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

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Bunuel
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If x and y are integers, is y an even integer? [#permalink] New post   10 Jul 2018, 20:59
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If x and y are integers, is y an even integer?


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

(2) x is an odd integer.
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niks18
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If x and y are integers, is y an even integer? [#permalink] New post   10 Jul 2018, 21:59
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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] New post   10 Jul 2018, 22:05
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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? [#permalink] New post   22 Nov 2018, 07:34
It is a Yes/No question, asking if Y is an even integer.

Let's start with statement 2, it says X is an odd integer, no information is provided regarding Y. Hence Insufficient.
You can eliminate options B and D.

Statement 1: -
2y−x=x^2 - y^2
y^2 + 2y = x^2 + x
y(y+2) = x(x+1)

Now we have four possibilities,
x,y - Both Even
x - Even, y - Odd
x - Odd, Y - Even
x,y - Both Odd

The equation works only in the first and the third case, that is when either both x and y are even or when x is odd and y is even. We are only concerned about y being even. Hence, Sufficient - A
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Re: If x and y are integers, is y an even integer? [#permalink] New post   30 Mar 2022, 04:44
niks18 wrote:
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



Hello,

How come is it x^2 + x + EVEN?

Shouldn't it be - EVEN?

Also, even - even = even.
This means y^2 is even.
Which then should mean, y is even.

Please let me know if I have inferred this correctly.
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Re: If x and y are integers, is y an even integer? [#permalink] New post   15 Sep 2022, 00:27
Statement 1 can be reduced to y(y+2) = x(x+1):
2y−x=x^2 - y^2
y^2 + 2y = x^2 + x
y(y+2) = x(x+1)

We can see x(x+1) will always be an even number, since one of the term here must be even. So y(y+2) must also be even. This allows us to confirm y is even since y and (y+2) will have the same even/odd properties.
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Re: If x and y are integers, is y an even integer? [#permalink]
15 Sep 2022, 00:27
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Bunuel
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