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If y and z are nonzero integers, is the square of (y + z) even?
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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.
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Re: If y and z are nonzero integers, is the square of (y + z) even?
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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? 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.yz 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?
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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 Answer  A
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Re: If y and z are nonzero integers, is the square of (y + z) even?
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04 May 2011, 20:40
yz = 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?
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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.yz 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 yz 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?
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07 May 2011, 17:00
1. Sufficient
yz 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.



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Re: If y and z are nonzero integers, is the square of (y + z) even?
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21 Oct 2014, 09:32



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Re: If y and z are nonzero integers, is the square of (y + z) even?
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21 Oct 2014, 20:52
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 yz =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?
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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 NONZERO 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 YZ = 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 YZ=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. 62=4); Odd  Odd = Even (e.g. 31=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 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If y and z are nonzero integers, is the square of (y + z) even?
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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: yz=O // Case 2. Sufficient. statement 2: yz = E // Either Case 1 or Case 2 Not sufficient. Answer A.
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Re: If y and z are nonzero integers, is the square of (y + z) even?
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12 Jul 2017, 23:56
In my opinion, this type of problem is super easy if you really understand what you're being asked, and then reframe 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.
answer is A



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