Consider three non-zero integers x, y, and z that satisfy all of the following conditions:… — GRE General Test

Consider three non-zero integers $x$ , $y$ , and $z$ that satisfy all of the following conditions:

I. $(-1)^{x^2 y + z} = -1$
II. $x y z^2 < 0$
III. $x + y$ is an even integer

Which of the following expressions MUST be an odd integer?

$(x + z)(y + z)$ Answer
B
$x y z$
C
$x + y + z$
D
$x(y + z)$
E
$x z + y$

Answer

The expression

(x + z)(y + z)

MUST be an odd integer.

Condition I dictates that

x^2 y + z

is odd. Condition III states that

x + y

is even, meaning

x

and

y

share the same parity. If

x

and

y

are both even,

x^2 y

is even, forcing

z

to be odd. If

x

and

y

are both odd,

x^2 y

is odd, forcing

z

to be even. Consequently,

z

always has the opposite parity of both

x

and

y

. Therefore,

(x + z)

is always odd and

(y + z)

is always odd. The product of two odd integers,

(x + z)(y + z)

, is guaranteed to be odd.

Step-by-Step Solution

Analyze Condition I for exponent parity

x^2 y + z

must be an odd integer

For

(-1)^k = -1

, the exponent

k

must be odd.

Analyze Condition III for shared parity of

x

and

y

x

and

y

are either both even or both odd

The sum of two integers is even if and only if they share the same parity.

Deduce parity relationship for

z

across cases

In Case 1 (

x, y

even),

x^2 y

is even, so

z

must be odd. In Case 2 (

x, y

odd),

x^2 y

is odd, so

z

must be even.

To satisfy

x^2 y + z = \text{odd}

x^2 y

and

z

must have opposite parities.

Evaluate the parity of

(x + z)(y + z)

(x + z)

is odd and

(y + z)

is odd, so their product is odd

In both cases,

z

has opposite parity to both

x

and

y

. Adding two integers of opposite parity always yields an odd integer, and the product of two odd integers is always odd.

Key Concept

Parity rules under exponentiation and algebraic combination

Estimated Time:2m 0s

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