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# Is x2 - y2 divisible by 8? 1. x and y are even integers 2. x

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Is x2 - y2 divisible by 8? 1. x and y are even integers 2. x [#permalink]  03 Nov 2007, 14:47
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Is x2 - y2 divisible by 8?

1. x and y are even integers
2. x + y is divisible by 8

Ok the original statement factors into (x+y)(x-y)

Statement 1 doesn't tell us anything
Statement 2 tells us X+Y us is multiple of 8

However, I was thinking of a situation where X and Y are both equal to 4..
If that is the case, then (X+Y) (X-Y) is equal to 0 and is not divisible by 8?
Am I wrong?

The OA is C
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Re: DS Problem-Help [#permalink]  03 Nov 2007, 15:28
jamesrwrightiii wrote:
However, I was thinking of a situation where X and Y are both equal to 4..
If that is the case, then (X+Y) (X-Y) is equal to 0 and is not divisible by 8?

I'm under the assumption that X and Y are different numbers. If they were the same number, they would both be X.

I initially got E, but after looking over the problem again, C makes perfect sense. When you check out statement 1 by itself, it doesn't say a whole lot. But when you combine statement 2, it makes sense.

For example:

X=5, Y=3
X+Y = 8
X-Y = 1
No dice

X=12, Y=4
X+Y = 16
X-Y = 8

So, you need X and Y to be even integers for the equation to work, that's why Statement 1 is important and that's why the OA is C.
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By the way, please don't post the OA in the original post. Some of us like to try and solve the problem on our own. Thanks!
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Re: DS Problem-Help [#permalink]  03 Nov 2007, 15:32
jamesrwrightiii wrote:
Is x2 - y2 divisible by 8?

1. x and y are even integers
2. x + y is divisible by 8

Ok the original statement factors into (x+y)(x-y)

Statement 1 doesn't tell us anything
Statement 2 tells us X+Y us is multiple of 8

However, I was thinking of a situation where X and Y are both equal to 4..
If that is the case, then (X+Y) (X-Y) is equal to 0 and is not divisible by 8?
Am I wrong?

The OA is C

B it is.

zero is divisible by any number, and the result is always zero, but no number is divisible by zero (in GMAT math).
Re: DS Problem-Help   [#permalink] 03 Nov 2007, 15:32
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