Question

If \( x \) is an even integer and \( y = 3x + 5 \), which of the following must be odd?

• A. \( x \)
• B. \( y \)
• C. \( 2x + y \)
• D. \( x + y \)
• E. \( 2y \)

Hint:

If \( x \) is even, any linear combination involving \( x \) and an odd number (like \( y = 3x + 5 \)) will result in an odd number.

Answer 1

The Correct Option is B

Step 1: Since \( x \) is an even integer, we can express it as \( x = 2k \), where \( k \) is an integer. Step 2: Now substitute \( x = 2k \) into the equation \( y = 3x + 5 \): \[ y = 3(2k) + 5 = 6k + 5. \] Since \( 6k \) is even (because it is a multiple of 2) and 5 is odd, the sum \( 6k + 5 \) must be odd. Therefore, \( y \) is always odd. Step 3: Let’s check the other options: - \( x \) is even by assumption. - \( 2x + y = 2(2k) + (6k + 5) = 4k + 6k + 5 = 10k + 5 \). Since \( 10k \) is even and 5 is odd, \( 2x + y \) is odd. - \( x + y = 2k + (6k + 5) = 8k + 5 \). Since \( 8k \) is even and 5 is odd, \( x + y \) is odd. - \( 2y = 2(6k + 5) = 12k + 10 \). Since \( 12k \) is even and 10 is even, \( 2y \) is even. Thus, \( y \) must be odd.