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Tricky even odd

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Source: — Data Sufficiency |

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by Mike@Magoosh » Tue Jan 14, 2014 5:11 pm
buoyant wrote:Is x^2-y^2 = EVEN ?

1) x+y = ODD
2) x-y = ODD
Dear buoyant,
I'm happy to respond. :-)

What's curious about this question is that we get no statement about the nature of x & y. It is usually a lethal mistake on the GMAT to assume, for example, that x & y are integers when that is not specified. Suppose they are pure numbers --- maybe positive or negative or zero, maybe integers or fractions or decimals. In that case, there a continuous infinity of decimals that are neither even nor odd.

For Statement #1
Try x = 1 and y = 0. Then x + y = 1, which is ODD, and x^2-y^2 = 1, which is ODD.
BUT, we could pick
x = y = 1.5. Then x + y = 1, which is ODD, and x^2-y^2 = 0, which is EVEN.
Different choices lead to different answers. This statement, by itself, is insufficient.

For Statement #2
Try x = 1 and y = 0. Then x - y = 1, which is ODD, and x^2-y^2 = 1, which is ODD.
With two integers, we have to have one even and one odd; the squares will remain even and odd, respectively, and the difference will always be ODD.
For non-integer values:
Try x = 2.5 and y = 1.5. Then x - y = 1, which is ODD, and x^2-y^2 = 6.25 - 2.25 = 4, which is EVEN.
Different choices lead to different answers. This statement, by itself, is insufficient.

For combined statements.
As above, a pair of integers, one even and one odd, will satisfy with statements and always produce an odd difference of squares --- recall that (x - y)*(x + y) = x^2-y^2. See:
https://magoosh.com/gmat/2012/gmat-quant ... o-squares/
https://magoosh.com/gmat/2013/three-alge ... -the-gmat/
Curiously, it is impossible to produce a pair of non-integer values that simultaneously have an ODD sum and and ODD difference. In other words, it's impossible to find a non-integer pair that simultaneously satisfies both statements. If both statements are true, then x & y must be integers, and the difference of squares is ODD. Thus, we would have a definitive answer to the prompt.

Together, the statements are sufficient. Answer = [spoiler](C)[/spoiler]

Does all this make sense?
Mike :-)
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by theCodeToGMAT » Tue Jan 14, 2014 9:27 pm
To find: (x-y)(x+y) = EVEN

Statement 1:
(x+y) = ODD
either x = ODD and y = EVEN or vice versa
in both cases
So, x - y = ODD
Hence, (x-y)(x+y) = ODDxODD = ODD
SUFFICIENT

Statement 2:
x-y = ODD
either x = Odd and y = EVEN or vice versa
So, (x+y) = ODD
Hence, (x-y)(x+y) = ODDxODD = ODD
SUFFICIENT

Answer [spoiler]{D}[/spoiler]?
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by GMATGuruNY » Tue Jan 14, 2014 10:43 pm
theCodeToGMAT wrote:To find: (x-y)(x+y) = EVEN

Statement 1:
(x+y) = ODD
either x = ODD and y = EVEN or vice versa
in both cases
So, x - y = ODD
Hence, (x-y)(x+y) = ODDxODD = ODD
SUFFICIENT

Statement 2:
x-y = ODD
either x = Odd and y = EVEN or vice versa
So, (x+y) = ODD
Hence, (x-y)(x+y) = ODDxODD = ODD
SUFFICIENT

Answer [spoiler]{D}[/spoiler]?
Be careful:
The portions in red assume that x and y are integers, a constraint not given in the problem.

Question stem, rephrased: Is (x+y)(x-y) even?

Statement 1: x+y is odd
It's possible that x=2 and y=1, since 2+1 = 3.
In this case, (x+y)(x-y) = (2+1)(2-1) = 3, which is ODD.

It's possible that x=5/2 and y=1/2, since 5/2 + 1/2 = 3.
In this case, (x+y)(x-y) = (5/2 + 1/2)(5/2 - 1/2) = 6, which is EVEN.
INSUFFICIENT.

Statement 2: x-y is odd
It's possible that x=2 and y=1, since 2-1 = 1.
In this case, (x+y)(x-y) = (2+1)(2-1) = 3, which is ODD.

It's possible that x=3/2 and y=1/2, since 3/2 - 1/2 = 1.
In this case, (x+y)(x-y) = (3/2 + 1/2)(3/2 - 1/2) = 2, which is EVEN.
INSUFFICIENT.

Statements combined:
(x+y)(x-y) = ODD * ODD = ODD.
SUFFICIENT.

The correct answer is C.
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by Matt@VeritasPrep » Wed Jan 15, 2014 11:53 am
Two key takeaways here:

1:: If (x + y) and (x - y) are both integers, (x + y) * (x - y) will be even if at least one of the two is even: even * even = even, and even * odd = even.

2:: NEVER assume that a number is an integer (especially a positive integer) unless that is stated or implied. This is one of the testwriters' favorite traps!
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by buoyant » Wed Jan 15, 2014 12:24 pm
Thank You Mike and Mitch for great explanations with two different approaches respectively!

Thank You Matt for summarizing the takeaways. This is a Veritas Prep question.

OA is C indeed.
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by buoyant » Wed Jan 15, 2014 12:25 pm
theCodeToGMAT wrote:To find: (x-y)(x+y) = EVEN

Statement 1:
(x+y) = ODD
either x = ODD and y = EVEN or vice versa
in both cases
So, x - y = ODD
Hence, (x-y)(x+y) = ODDxODD = ODD
SUFFICIENT

Statement 2:
x-y = ODD
either x = Odd and y = EVEN or vice versa
So, (x+y) = ODD
Hence, (x-y)(x+y) = ODDxODD = ODD
SUFFICIENT

Answer [spoiler]{D}[/spoiler]?
Hi Rahul,

You have committed the same mistake as i did.

So, we both have great takeaways from this question.

Thanks for pitching in!
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