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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, 06:42
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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, 07:03
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(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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14 Apr 2017, 01:05
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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
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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13 Apr 2017, 09:53
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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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19 Jan 2019, 02:25
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