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For integers x and y, is x + y even?

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

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Re: For integers x and y, is x + y even? [#permalink]

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Answer is D.
There are only two situations when Addition/Subtraction of two integers yields an Even Output.
1) Odd+/-Odd=Even
2) Even+/-Even=Even

As per this logic, Statement (1) is an Even and hence X+Y has to be even. This is sufficient to answer the question.

Statement (2) is also an Even hence X+Y has to be even.This is sufficient to answer the question.
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For integers x and y, is x + y even? [#permalink]

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New post Updated on: 08 Feb 2017, 02:29
Bunuel wrote:
For integers x and y, is x + y even?

(1) x^2 − y^2 is even

(2) x − y is even


St I \(x^2 − y^2\) is even => (x-y)(x+y) is even;
If (x + y) or (x - y) is even then x and y are both even or odd. So, (x + y) has to be even. ---------Sufficient

St II.
x-y is even; If you take any 2 Integers whose difference is even their sum will always be even, Hence (x+y) is even -------Sufficient
eg; 9-5=4 even and 9+5=14 even
10-2=8 even and 10+2=12 even

Hence Option D is correct.
Hit Kudos if you liked it 8-).

Originally posted by 0akshay0 on 08 Feb 2017, 02:03.
Last edited by 0akshay0 on 08 Feb 2017, 02:29, edited 1 time in total.
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Re: For integers x and y, is x + y even? [#permalink]

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New post 08 Feb 2017, 02:13
0akshay0 wrote:

St I \(x^2 − y^2\) is even => (x-y)(x+y) is even;
In this case either one term (x-y) can be even and the other term (x+y) can be even or odd or
one term (x+y) can be even and the other term (x-y) can be even or odd or vice versa. --------Insufficient


If (x + y) or (x - y) is even then x and y are both even or odd. So, (x + y) has to be even.

Answer: D
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For integers x and y, is x + y even? [#permalink]

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New post 08 Feb 2017, 02:27
Vyshak wrote:
0akshay0 wrote:

St I \(x^2 − y^2\) is even => (x-y)(x+y) is even;
In this case either one term (x-y) can be even and the other term (x+y) can be even or odd or
one term (x+y) can be even and the other term (x-y) can be even or odd or vice versa. --------Insufficient


If (x + y) or (x - y) is even then x and y are both even or odd. So, (x + y) has to be even.

Answer: D



:oops guess I need some coffee now.
Thanks Vyshak, have edited my reply to the Question
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Re: For integers x and y, is x + y even? [#permalink]

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New post 08 Feb 2017, 10:38
Bunuel wrote:
For integers x and y, is x + y even?

(1) x^2 − y^2 is even

(2) x − y is even


1) Tells us that x and y both have to have the same sign. SUFF

2) Tells us that x and y both have to have the same sign. SUFF

D
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Re: For integers x and y, is x + y even? [#permalink]

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Bunuel wrote:
For integers x and y, is x + y even?

(1) x^2 − y^2 is even

(2) x − y is even


We need to determine whether x + y is even. Before jumping into the statements, there are two important rules to consider.

Rule #1:

When an odd integer is raised to a positive integer exponent, the result will still be odd, and when an even number is raised to a positive integer exponent, the result will still be even.

Rule #2:

Addition rules for even and odd numbers are the same as subtraction rules for even and odd numbers.

Statement One Alone:

x^2 - y^2 is even.

Since x^2 - y^2 is even, following rules one and two, we see that x + y will also be even. Statement one alone is sufficient to answer the question.

Statement Two Alone:

x - y is even.

Following rule two, we see that if x - y is even, then x + y will also be even. Statement two alone is also sufficient to answer the question.

Answer: D
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For integers x and y, is x + y even? [#permalink]

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Bunuel wrote:
For integers x and y, is \(x + y\) even?

(1) \(x^2 − y^2\) is even

(2) \(x − y\) is even


Official solution from Veritas Prep.

D. This question is a good example of the GMAT's classic "C trap", in which the statements are written to bait you toward selecting C. Yes, you can deconstruct statement 1 into:

\((x + y)(x - y)\) is even

But at this point you don't need to use statement 2 along with it. Your options now with statement 1 alone are:

\((Even + Even)(Even - Even) = Even\)

and

\((Odd + Odd)(Odd - Odd) = Even\)

It cannot be \((Even + Odd)(Even - Odd)\) because then the two parentheticals would both be odd, making the product odd. So statement 1 guarantees that each of \((x + y)\) and \((x - y)\) are even.

Similarly, statement 2 guarantees that \(x + y\) is even. For \(x - y\) to be even, it's either Odd - Odd or Even - Even. Change that minus to a plus in either case and the result will still be even, so statement 2 is sufficient, also. The correct answer is D.
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Re: For integers x and y, is x + y even? [#permalink]

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Re: For integers x and y, is x + y even?   [#permalink] 08 Apr 2018, 07:03
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