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If x and y are integers, and N = (x + y)(2x – y)(x + 2y
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12 Apr 2017, 07:42

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5

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A

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Difficulty:

85% (hard)

Question Stats:

54% (02:46) correct 46% (02:35) wrong based on 193 sessions

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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

Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y
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12 Apr 2017, 08:03

2

(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)

Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y
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13 Apr 2017, 10:53

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GMATPrepNow wrote:

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 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.

Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y
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14 Apr 2017, 02:05

2

GMATPrepNow wrote:

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

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

Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y
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19 Jan 2019, 03:25

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