Suppose a, b, and c are integers such that a < 0 < b < c. If a(b - c) is an odd integer an… — GRE General Test

Suppose $a$ , $b$ , and $c$ are integers such that $a < 0 < b < c$ . If $a(b - c)$ is an odd integer and $a^2 b + b c$ is an even integer, which of the following expressions MUST be a positive even integer?

$b(c - a)$ Answer
B
$a b c$
C
$a(b - c)$
D
$(a - c) b$
E
$(a + c) b$

Answer

b(c - a)

is guaranteed to be a positive even integer.

The expression

b(c - a)

consists of

b

(which is positive and even) multiplied by

(c - a)

(which is positive and even, as subtracting a negative odd number from a positive odd number yields a positive even number). The product of two positive even integers is always a positive even integer.

Step-by-Step Solution

Determine the signs of variables

a

b

, and

c

a < 0

(negative),

b > 0

(positive), and

c > 0

(positive).

Directly given by the inequality

a < 0 < b < c

Analyze parity from the given condition that

a(b - c)

is odd.

a

is odd and

(b - c)

is odd.

A product of two integers is odd if and only if both factors are odd.

Analyze parity from the second condition

a^2 b + b c = b(a^2 + c)

being even.

b

must be even, and

c

must be odd.

Since

a

is odd,

a^2

is odd. If

b

were odd, then

a^2 + c

would need to be even (making

c

odd), but if both

b

and

c

were odd,

b - c

would be even, contradicting step 2. Therefore,

b

must be even. Since

b

is even and

b - c

is odd,

c

must be odd.

Evaluate the sign and parity of

b(c - a)

c - a = \text{odd} - (\text{negative odd}) = \text{positive even}

. Since

b

is positive even,

b(c - a) = \text{positive even} \times \text{positive even} = \text{positive even}

Subtracting a negative number yields addition (

c - a > 0

), and subtracting an odd integer from an odd integer produces an even integer.

Key Concept

Even-Odd Parity and Integer Sign Properties under Multiplication and Subtraction

Question

Answer

Step-by-Step Solution

Key Concept

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