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

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

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New post 05 Jun 2015, 06:20
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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 05 Jun 2015, 10:09
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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

Answer: Option
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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 05 Jun 2015, 07:30
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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

Answer: C
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If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 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 \(p^2\neq{16}\)

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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 05 Jun 2015, 19:21
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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
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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 06 Jun 2015, 11:31
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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:

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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 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:

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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?

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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 07 Jun 2015, 15:10
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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.

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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 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.
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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 08 Jun 2015, 04:09
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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.

The correct answer is C.
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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 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.

The correct answer is C.
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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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New post 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

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Re: If p is a positive integer, is p^2 divisible by 96?  [#permalink]

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Re: If p is a positive integer, is p^2 divisible by 96?   [#permalink] 03 Jan 2019, 07:54
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