Question

If \( m \) and \( n \) are both odd integers, which of the following is not necessarily odd?

• A. \( m + n^2 \)
• B. \( m - 2n \)
• C. \( mn \)
• D. \( m \cdot n \)
• E. \( 2m - n \)

Hint:

When dealing with odd/even integers, remember: - Odd + Odd = Even - Odd - Odd = Even - Odd ± Even = Odd - Odd × Odd = Odd - Even × Odd = Even

Answer 1

The Correct Option is B

Step 1: Check each option separately.
- (A) \( m + n^2 \): Since \( n \) is odd, \( n^2 \) is odd. Odd + Odd = Even. Hence, this will \emph{always} be even.
- (B) \( m - 2n \): Here, \( 2n \) is even (since any integer multiplied by 2 is even). Odd - Even = Odd. But, depending on values, it could also be even. Thus, it is \emph{not necessarily odd}.
- (C) \( mn \): Odd × Odd = Odd. Always odd.
- (D) \( m \cdot n \): Same as above, always odd.
- (E) \( 2m - n \): Since \( 2m \) is even and \( n \) is odd, Even - Odd = Odd. Always odd.

Step 2: Identify the exception.
Among the given options, only \( m - 2n \) can fail to be odd depending on values of \( m \) and \( n \).
Final Answer: \[ \boxed{m - 2n} \]