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

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.

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.

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.

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

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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.

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.

answer is A

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