Is p^2 - 1 divisible by 12?

Question

(1) p > 3

(2) p is a prime number

Source: 4gmat

Bunuel · Answer

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.

sunita123 · Answer

Hi Bunuel,
for P = 5 or 7 , p^2-1 is divisible by 12.
Should not answer be E?

Bunuel · Answer

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.

vad3tha · Answer

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

annie2014 · Answer

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?

Bunuel · Answer

It should be >=1.

5 = 6*1 - 1;
7 = 6*1 + 1.

annie2014 · Answer

Thank you Bunnel for your prompt response! You are truly incredible!

KanishkM · Answer

p^2 - 1 
can be written as (p-1) (p+1)

Statement 1 -> p=4,5,6
Case 1: 4,  \frac{5*6}{12}  , Question will be answered by a No

Case 2 : 5,  \frac{4*6}{12} , Question will be answered by a Yes

Statement 2-> p =2,5
Case 1: 2,  \frac{1*3}{12} , Question will be answered by a No

Case 2 : 5,  \frac{4*6}{12} , Question will be answered by a Yes

Combine, you will get a Yes if you take the case as p = 5, 7
Case 1 : 5,  \frac{4*6}{12}

Case 2: 7,  \frac{8*6}{12}

Correct Answer C

Is p^2 - 1 divisible by 12?

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