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y and z are non zero integers ,is the square of (y+z) even?

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y and z are non zero integers ,is the square of (y+z) even? [#permalink] New post 04 May 2011, 17:20
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y and z are non zero integers ,is the square of (y+z) even?

(1) y-z is odd
(2) yz is even
[Reveal] Spoiler: OA

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Re: odds and evens! [#permalink] New post 04 May 2011, 17:56
The question can rephrased as - Is y even and z odd or vice versa?

or

Are both even or both odd?

Answering any of these questions will answer our question.

1) Sufficient
y may be even or z odd and vice versa. The answer is always NO

PS : It is given that y and z are NON ZERO.

2) Insufficient

Both may be even or just one of them may be even. In the first case y + z = even. The answer is YES
In the second case y + z = odd. The answer is NO

Answer A

AnkitK wrote:
y and z are nin zero integers ,is the square of (y+z) even?
1.y-z is odd
2.yz is even
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Re: odds and evens! [#permalink] New post 04 May 2011, 18:04
(1)

One of y or z is even, and one of y or z is odd

So y+z is odd

hence (y+z)^2 = odd

Sufficient


(2)

y can be even, z can be odd

y can be odd, z can be even

y or z can both be even

So square of (y+z) is even.

Not Sufficient


Answer - A
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Re: odds and evens! [#permalink] New post 04 May 2011, 19:40
y-z = odd means either y is even and z is odd or vice versa. Hence A.
yz = even means y,z even or y is even and z odd and viceversa.

Thus A.
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Re: odds and evens! [#permalink] New post 05 May 2011, 14:00
AnkitK wrote:
y and z are nin zero integers ,is the square of (y+z) even?
1.y-z is odd
2.yz is even


As I posted in another thread a moment ago, positive integer exponents never matter in an even/odd question, so we can just ignore the 'square of' part of the question: it's just asking if y+z is even. Addition and subtraction follow the same odd/even rules, so if y-z is odd, then y+z is odd, and Statement 1 is sufficient. From Statement 2, y and z can both be even, in which case y+z is even, or one can be even and the other odd, in which case y+z is odd. So the answer is A.
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Re: odds and evens! [#permalink] New post 07 May 2011, 16:00
1. Sufficient

y-z is odd
=> either
y is even,z is odd or
y is odd ,z is even

in both the scenarios mentioned above y+z is odd = > (y+z)^ 2 is odd

2. Not sufficient
yz is even

atleast one of the above is even

when both y and z are even , given expression is even
but when y is odd and z is even , given expression is odd

Answer is A.
Re: odds and evens!   [#permalink] 07 May 2011, 16:00
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