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For integers x and y, is x + y even?
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08 Feb 2017, 00:36
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75% (01:24) correct 25% (01:05) wrong based on 107 sessions
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Re: For integers x and y, is x + y even?
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08 Feb 2017, 00:50
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?
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Updated on: 08 Feb 2017, 01:29
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 => (xy)(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. xy 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; 95=4 even and 9+5=14 even 102=8 even and 10+2=12 even Hence Option D is correct. Hit Kudos if you liked it .
Originally posted by 0akshay0 on 08 Feb 2017, 01:03.
Last edited by 0akshay0 on 08 Feb 2017, 01:29, edited 1 time in total.



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Re: For integers x and y, is x + y even?
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08 Feb 2017, 01:13
0akshay0 wrote: St I \(x^2 − y^2\) is even => (xy)(x+y) is even; In this case either one term (xy) 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 (xy) 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?
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08 Feb 2017, 01:27
Vyshak wrote: 0akshay0 wrote: St I \(x^2 − y^2\) is even => (xy)(x+y) is even; In this case either one term (xy) 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 (xy) 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 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?
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08 Feb 2017, 09:38
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?
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13 Feb 2017, 07:43
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?
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23 Feb 2017, 00:46
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?
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08 Apr 2018, 06:03
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