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If y and z are nonzero integers, is the square of (y + z) even?

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If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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04 May 2011, 18:20
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If y and z are nonzero 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: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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04 May 2011, 18:56
The question can rephrased as - Is y even and z odd or vice versa?

or

Are both even or both odd?

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

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: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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04 May 2011, 19:04
(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

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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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04 May 2011, 20:40
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: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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05 May 2011, 15:00
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: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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07 May 2011, 17:00
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

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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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21 Oct 2014, 09:32

Tough and Tricky questions: properties of numbers.

If y and z are nonzero integers, is the square of (y + z) even?

(1) y – z is odd.
(2) yz is even.
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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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21 Oct 2014, 20:52
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Bunuel wrote:

Tough and Tricky questions: properties of numbers.

If y and z are nonzero integers, is the square of (y + z) even?

(1) y – z is odd.
(2) yz is even.

We know that even +/- even= Even
Even+/-Odd=Odd

Odd+/-Odd= Even..

St 1 tells us that y-z =odd : So one is even and other is odd..Sum of even+odd=Odd..Therefore (Odd)^2=Odd..St1 is sufficient
St2 says yz=even : this implies either both are even or atleast one is even...

We can have 2 cases: Even+Even =Even or Odd+Even=Odd

Ans is A
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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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03 Feb 2015, 21:28
Hi All,

It looks like everyone who posted in this thread was comfortable with the Number Property rules that this DS question was built on. If you don't recognize those rules when you look at this question, then you can still answer it by TESTing VALUES.

We're told that Y and Z are NON-ZERO INTEGERS. We're asked if (Y+Z)^2 is even. This is a YES/NO question.

Fact 1: Y - Z = ODD

Let's TEST VALUES

IF....
Y = 3
Z = 2
Y-Z = 1
(3+2)^2 = 25 and the answer to the question is NO.

Notice how in the first example, we used an odd number for Y and an even number for Z. Let's try something different next...

IF....
Y=6
Z=1
Y-Z=5
(6+1)^2 = 49 and the answer to the question is NO.

The 'restriction' that Fact 1 places on us means that Y and Z CANNOT have the same sign. Even - Even = Even (e.g. 6-2=4); Odd - Odd = Even (e.g. 3-1=2). But since we're supposed to have an ODD number as a result, neither of these options is a possibility. With the information from Fact 1, the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT

Fact 2: YZ = EVEN

IF....
Y=2
Z=2
YZ = 4
(2+2)^2 = 16 and the answer to the question is YES.

IF...
Y=2
Z=3
YZ=6
(2+3)^2 = 25 and the answer to the question is NO
Fact 2 is INSUFFICIENT

[Reveal] Spoiler:
A

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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]

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04 Feb 2015, 02:00
AnkitK wrote:
If y and z are nonzero integers, is the square of (y + z) even?

(1) y – z is odd.
(2) yz is even.

square (y+z) = y^2+2yz+z^2 --> mixed term will always be even, so Case 1: y and z are both odd or even the whole expression will be even; Case 2 y and z are even/odd or viceversa the whole expression will be odd.

statement 1: y-z=O // Case 2. Sufficient.
statement 2: yz = E // Either Case 1 or Case 2 Not sufficient.

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Re: If y and z are nonzero integers, is the square of (y + z) even?   [#permalink] 04 Feb 2015, 02:00
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