GMAT PREP-Someone please help me with this math questionv3

[SB]

Remainder problem - Does anyone know the easiest way to solve the following problem.. and also please let me know how you got the answer

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

Answer: E

[StaceyKoprince]

Please read (and follow!) the guidelines in the stickies. GMATPrep problems have their own folders. Also, your subject line should be the first 5-8 words of the problem itself. I'll move this over to the right folder for you this time but please remember for next time! Thanks. :)

[mdh3000]

I find these types of questions (remainder) difficult, so I'll give it a shot.

Remember that...
p^2-n^2 = (p+n)(p-n)

1) Since p + n are positive integers, this tells us p+n could be: 6, 11, 16, 21, 26, 31.... (Insufficient)

2) Since we know p>n, this tells us p-n could be: 7, 4, 1, -1, -4, -7.... (Insufficient)

Taken together, we might get somewhere because (p+n)(p-n) = p^2-n^2; however, if you try a couple combinations, you'll see that a bunch of different remainders result....

(1)(11) = 11/15 = 0 +R 11
(-1)(16) = -16/16 = -1 +R -1
(-4)(11) = -44/16 = -2 +R -12

Using both of the statements together doesn't narrow down the remainder of (p^2-n^2)/15 to a single number, so the answer is E.

mdh

[Nagm]

(1) The remainder when p+n is divided by 5 is 1
(2) The remainder when p-n is divided by 3 is 1

If remainder is 1 when p+n is divided by 5 then p+n = 1 or multiple of 6 (5+1)

If remainder is 1 when p+n is divided by 3 then p+n = 1 or multiple of 4 (3+1)

Combining these together you will get the answer

[Nagm]

OOPS
I mean to say p+n is 5K+1, where K=0,1,2,3
I mean to say p+n is 3K+1, where K=0,1,2,3

[RonPurewal]

I find these types of questions (remainder) difficult, so I'll give it a shot.

Remember that...
p^2-n^2 = (p+n)(p-n)

1) Since p + n are positive integers, this tells us p+n could be: 6, 11, 16, 21, 26, 31.... (Insufficient)

2) Since we know p>n, this tells us p-n could be: 7, 4, 1, -1, -4, -7.... (Insufficient)

Taken together, we might get somewhere because (p+n)(p-n) = p^2-n^2; however, if you try a couple combinations, you'll see that a bunch of different remainders result....

(1)(11) = 11/15 = 0 +R 11
(-1)(16) = -16/16 = -1 +R -1
(-4)(11) = -44/16 = -2 +R -12

Using both of the statements together doesn't narrow down the remainder of (p^2-n^2)/15 to a single number, so the answer is E.

mdh

wo wo, no, there are 2 significant errors here.

first of all, you know that p is greater than n. this means that (p - n) MUST be a positive number, so all the negative values you've listed are impossible.
the proper interpretation is that (p - n) does, indeed, have to be one of 1, 4, 7, ..., but also must be LESS than the listed value for p + n.
specifically, the difference between (p + n) and (p - n) is 2n, which is an even number (because n is an integer). so the two values you choose must be either both even or both odd; for instance, you can't choose p + n = 11 and p - n = 4.

still, you can prove that the answer is (e) by finding the right pair of numbers:
p + n = 11, p - n = 7 --> p^2 - n^2 = 77, remainder = 2
p + n = 6, p - n = 4 --> p^2 - n^2 = 24, remainder = 9
these two examples alone, which satisfy both (1) and (2) but yield different results, constitute conclusive proof that the answer to this problem is (e).

--

second of all (this is not important for the gmat, but i figured you should have an 'fyi'):
if you're going to find remainders with negative numbers, the remainders are still positive.
this is an absolutely necessary rule; patterns don't just flip on their heads.
here's how it works for division by 3:
5 / 3: quotient = 1, remainder = 2
4 / 3: quotient = 1, remainder = 1
3 / 3: quotient = 1, remainder = 0
2 / 3: quotient = 0, remainder = 2
1 / 3: quotient = 0, remainder = 1
0 / 3: quotient = 0, remainder = 0
-1 / 3: quotient = -1, remainder = 2
-2 / 3: quotient = -1, remainder = 1
-3 / 3: quotient = -1, remainder = 0
-4 / 3: quotient = -2, remainder = 2
-5 / 3: quotient = -2, remainder = 1
-6 / 3: quotient = -2, remainder = 0
etc.

this is pretty weird, but it's the only way to define remainders that makes any sense. (it's a continuation of something called the euclidean algorithm, which is the method formally used in number theory to define remainders.)

[guest12]

hi ron. can you please explain the following to your answer:

the proper interpretation is that (p - n) does, indeed, have to be one of 1, 4, 7, ..., but also must be LESS than the listed value for p + n.
specifically, the difference between (p + n) and (p - n) is 2n, which is an even number (because n is an integer). so the two values you choose must be either both even or both odd;

[rfernandez]

the proper interpretation is that (p - n) does, indeed, have to be one of 1, 4, 7, ..., but also must be LESS than the listed value for p + n.
specifically, the difference between (p + n) and (p - n) is 2n, which is an even number (because n is an integer). so the two values you choose must be either both even or both odd;

If p and n are positive integers such that p > n, then it must be true that their difference (p - n) is less than their sum (p + n).

(p + n) - (p - n) = p + n - p + n = 2n. Therefore, the difference between p+n and p-n must be an even number. So if p+n is even, then p-n is even (because they are separated by an even number). By similar reasoning, if p+n is odd, then p-n is odd.

[rockrock]

For this problem, I came up with E based on the fact that we know the remainders for the two variables (p+n) and (p-n) for different divisors. Thus, if (p-n) is divisible by 3 with a remainder of 1 - that tells us nothing about its divisibilty of 5.

If they were the same divisors, can't we just add the remainders - making the answer C?

[RonPurewal]

For this problem, I came up with E based on the fact that we know the remainders for the two variables (p+n) and (p-n) for different divisors. Thus, if (p-n) is divisible by 3 with a remainder of 1 - that tells us nothing about its divisibilty of 5.

correct, although you have to be careful with this sort of reasoning -- if the two divisors are related (specifically, if one of the divisors is a factor of the other one), then knowing one of the remainders may allow you to find the other.
for instance, if dividing a number N by 8 gives a remainder of 5, then dividing the same number N by 4 (which is a factor of 8) must give a remainder of 1.

so, your reasoning is sound, but only because neither of the divisors in question (3 and 5) is a factor of the other.

If they were the same divisors, can't we just add the remainders - making the answer C?

yeah -- if you were dividing by the same number in both stated facts AND in the question itself, then, yes, you could do this.
the remainders are additive, but you may have to "roll over". for instance, if dividing M by 8 gives a remainder of 6, and if dividing N by 8 gives a remainder of 5, then dividing (M + N) by 8 doesn't give a remainder of 11; you can take another 8 out of that 11, yielding a remainder of 3.
otherwise, yes, this logic is sound.