therefore, p^2 has (2^3)^2 = 2^6 in its prime box, and therefore has the required five 2's. However, it is uncertain whether p^2 has at least one 3 in its prime box.
Alternatively, we could list numbers:
p = 0, 8, 16, 24, etc.
p^2 = 0, 64, 256, 576, etc.
Divisible by 96? Yes, No, No, Yes

(2) INSUFFICIENT: If p^2 is a multiple of 12 = (2)(2)(3), p^2 has two 2's and one 3 in its prime box. It is uncertain whether p^2 has at least five 2's total, as there may or may not be three more 2's in the prime box.
Alternatively, we could list numbers:
p^2 = 0, 36, 144, 962, etc. (perfect squares that are multiples of 12)
Divisible by 96? Yes, No, No, Yes
(1) AND (2) SUFFICIENT: We know from (1) that p^2 has 2^6 in its prime box, and we know from (2) that p^2 has a 3 in its prime box. Therefore, it is certain that p^2 has at least five 2's and one 3 in its prime box.

_________________

Thanks & Regards,
Anaira Mitch

Director
Joined: 12 Nov 2016
Posts: 724
Location: United States
Schools: Yale '18
GMAT 1: 650 Q43 V37
GRE 1: Q157 V158
GPA: 2.66
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

19 Jul 2017, 22:07
Bunuel wrote:
If p is a positive integer, is p^2 divisible by 96?

(1) p is a multiple of 8.
(2) p^2 is a multiple of 12.

Just gotta be careful with the distinction between the two statements

P= 3 x 2^5 x some integer k? let's find out

St 1

P=2^3 x some integer k clearly insufficient

St 2

P^2= 3 x 2^2 x some integer k

St 1 and St 2

Each of P must be divisible by 2^3 x some integer k so P^2 must be divisible by 2^6 which is enough 2's

C
Non-Human User
Joined: 09 Sep 2013
Posts: 10151
Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

### Show Tags

03 Jan 2019, 07:54
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 p is a positive integer, is p^2 divisible by 96?   [#permalink] 03 Jan 2019, 07:54
Display posts from previous: Sort by