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If x and y are integers, is x + y even?

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If x and y are integers, is x + y even?  [#permalink]

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New post 28 Jul 2018, 20:33
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Question Stats:

76% (00:46) correct 24% (01:02) wrong based on 54 sessions

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If x and y are integers, is x + y even?

(1) x + 2y is odd.
(2) xy is odd.
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Re: If x and y are integers, is x + y even?  [#permalink]

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New post 28 Jul 2018, 20:38
This one (from GMATPrep Exam 5) has me pulling my hair out. According to the OA, only (2) is sufficient. I think (1) is sufficient alone too, however.

My logic for (1):
- x + 2y -> O
- 2y must be even (any integer times 2 is even)
- thus x must be odd
- "x + y" is only odd if either x or y is odd
- thus (1) indicates that "x + y" is odd
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Re: If x and y are integers, is x + y even?  [#permalink]

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New post 28 Jul 2018, 20:53
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To determine whether x + y is even or odd

Statement 1

x + 2y is odd

=> 2y is always even irrespective of even/odd nature of y (any integer multiplied by even results in an even integer)

=> x + 2y = odd

=> x = odd - 2y = odd - even = odd

So, x is odd and y can either be odd or even

if x is odd and y is odd => x + y = odd + odd = even

if x is odd and y is even => x + y = odd + even = odd

As we have two possible cases, statement 1 is insufficient

Statement 2

xy is odd

=> xy is odd if and only if x is odd and y is odd

=> x + y = odd + odd = even

Statement 2 is sufficient

Hence option B
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Re: If x and y are integers, is x + y even?  [#permalink]

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New post 28 Jul 2018, 20:54
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emanresu1 wrote:
This one (from GMATPrep Exam 5) has me pulling my hair out. According to the OA, only (2) is sufficient. I think (1) is sufficient alone too, however.

My logic for (1):
- x + 2y -> O
- 2y must be even (any integer times 2 is even)
- thus x must be odd
- "x + y" is only odd if either x or y is odd
- thus (1) indicates that "x + y" is odd


emanresu1, 'y' could be even OR odd for 'x' + '2y' to result in an odd integer. Therefore, statement (a) does not give us sufficient information as to whether integer 'y' is indeed an odd integer. Therefore, plugging it into the question stem (x+y) may result in an even OR an odd expression, hence insufficient. Hope that helps :)
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Re: If x and y are integers, is x + y even?  [#permalink]

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New post 28 Jul 2018, 20:56
emanresu1 wrote:
This one (from GMATPrep Exam 5) has me pulling my hair out. According to the OA, only (2) is sufficient. I think (1) is sufficient alone too, however.

My logic for (1):
- x + 2y -> O
- 2y must be even (any integer times 2 is even)
- thus x must be odd
- "x + y" is only odd if either x or y is odd
- thus (1) indicates that "x + y" is odd


If both x and y are odd then x + y becomes even. You only know the even/odd nature of x and y can be either even or odd. Hence statement 1 is insufficient

Example

x = 3 and y = 6 => x + y = 3 + 6 = 9 odd

x = 3 and y = 7 => x + y = 3 + 7 = 10 even
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Re: If x and y are integers, is x + y even?  [#permalink]

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New post 28 Jul 2018, 21:03
workout wrote:
To determine whether x + y is even or odd

Statement 1

x + 2y is odd

=> 2y is always even irrespective of even/odd nature of y (any integer multiplied by even results in an even integer)

=> x + 2y = odd

=> x = odd - 2y = odd - even = odd

So, x is odd and y can either be odd or even

if x is odd and y is odd => x + y = odd + odd = even

if x is odd and y is even => x + y = odd + even = odd

As we have two possible cases, statement 1 is insufficient

Statement 2

xy is odd

=> xy is odd if and only if x is odd and y is odd

=> x + y = odd + odd = even

Statement 2 is sufficient

Hence option B

THANK YOU! This Quant pressure has me flubbing even simple things like thinking 7+7 is odd (yes, I really thought that during my practice exam, and thus the ridiculous confusion)
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Re: If x and y are integers, is x + y even? &nbs [#permalink] 28 Jul 2018, 21:03
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