If a, b, and c are non-zero integers such that a^3 b^2 c < 0, a + b is even, and b + c is … — GRE General Test

If $a$ , $b$ , and $c$ are non-zero integers such that $a^3 b^2 c < 0$ , $a + b$ is even, and $b + c$ is odd, which of the following expressions MUST be a negative even integer?

$a c (b^2 + 1)$ Answer
B
$a b c$
C
$(a + c) b^2$
D
$a^2 c (b + 1)$
E
$a c + b$

Answer

The expression

a c (b^2 + 1)

must be a negative even integer.

The expression

a c (b^2 + 1)

is guaranteed to be negative because

a c < 0

(derived from

a^3 b^2 c < 0

) and

b^2 + 1 > 0

for all non-zero integers

b

. It is guaranteed to be even because

a

and

c

have opposite parity, meaning one of them must be even, rendering

a c

(and thus any integer multiple of

a c

) even.

Step-by-Step Solution

Analyze the sign constraint

a^3 b^2 c < 0

a c < 0

, which means

a

and

c

have opposite signs.

Since

b \neq 0

b^2

is strictly positive. Dividing

a^3 b^2 c < 0

b^2

yields

a^3 c < 0

. Since

a^3

has the same sign as

a

a c < 0

Analyze the parity constraints

a + b

is even and

b + c

is odd.

a

and

c

have opposite parity (one is even, the other is odd).

a + b

is even,

a

and

b

share the same parity. If

b + c

is odd,

b

and

c

have opposite parity. Substituting the parity of

a

for

b

shows that

a

and

c

must have opposite parity.

Determine the sign and parity of

a c (b^2 + 1)

a c (b^2 + 1)

is strictly negative and even.

Since

a

and

c

have opposite parity, at least one of them is even, making the product

a c

an even integer. Since

a c < 0

and

b^2 + 1 \ge 1 > 0

, the product of negative

a c

and positive

(b^2 + 1)

is negative and even.

Key Concept

Combining sign rules (

a c < 0

) with even/odd parity logic across multiple variables.

Question

Answer

Step-by-Step Solution

Key Concept

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