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Manager  Joined: 11 Feb 2011
Posts: 108
If y and z are nonzero integers, is the square of (y + z) even?  [#permalink]

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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.

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

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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

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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(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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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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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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1
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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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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1
1
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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If y and z are nonzero integers, is the square of (y + z) even?  [#permalink]

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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

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

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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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Intern  Joined: 24 Nov 2015
Posts: 16
Re: If y and z are nonzero integers, is the square of (y + z) even?  [#permalink]

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In my opinion, this type of problem is super easy if you really understand what you're being asked, and then re-frame the question to be more straightforward. It's more about the question than the math.

Given: (y,z) ≠ 0, y and z are integers

Original Question: Is (y + z) even?

We know that EVEN + EVEN = EVEN, and ODD + ODD = EVEN...

In other words, we know that when two numbers of the same parity are added together, the result is always even. When two numbers of different parity are added together (eg. odd + even), the answer is always odd. ("Parity" is the word that describes whether a number is odd or even)

So the real question being asked is: Are "y" and "z" of the same parity?

A.) (y - z) is odd.

(y - z) can only be odd if they are different parities. Since we now know that Y and Z are of different parities, we know that (y + z) is odd. SUFFICIENT.

B.) yz is even.

For yz to be even, either y, or z, or both can be even. We can't answer the question "Are 'y' and 'z' of the same parity?" with the information given here. INSUFFICIENT.

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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] 24 Aug 2018, 12:55
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