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If x and y are integers, is x+y an odd number?

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If x and y are integers, is x+y an odd number? [#permalink]

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New post 16 May 2017, 00:32
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If x and y are integers, is x+y an odd number?

1) y=3x+1
2) y=2x+3
[Reveal] Spoiler: OA

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Re: If x and y are integers, is x+y an odd number? [#permalink]

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New post 16 May 2017, 02:52
St1: y = 3x + 1
x = even --> y = odd
x = odd --> y = even
In both cases x + y = odd
Sufficient

St2: y = 2x+3
x = even --> y = odd --> x + y = odd
x = odd --> y = odd --> x + y = even
Not Sufficient

Answer: A
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Re: If x and y are integers, is x+y an odd number? [#permalink]

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New post 16 May 2017, 03:23
MathRevolution wrote:
If x and y are integers, is x+y an odd number?

1) y=3x+1
2) y=2x+3

1)
x+y = 4x+1 is odd. Sufficient

2)
x+y = 3x+3

If x is odd, x+y is even.
If x is even, x+y is odd.
Insufficient
The answer is A
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Re: If x and y are integers, is x+y an odd number? [#permalink]

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New post 16 May 2017, 11:01
St1: y = 3x + 1
x = even => y = odd (2,7)
x = odd => y = even (1,4)
In both cases x + y = odd
Sufficient

St2: y = 2x+3
x = even => y = odd; x + y = odd
x = odd => y = odd ; x + y = even
Not Sufficient

Option A is correct
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Re: If x and y are integers, is x+y an odd number? [#permalink]

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New post 18 May 2017, 00:06
==> In the original condition, there are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from 3x+1=2x+3, you get x=2 y=7. Then, you get x+y=2+7=9=odd, hence yes, it is sufficient. The answer is C. However, this is an integer question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), from y=3x+1, you get (x,y)=(even,odd) or (odd,even), which always becomes x+y=odd, hence yes, it is sufficient. For con 2), you get (x,y)=(1,5) no, (2,7) yes, it is not sufficient.

Therefore, the answer is A.
Answer: A
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Re: If x and y are integers, is x+y an odd number?   [#permalink] 18 May 2017, 00:06
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