Let m and n be integers such that m < 0, n > 0, (-1)^m = -1, and m + n is an even integer.… — GRE General Test

Let $m$ and $n$ be integers such that $m < 0$ , $n > 0$ , $(-1)^m = -1$ , and $m + n$ is an even integer. Which of the following expressions must be a positive even integer?

A
$m \cdot n$
B
$(-m)^n$
$n - m$ Answer
D
$(m + n)^2$
E
$m^2 - n$

Answer

The expression

n - m

must be a positive even integer.

The expression

n - m

subtracts a negative odd integer from a positive odd integer, which equals adding two positive odd integers. The sum of two positive odd integers is always a positive even integer.

Step-by-Step Solution

Determine the parity and sign of

m

Since

m < 0

and

(-1)^m = -1

m

must be a negative odd integer.

An odd exponent on

-1

yields

-1

Determine the parity and sign of

n

Since

n > 0

and

m + n

is even,

n

must be a positive odd integer.

The sum of two integers is even if and only if both integers have the same parity. Since

m

is odd,

n

must also be odd.

Evaluate the sign and parity of

n - m

n - m = n + (-m)

. Since

n \ge 1

and

-m \ge 1

n - m \ge 2

(strictly positive). Also, odd minus odd is always even.

Combining the sign rules (

n > 0

and

-m > 0

) with the even-odd subtraction rule confirms

n - m

is always a positive even integer.

Key Concept

Even-Odd Properties and Sign Rules

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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