If k and m are integers such that (-k)^5 m < 0 and k - m is an odd integer, which of the f… — GRE General Test

If $k$ and $m$ are integers such that $(-k)^5 m < 0$ and $k - m$ is an odd integer, which of the following expressions must be a positive even integer?

$k^2 m^2$ Answer
B
$(-k)^2 m$
C
$(k - m)^2$
D
$-k m$
E
$k^3 + m^3$

Answer

The expression

k^2 m^2

must be a positive even integer.

Simplifying

(-k)^5 m < 0

gives

-k^5 m < 0 \implies k^5 m > 0

. This confirms

k \neq 0

and

m \neq 0

, so both

k^2

and

m^2

are positive integers, making

k^2 m^2 > 0

. Additionally,

k - m

being odd requires one variable to be even and the other odd. Squaring an even integer yields an even integer, and multiplying by any integer keeps it even. Hence, the expression

k^2 m^2

is guaranteed to be a positive even integer.

Step-by-Step Solution

Analyze the sign condition

(-k)^5 m < 0

Since

(-k)^5 = -k^5

, the inequality becomes

-k^5 m < 0

, which means

k^5 m > 0

. This implies that neither

k

nor

m

is zero, and both

k

and

m

have the same sign (either both positive or both negative).

Raising a negative quantity to an odd power retains the negative sign.

Analyze the parity condition

k - m

is odd.

The difference between two integers is odd if and only if one integer is even and the other is odd.

Even minus odd (or odd minus even) produces an odd result.

Evaluate the sign and parity of

k^2 m^2

Since

k \neq 0

and

m \neq 0

k^2 > 0

and

m^2 > 0

, so

k^2 m^2 > 0

(positive). Since one of

k

m

is even, its square is also even, so the product

k^2 m^2

must be even. Thus,

k^2 m^2

is guaranteed to be a positive even integer.

The product of non-zero squares is positive, and any integer multiple of an even number is even.

Key Concept

Parity rules under subtraction/multiplication and sign rules under odd powers.

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