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Re: If p is a positive integer, is p^2 divisible by 96?
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05 Jun 2015, 18:21
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 )
Re: If p is a positive integer, is p^2 divisible by 96?
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06 Jun 2015, 10:31
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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
Re: If p is a positive integer, is p^2 divisible by 96?
[#permalink]
07 Jun 2015, 02:59
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
Re: If p is a positive integer, is p^2 divisible by 96?
[#permalink]
07 Jun 2015, 14:10
1
Expert Reply
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.
Re: If p is a positive integer, is p^2 divisible by 96?
[#permalink]
07 Jun 2015, 14:27
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.
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.
Re: If p is a positive integer, is p^2 divisible by 96?
[#permalink]
12 Mar 2017, 05:14
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.
Re: If p is a positive integer, is p² divisible by 96?
[#permalink]
25 Apr 2020, 10:48
1
Top Contributor
Expert Reply
Bunuel wrote:
If p is a positive integer, is p² divisible by 96?
(1) p is a multiple of 8. (2) p² is a multiple of 12.
Target question:Is p² divisible by 96? This is a good candidate for rephrasing the target question.
-----ASIDE--------------------- A lot of integer property questions can be solved using prime factorization. For questions involving divisibility, divisors, factors and multiples, we can say: If N is divisible by k, then k is "hiding" within the prime factorization of N
Consider these examples: 24 is divisible by 3 because 24 = (2)(2)(2)(3) Likewise, 70 is divisible by 5 because 70 = (2)(5)(7) And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7) And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7) --------------------------
Since 96 = (2)(2)(2)(2)(3), we can rephrase the target question as: REPHRASED target question:Are there four 2's and one 3 hiding in the prime factorization of p² ?
Aside: the video below has tips on rephrasing the target question
Statement 1: p is a multiple of 8 In other words, p is divisible by 8 8 = (2)(2)(2) So, we know that there are at least three 2's hiding in the prime factorization of p This also tells us that there are SIX 2's hiding in the prime factorization of p² Unfortunately this information is not sufficient to answer the target question. Consider these two possible cases: Case a: p = 8, in which case p² = 64. In this case, the answer to the target question is NO, p² is NOT divisible by 96 Case b: p = 24, in which case p² = 576. In this case, the answer to the target question is YES, p² IS divisible by 96 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: p² is a multiple of 12 12 = (2)(2)(3) So, we know that there are at least two 2's and one 3 hiding in the prime factorization of p² Unfortunately this information is not sufficient to answer the target question. Consider these two possible cases: Case a: p = 12, in which case p² = 144 (which is divisible by 12). In this case, the answer to the target question is NO, p² is NOT divisible by 96 Case b: p = 24, in which case p² = 576 (which is divisible by 12). In this case, the answer to the target question is YES, p² IS divisible by 96 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Statement 1 tells us that there are SIX 2's hiding in the prime factorization of p² Statement 2 tells us that there is at least ONE 3 hiding in the prime factorization of p² So, when we combine the two statements, we can be certain that there are at least four 2's and one 3 hiding in the prime factorization of p² Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers, Brent RELATED VIDEO FROM MY COURSE
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Re: If p is a positive integer, is p^2 divisible by 96?
[#permalink]
25 Apr 2020, 13:37
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.
96 = 3 * 32 = 2^5 * 3. So, to be divisible by 96, p has to be expresses as 2^5* 3 1) p = 2^3 k . So, p = 2^6 *k^2. Depending on the value of k, p^2 will be divisible by 96 or not. not sufficient
2) p^2 = k*(2^2 * 3). Not sufficient. Together, p^2 = 2^6 *3^2, it will be divisible by 96. Sufficient C is the answer
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