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Is p^2 - 1 divisible by 12?

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Is p^2 - 1 divisible by 12? [#permalink]

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Is p^2 - 1 divisible by 12?

(1) p > 3

(2) p is a prime number

Source: 4gmat
[Reveal] Spoiler: OA

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? [#permalink]

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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 \(p=6n+5\) (\(p=6n-1\)), 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 vise-versa 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 \(p=6n-1\). If \(p=6n+1\), then \(p^2 - 1 = 36n^2 +12n=12(3n^2+1)\) and if \(p=6n-1\), then \(p^2 - 1 = 36n^2 -12n=12(3n^2-1)\). In both cases p is a multiple of 12. Sufficient.

Answer: C.
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Re: Is p^2 - 1 divisible by 12? [#permalink]

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New post 16 Aug 2014, 19:27
Hi Bunuel,
for P = 5 or 7 , p^2-1 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 \(p=6n+5\) (\(p=6n-1\)), 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 vise-versa 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 \(p=6n-1\). If \(p=6n+1\), then \(p^2 - 1 = 36n^2 +12n=12(3n^2+1)\) and if \(p=6n-1\), then \(p^2 - 1 = 36n^2 -12n=12(3n^2-1)\). In both cases p is a multiple of 12. Sufficient.

Answer: C.

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Re: Is p^2 - 1 divisible by 12? [#permalink]

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New post 17 Aug 2014, 04:05
sunita123 wrote:
Hi Bunuel,
for P = 5 or 7 , p^2-1 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 \(p=6n+5\) (\(p=6n-1\)), 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 vise-versa 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 \(p=6n-1\). If \(p=6n+1\), then \(p^2 - 1 = 36n^2 +12n=12(3n^2+1)\) and if \(p=6n-1\), then \(p^2 - 1 = 36n^2 -12n=12(3n^2-1)\). 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? [#permalink]

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New post 17 Aug 2014, 16:24
ANY prime number \(p\) greater than 3 can be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n-1\)), 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 vise-versa 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 :-D :-D
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Is p^2 - 1 divisible by 12? [#permalink]

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New post 05 Oct 2014, 12:17
ANY prime number p greater than 3 can be expressed as p=6n+1 or p=6n+5 (p=6n-1), 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? [#permalink]

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New post 06 Oct 2014, 00:31
annie2014 wrote:
ANY prime number p greater than 3 can be expressed as p=6n+1 or p=6n+5 (p=6n-1), 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?


It should be >=1.

5 = 6*1 - 1;
7 = 6*1 + 1.
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Collection of Questions:
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Re: Is p^2 - 1 divisible by 12? [#permalink]

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New post 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? [#permalink]

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Re: Is p^2 - 1 divisible by 12?   [#permalink] 28 Sep 2017, 07:37
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