Author 
Message 
TAGS:

Hide Tags

Current Student
Joined: 22 Jul 2014
Posts: 125
Concentration: General Management, Finance
WE: Engineering (Energy and Utilities)

Is p^2  1 divisible by 12? [#permalink]
Show Tags
Updated on: 15 Aug 2014, 09:13
Question Stats:
68% (01:10) correct 32% (00:54) wrong based on 134 sessions
HideShow timer Statistics
Is p^2  1 divisible by 12? (1) p > 3 (2) p is a prime number Source: 4gmat
Official Answer and Stats are available only to registered users. Register/ Login.
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.
Edited the question



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: Is p^2  1 divisible by 12? [#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 13 Oct 2013
Posts: 130
Concentration: Strategy, Entrepreneurship

Re: Is p^2  1 divisible by 12? [#permalink]
Show Tags
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.
_________________
 Kindly press +1 Kudos if my post helped you in any way



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: Is p^2  1 divisible by 12? [#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 22 Feb 2009
Posts: 194

Re: Is p^2  1 divisible by 12? [#permalink]
Show Tags
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
_________________
......................................................................... +1 Kudos please, if you like my post



Intern
Joined: 05 Sep 2014
Posts: 8

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



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: Is p^2  1 divisible by 12? [#permalink]
Show Tags
06 Oct 2014, 00:31



Intern
Joined: 05 Sep 2014
Posts: 8

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



NonHuman User
Joined: 09 Sep 2013
Posts: 7040

Re: Is p^2  1 divisible by 12? [#permalink]
Show Tags
28 Sep 2017, 07:37
Hello from the GMAT Club BumpBot! 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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




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






