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#### p and n are positive integers

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 Post subject: p and n are positive integers  Posted: Mon Sep 15, 2008 1:51 pm
 If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15 The remainder when p+q is divided by 5 is 1 The remainder when p-q is divided by 3 is 1

 Post subject:   Posted: Tue Sep 16, 2008 2:23 am
 Use listing method.. p+n=11 p-n=1 R=11 p+n=6 p-n=1 R=6 So answer is E.

 Post subject:   Posted: Thu Oct 09, 2008 7:58 am
 ManhattanGMAT Staff

Posts: 8087
 lionkiNg wrote:Use listing method.. p+n=11p-n=1R=11p+n=6p-n=1R=6So answer is E. good enough. remember, if you aren't getting anywhere by using theory, you should IMMEDIATELY start using secondary methods (such as listing possibilities). it can sometimes take a relatively long time to find the possibilities you need, so you shouldn't delay. do not deliberate.

 Post subject:   Posted: Sat Oct 11, 2008 10:54 am
 However there are no positive integer p and n that satisfy the below example. p+n=6 p-n=1 p = 7/2, n = 5/2 These are not positive integers! Therefore p+n=6 p-n=1 is a bad example. p+n=16 p-n=4 works better since r = 4 and (p,n) = (10,6).

 Post subject:   Posted: Sat Oct 11, 2008 11:54 am
 lionkiNg wrote:Use listing method.. p+n=11p-n=1R=11p+n=6p-n=1R=6So answer is E. I don't understand a thing :mrgreen: ! Please explain

 Post subject:   Posted: Sun Oct 12, 2008 4:02 pm
 ManhattanGMAT Staff

Posts: 8087
 amit wrote:lionkiNg wrote:Use listing method.. p+n=11p-n=1R=11p+n=6p-n=1R=6So answer is E.I don't understand a thing :mrgreen: ! Please explain ok, i'll explain. by the way, the problem statement contains 'q' but the rest of everything contains 'n', so i'm going to go with 'n' just for the sake of consistency (and because it's easier to type). the essence of this solution is that lionking is just finding different examples of numbers that satisfy the conditions, and just trying those numbers out. essential fact: this problem features the DIFFERENCE OF SQUARES, p^2 - n^2 = (p - n)(p + n). if you don't make this realization IMMEDIATELY, then you're going to be mired forever in trying to solve for p and n individually. (remember that, if a problem is stated entirely in terms of combinations, then the problem can almost certainly be solved entirely in terms of combinations.) lionking's first example chooses p + n = 11 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required). when 11 x 1 = 11 is divided by 15, the remainder is 11. lionking's second example chooses p + n = 6 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required). when 6 x 1 = 11 is divided by 15, the remainder is 6. these are two different remainders, so, insufficient.

 Post subject: Re:  Posted: Sun May 06, 2012 9:03 am
 Students

Posts: 5
 RonPurewal wrote:ok, i'll explain. by the way, the problem statement contains 'q' but the rest of everything contains 'n', so i'm going to go with 'n' just for the sake of consistency (and because it's easier to type).the essence of this solution is that lionking is just finding different examples of numbers that satisfy the conditions, and just trying those numbers out.essential fact: this problem features the DIFFERENCE OF SQUARES, p^2 - n^2 = (p - n)(p + n). if you don't make this realization IMMEDIATELY, then you're going to be mired forever in trying to solve for p and n individually. (remember that, if a problem is stated entirely in terms of combinations, then the problem can almost certainly be solved entirely in terms of combinations.)lionking's first example chooses p + n = 11 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required).when 11 x 1 = 11 is divided by 15, the remainder is 11.lionking's second example chooses p + n = 6 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required).when 6 x 1 = 11 is divided by 15, the remainder is 6.these are two different remainders, so, insufficient.I don't mean to rain on the parade...but p + n = 6p - n = 1 do not yield integers for p and n.(that is required as per problem)I guess we can choose ..for the second casep + n = 6p - n = 4 and we get- insufficient.

 Post subject: Re: Re:  Posted: Mon May 07, 2012 2:15 am
 ManhattanGMAT Staff

Posts: 8087
 chandra, good catch. _________________Being well-dressed gives a feeling of inward tranquillity [that] religion is powerless to bestow.C.F. Forbes

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