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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]
05 May 2011, 14:00

Expert's post

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]
21 Oct 2014, 19:52

1

This post received KUDOS

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This post was BOOKMARKED

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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]
03 Feb 2015, 15:42

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]
03 Feb 2015, 20:28

Expert's post

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

Re: If y and z are nonzero integers, is the square of (y + z) even? [#permalink]
04 Feb 2015, 01:00

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.

Answer A. _________________

learn the rules of the game, then play better than anyone else.

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Re: If y and z are nonzero integers, is the square of (y + z) even?
[#permalink]
04 Feb 2015, 01:00

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