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If x and y are integers, is y an even integer?

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

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New post 10 Jul 2018, 20:59
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

47% (01:43) correct 53% (01:31) wrong based on 30 sessions

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

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New post 10 Jul 2018, 21:59
1
Bunuel wrote:
If x and y are integers, is y an even integer?


(1) \(2y - x = x^2 - y^2\)

(2) x is an odd integer.


Statement 1: \(2y - x = x^2 - y^2\). \(2y\) will always be Even

\(=>y^2=x^2+x+Even\). Now irrespective of the value of \(x\), \(x^2+x\) will always be Even, because \(x\) is an integer.

Hence \(y^2=Even+Even=Even\). Sufficient

Statement 2: nothing mentioned about \(y\). Insufficient

Option A
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Re: If x and y are integers, is y an even integer?  [#permalink]

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New post 10 Jul 2018, 22:05
GIven X and Y are integers

We need to find if Y is an even integer

Statement 1

\(2y - x = x^{2}-y^{2}\)

=> \(2y = x^{2}-y^{2} + x\)

The left hand side of above equation is always even

=> \(x^{2}-y^{2} + x\) is even

Lets see the result for different combinations of X and Y

X | Y | \(x^{2}-y^{2} + x\)

Even | Even | Even
Odd | Even | Even
Odd | Odd | Odd
Even | Odd | Odd

We can see that Y is an even integer whenever \(x^{2}-y^{2} + x\) is even

Statement 1 is sufficient

Statement 2

X is an odd integer => doesn't tell anything about Y

Statement 2 not sufficient

Analysis of statement 1 and 2 together is not required

Hence option A
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Re: If x and y are integers, is y an even integer? &nbs [#permalink] 10 Jul 2018, 22:05
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