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If p is a positive integer, is p^2 divisible by 96?
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05 Jun 2015, 06:20
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Re: If p is a positive integer, is p^2 divisible by 96?
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05 Jun 2015, 10:09
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 8This 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 SUFFICIENTStatement 2: p^2 is a multiple of 12This 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 SUFFICIENTCombining the two statementsStatement 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 \(2^3 * 3\) SUFFICIENTAnswer: Option
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Re: If p is a positive integer, is p^2 divisible by 96?
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05 Jun 2015, 07:30
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?
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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?
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05 Jun 2015, 19:21
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?
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06 Jun 2015, 11:31
Hi All, This question can be solved with a mix of primefactorization, 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 primefactorization. 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 primefactorization... 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
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Re: If p is a positive integer, is p^2 divisible by 96?
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07 Jun 2015, 03:59
EMPOWERgmatRichC wrote: Hi All, This question can be solved with a mix of primefactorization, 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 primefactorization. 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 primefactorization... 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



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Re: If p is a positive integer, is p^2 divisible by 96?
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07 Jun 2015, 15:10
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
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Re: If p is a positive integer, is p^2 divisible by 96?
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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?
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08 Jun 2015, 04:09
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?
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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?
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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



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