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16 Sep 2014, 00:17



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18 Apr 2015, 15:16
Here, if x=y (Question didn't say that x and y are different, so we can consider this use case as well), then option (D) becomes 0/2 =0, which is not an even integer. Can you please let me know your thoughts on this?



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19 Apr 2015, 02:47



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15 Aug 2017, 15:10
Option 2 in the question looks like a Mixed fraction and not 2*(x/y)



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07 Feb 2018, 05:53
Bunuel wrote: Official Solution:
If \(x\) and \(y\) are even integers, which of the following must also be an even integer?
A. \(x^y\) B. \(2\frac{x}{y}\) C. \(\frac{x  y}{x + y}\) D. \(\frac{x^2  y^2}{2}\) E. \((x + 1)(y  1)\)
\(\frac{x^2  y^2}{2} = \frac{(x  y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\). \(x^y\) is 1 if \(y = 0\) and \(x\) is positive. \(2*\frac{x}{y}\) is 1 if \(x = 2\) and \(y = 4\). \(\frac{x  y}{x + y}\) might not be an integer at all. \((x + 1)(y  1)\) is always odd.
Answer: D Hello. How can I know the difficulty level of this or any other quant question (part of gmat club test)? Sent from my Redmi Note 4 using GMAT Club Forum mobile app



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07 Feb 2018, 05:57
HarshBazooka wrote: Bunuel wrote: Official Solution:
If \(x\) and \(y\) are even integers, which of the following must also be an even integer?
A. \(x^y\) B. \(2\frac{x}{y}\) C. \(\frac{x  y}{x + y}\) D. \(\frac{x^2  y^2}{2}\) E. \((x + 1)(y  1)\)
\(\frac{x^2  y^2}{2} = \frac{(x  y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\). \(x^y\) is 1 if \(y = 0\) and \(x\) is positive. \(2*\frac{x}{y}\) is 1 if \(x = 2\) and \(y = 4\). \(\frac{x  y}{x + y}\) might not be an integer at all. \((x + 1)(y  1)\) is always odd.
Answer: D Hello. How can I know the difficulty level of this or any other quant question (part of gmat club test)? Sent from my Redmi Note 4 using GMAT Club Forum mobile appYou can check the stats in the original post: Attachment: 20180207_1755.png
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Re: M2233
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07 Feb 2018, 10:42
I used test cases and got down to two answer choices: A and D. I couldn't find an example of two even integers that disproved A?
Any Advice?



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08 Feb 2018, 19:51
After initial analysis , answer choices narrowed down to A and D. The option "0" was the caveat in the choice A which made the only possible option is D. Thanks Bunuel for clear explanation as always
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17 Nov 2018, 07:03
I think this is a highquality question and I agree with explanation.



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Hi Bunuel,
Using the same logic( i.e. 0 is neither even nor odd) shouldn't A also satisfy the conditions?
x^y would always be even as y can never be 0(as 0 is not an even number).
Thanks, Sam



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13 Jan 2019, 02:35
Thanks, Bunuel. That was rather a stupid query from my end.










