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If x and y are integers, and N = (x + y)(2x – y)(x + 2y + 1), which of the following is true?
A) If N is even, then x must be even
B) If N is odd, then x must be odd
C) If N is even, then x and y must both be odd
D) If N is odd, then y must be odd
E) If N is even, then y must be even
*kudos for all correct solutions
Given the various answer choices, it might be useful to take a systematic approach and first look at all 4 possible cases.
Case i) x is EVEN and y is EVEN
Case ii) x is EVEN and y is ODD
Case iii) x is ODD and y is EVEN
Case iv) x is ODD and y is ODD
For each case, we can EITHER apply the rules for even and odd numbers (e.g., EVEN + ODD = ODD) OR we can just plug in some easy numbers.
I’ll go the plugging in route.
For even numbers, I’ll plug in 0, and for odd numbers I’ll plug in 1. Here’s what we get:
Case i) x = 0, y = 0, so N = (0)(0)(1) = 0. Result: N is EVEN
Case ii) x = 0, y = 1, so N = (1)(-1)(3) = -3. Result: N is ODD
Case iii) x = 1, y = 0, so N = (1)(2)(2) = 4. Result: N is EVEN
Case iv) x = 1, y = 1, so N = (2)(1)(4) = 8. Result: N is EVEN
So, we have:
Case i) x is EVEN and y is EVEN. N is EVEN
Case ii) x is EVEN and y is ODD. N is ODD
Case iii) x is ODD and y is EVEN. N is EVEN
Case iv) x is ODD and y is ODD. N is EVENNow check the answer choices….
A) If N is even, then x must be even. FALSE. Cases iii and iv refute this statement.
B) If N is odd, then x must be odd. FALSE. Case ii refutes this statement.
C) If N is even, then x and y must both be odd. FALSE. Cases i and iii refute this statement.
D) If N is odd, then y must be odd. TRUE. See case ii
E) If N is even, then y must be even. FALSE. Case iv refutes this statement.
Answer: D
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