Let x, y, and z be non-zero integers that satisfy the following three conditions: I. (-1)^… — GRE General Test

Let $x$ , $y$ , and $z$ be non-zero integers that satisfy the following three conditions:

I. $(-1)^{x y + z} = -1$
II. $(-1)^{x z + y} = 1$
III. $x^3 y z^2 < 0$

Which of the following statements MUST be true?

$x$ and $y$ are both even integers, and $x y < 0$ Cevap
B
$x$ is an odd integer, and $z$ is an even integer
C
$y$ and $z$ are both odd integers
D
$x + z$ is an even integer
E
$x y z > 0$

Cevap

x

and

y

are both even integers, and

x y < 0

From Condition I,

(-1)^{xy+z} = -1

implies

xy + z

is odd, so

xy

and

z

have opposite parity. From Condition II,

(-1)^{xz+y} = 1

implies

xz + y

is even, so

xz

and

y

have the same parity. If

x

were odd,

xy

would share the parity of

y

, and

xz

would share the parity of

z

. This would mean

y

and

z

must have opposite parity (from Condition I) and the same parity (from Condition II), which is impossible. Thus,

x

must be even. With

x

even,

xy

is even, so

z

must be odd for

xy + z

to be odd. Likewise,

xz

is even, so

y

must be even for

xz + y

to be even. Finally, from Condition III, since

z^2 > 0

for non-zero

z

x^3 y z^2 < 0

reduces to

x^3 y < 0

, which means

xy < 0

. Thus,

x

and

y

are both even integers, and

xy < 0

Adım Adım Çözüm

Analyze Condition I for parity requirements

x y + z

must be an odd integer

Since

(-1)^k = -1

if and only if

k

is an odd integer,

x y + z

is odd. Thus,

x y

and

z

must have opposite parity.

Analyze Condition II for parity requirements

x z + y

must be an even integer

Since

(-1)^k = 1

if and only if

k

is an even integer,

x z + y

is even. Thus,

x z

and

y

must have the same parity.

Determine the parity of

x

y

, and

z

using proof by contradiction

x

is even,

y

is even, and

z

is odd

x

were odd, then

x y

would have the same parity as

y

, and

x z

would have the same parity as

z

. Condition I would require

y

and

z

to have opposite parity, while Condition II would require

y

and

z

to have the same parity, a contradiction. Therefore,

x

must be even. Since

x

is even,

x y

is even, making

z

odd (from Condition I). Similarly,

x z

is even, making

y

even (from Condition II).

Analyze Condition III for sign requirements

x y < 0

Since

z

is a non-zero integer,

z^2 > 0

. The inequality

x^3 y z^2 < 0

simplifies to

x^3 y < 0

. Since

x^3

has the same sign as

x

x^3 y < 0

implies

x y < 0

Anahtar Kavram

Parity rules for exponentiation and basic arithmetic operations, combined with sign rules for products

Soru

Cevap

Adım Adım Çözüm

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