If x and y are integers, and N = (x + y)(2x – y)(x + 2y

Question

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

BrentGMATPrepNow · Accepted Answer

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 EVEN

Now 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

LeonidK · Answer

(x+y)*(2x-y)*(x+2y+1)

Result of multiplication is Odd in case all multiplicands are Odd

1) x+y = Odd (x or y should be Even, and y or x should be Odd)
2) 2x-y = Odd (2x Even, then y must be Odd)
3) x+2y-1 = Odd (2y - Even, 1 Odd, then x must be Even)

Therefore answer D

Mo2men · Answer

Let's simplify N first. 2x=Even & 2y+1=Even + ODD=ODD Hence.... N= (x + y)(E – y)(x + O) Let see when is either Even or Odd..Here I will borrow Brent's table above

Case i) x is EVEN and y is EVEN.............N= (E+E) (E-E) (E+O)= EVEN Case ii) x is EVEN and y is ODD..............N = (E+O) (E-O) (E+O)= ODD Case iii) x is ODD and y is EVEN.............N = (O+E) (E-E) (O+O) = Even Case iv) x is ODD and y is ODD............. N = (O+O) (E-O) (O+O) = Even So we have one 3 cases with Even and 1 case with ODD.. So I will focus my effort when N is ODD....Eliminate A, C & E To be ODD, y has to be ODD Answer: D

Kinshook · Answer

Asked: If x and y are integers, and N = (x + y)(2x – y)(x + 2y + 1), which of the following is true?

If N is odd
x+y must be odd -> If x is even, y is odd, or If x is odd, y is even
2x-y must be odd -> y is odd
x+2y+1 must be odd - > x is even
x is even, y is odd

IMO D

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x and y are integers, and N = (x + y)(2x – y)(x + 2y

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If x and y are integers, and N = (x + y)(2x – y)(x + 2y

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards