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Is p^2  1 divisible by 12?
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Updated on: 15 Aug 2014, 09:13
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Is p^2  1 divisible by 12? (1) p > 3 (2) p is a prime number Source: 4gmat
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Originally posted by alphonsa on 15 Aug 2014, 08:13.
Last edited by Bunuel on 15 Aug 2014, 09:13, edited 1 time in total.
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Re: Is p^2  1 divisible by 12?
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15 Aug 2014, 09:23
Is p^2  1 divisible by 12?(1) p > 3. This one is clearly insufficient: if p = 4, then the answer is NO but if p = 5, then the answer is YES. Not sufficient. (2) p is a prime number. If p = 2, then the answer is NO but if p = 5, then the answer is YES. Not sufficient. (1)+(2) Important property: ANY prime number \(p\) greater than 3 can be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n1\)), where \(n\) is an integer >1.That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3). But:Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so viseversa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number. So, according to the above, p can be expressed as \(p=6n+1\) or \(p=6n1\). If \(p=6n+1\), then \(p^2  1 = 36n^2 +12n=12(3n^2+1)\) and if \(p=6n1\), then \(p^2  1 = 36n^2 12n=12(3n^21)\). In both cases p is a multiple of 12. Sufficient. Answer: C.
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Re: Is p^2  1 divisible by 12?
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16 Aug 2014, 19:27
Hi Bunuel, for P = 5 or 7 , p^21 is divisible by 12. Should not answer be E? Bunuel wrote: Is p^2  1 divisible by 12?
(1) p > 3. This one is clearly insufficient: if p = 4, then the answer is NO but if p = 5, then the answer is YES. Not sufficient.
(2) p is a prime number. If p = 2, then the answer is NO but if p = 5, then the answer is YES. Not sufficient.
(1)+(2) Important property: ANY prime number \(p\) greater than 3 can be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n1\)), where \(n\) is an integer >1.
That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).
But: Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so viseversa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.
So, according to the above, p can be expressed as \(p=6n+1\) or \(p=6n1\). If \(p=6n+1\), then \(p^2  1 = 36n^2 +12n=12(3n^2+1)\) and if \(p=6n1\), then \(p^2  1 = 36n^2 12n=12(3n^21)\). In both cases p is a multiple of 12. Sufficient.
Answer: C.
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Re: Is p^2  1 divisible by 12?
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17 Aug 2014, 04:05
sunita123 wrote: Hi Bunuel, for P = 5 or 7 , p^21 is divisible by 12. Should not answer be E? Bunuel wrote: Is p^2  1 divisible by 12?
(1) p > 3. This one is clearly insufficient: if p = 4, then the answer is NO but if p = 5, then the answer is YES. Not sufficient.
(2) p is a prime number. If p = 2, then the answer is NO but if p = 5, then the answer is YES. Not sufficient.
(1)+(2) Important property: ANY prime number \(p\) greater than 3 can be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n1\)), where \(n\) is an integer >1.
That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).
But: Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so viseversa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.
So, according to the above, p can be expressed as \(p=6n+1\) or \(p=6n1\). If \(p=6n+1\), then \(p^2  1 = 36n^2 +12n=12(3n^2+1)\) and if \(p=6n1\), then \(p^2  1 = 36n^2 12n=12(3n^21)\). In both cases p is a multiple of 12. Sufficient.
Answer: C. For any prime number p greater than 3, p^2  1 IS divisible by 12. So, taken together the statements are sufficient to get a definite YES answer to the question. Which means that the answer is C.
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Re: Is p^2  1 divisible by 12?
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17 Aug 2014, 16:24
ANY prime number \(p\) greater than 3 can be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n1\)), where \(n\) is an integer >1.
That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).
But: Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so viseversa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.Again, I have never known the concept before. Thank you, Bunuel
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Is p^2  1 divisible by 12?
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05 Oct 2014, 12:17
ANY prime number p greater than 3 can be expressed as p=6n+1 or p=6n+5 (p=6n1), where n is an integer >1.
I have a problem with highlighted part. How can we represent 5 or 7 in the forms 6n +1 or 6n +5. I think the above property holds good for n>=0. Can somebody clarify?



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Re: Is p^2  1 divisible by 12?
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06 Oct 2014, 00:31



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Re: Is p^2  1 divisible by 12?
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08 Oct 2014, 09:44
Thank you Bunnel for your prompt response! You are truly incredible!



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Re: Is p^2  1 divisible by 12?
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28 Sep 2017, 07:37
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Re: Is p^2  1 divisible by 12? &nbs
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