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

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If x and y are integers, and N = (x + y)(2x – y)(x + 2y  [#permalink]

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New post 12 Apr 2017, 07:42
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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

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Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y  [#permalink]

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New post 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)

Therefore answer D
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Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y  [#permalink]

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New post 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.

Answer:

RELATED VIDEO (testing all possible cases)

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Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y  [#permalink]

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New post 14 Apr 2017, 02:05
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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

To be ODD, y has to be ODD

Answer: D
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Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y  [#permalink]

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New post 19 Jan 2019, 03:25
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Re: If x and y are integers, and N = (x + y)(2x – y)(x + 2y   [#permalink] 19 Jan 2019, 03:25
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