VyDinh wrote:If p and n are positive integers and p>n, what is the remainder when p² - n² is divided by 15?
1) The remainder when p+n is divided by 5 is 1
2) The remainder when p-n is divided by 3 is 1
NOTE: You have transcribed the original GMATPrep question incorrectly in a few places. The corrections are made in red.
Target question: What is the remainder when p² - n² is divided by 15
Rephrased target question: What is the remainder when (p + n)(p - n) is divided by 15?
For this question, we're going to use a nice rule that says:
If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
Okay, onto the question . . .
Statement 1: The remainder when p+n is divided by 5 is 1
Since there's no information about p-n, statement 1 is NOT SUFFICIENT.
Statement 2: The remainder when p-n is divided by 3 is 1
Since there's no information about p+n, statement 2 is NOT SUFFICIENT.
Statements 1 and 2 combined
Statement 1: Applying the above
rule, some possible values of p+n are 6, 11, 16, 21, 26, etc.
Aside: you'll notice that I didn't include 1 as a possible value since we're told that p and n are positive integers, and we can't get a sum of 1 if both are positive
Statement 2: Applying the above
rule, some possible values of p-n are 1, 4, 7, 10, 13, etc
Let's examine two cases with conflicting results.
case a: p+n = 11 and p-n = 1
Add the equations to get 2p = 12, which means p = 6 and n = 5 (perfect, we have positive integer values for p and n)
In this case,
when (p + n)(p - n) is divided by 15, the remainder is 11
case b: p+n = 6 and p-n = 4
Add the equations to get 2p = 10, which means p = 5 and n = 1 (perfect, we have positive integer values for p and n)
In this case,
when (p + n)(p - n) is divided by 15, the remainder is 9
Since we cannot answer the
target question with certainty, the combined statements are NOT SUFFICIENT
Answer =
E
Cheers,
Brent
