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(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.

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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--------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way

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.
_________________

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
_________________

......................................................................... +1 Kudos please, if you like my post

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?

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?

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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