BANON wrote:
If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 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.
Hello, everyone. I came across this question today in a session with a student and used a number-sense approach to disqualify any hopes of pinning down an answer. I can see that others have started along the same lines—spotting the difference of squares, brainstorming numbers—but not quite pursued the same line of reasoning thereafter, so I thought I would share my approach, just in case it may help someone else. I am going to pick up my analysis from the (C) versus (E) perspective:
Statement (1):
First, (p + n) CANNOT be 1 because
p and
n must EACH be
positive integers (0 does not qualify). So, we are looking at
6, 11, 16, 21, 26, 31, 36, 41, 46, 51... (just as
GMATGuruNY has outlined in an earlier post).
Statement (2):
(p - n) CAN be 1 because two positive integers could conceivably produce such a difference. So, we are looking at
1, 4, 7, 10, 13, 16, 19, 22, 25, 28...All we have to do is figure out valid combinations of
positive integers that could both satisfy the statements and conform to the information given in the problem. Start small:
(p + n) = 6
LET p = 5 and n = 1.
p > n
√(p - n) = (5 - 1) = 4
√This value appears in the second list of numbers from above, so p = 5, n = 1 is a VALID combination to test within the question.
\(\frac{(p^2 - n^2)}{15}\)
\(\frac{(5^2 - 1^2)}{15}\)
\(\frac{(25 - 1)}{15}\)
\(\frac{24}{15}\)
The remainder is 9.Now, our task is to check whether any other valid combinations of
p and
n values will yield the same remainder. We can even go back to (p + n) = 6 and check other values.
LET p = 4 and n = 2.
p > n
√(p - n) = (4 - 2) = 2
XSince 2 is NOT in the second list of numbers from earlier, we can disqualify this combination of values. Furthermore, we CANNOT use 3 and 3 since they would not satisfy the given inequality p > n. We need to test different numbers. How about 11, the next number up, in our sum? We have 10 + 1, 9 + 2, 8 + 3, 7 + 4, and 6 + 5 to run through, potentially.
LET p = 10 and n = 1.
p > n
√(p - n) = (10 - 1) = 9
XLET p = 9 and n = 2.
p > n
√(p - n) = (9 - 2) = 7
√We have another VALID combination of values to test. Go back to the question and substitute:
\(\frac{(p^2 - n^2)}{15}\)
\(\frac{(9^2 - 2^2)}{15}\)
\(\frac{(81 - 4)}{15}\)
\(\frac{77}{15}\)
The remainder is 2.Now, you can see that we CANNOT tell what the remainder will be, since we have created two perfectly valid cases that produced different remainders. Thus,
the answer must be (E).
You might see the above approach as sloppy or less refined than the methods outlined above, but it proved pretty efficient in my case, and I was
certain I had the answer. (And trust me, GMAC™ does not reward bonus points for elegant work.)
Good luck with your studies, however you tackle these tough Quant questions.
- Andrew