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# Is (x-y)*(x+y) an even integer?

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Is (x-y)*(x+y) an even integer? [#permalink]

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26 May 2011, 09:19
1
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Is (x-y)*(x+y) an even integer?

1)x and y are integers
2) x+y is even

OA C
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Last edited by InspiredAnimal on 26 May 2011, 11:57, edited 1 time in total.
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Re: Is (x-y)*(x+y) an even integer? [#permalink]

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26 May 2011, 09:46
hi can you correct your st-1
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Re: Is (x-y)*(x+y) an even integer? [#permalink]

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26 May 2011, 09:50
if your statement 1 means x and y are integers than

st-1 ==> not sufficient. x and y could be integer like x = 4 and y = 1 then (x+y)*(x-y) is odd integer and x =4 and y =2 makes the expression even integer
st-2 ==>not sufficient; x+y just says even... we dont know if its integer.like x = 4 and y=0.2 then (x+y) * (x-y) is even number (but not integer) and x=4 and y=2 makes the expression even integer.

together we know x and y are integers and x+y is even so the exp wil be even integer.
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Re: Is (x-y)*(x+y) an even integer? [#permalink]

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26 May 2011, 20:16
InspiredAnimal wrote:
Is (x-y)*(x+y) an even integer?

1)x and y are integers
2) x+y is even

OA C

Set and question number?
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Re: Is (x-y)*(x+y) an even integer? [#permalink]

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26 May 2011, 20:46
a x and y can both be either even or odd giving different values. x,y = 2,4 or 5,2. Not sufficient.

b x=1.6 y = 0.4 giving non integer value for the equation. Not sufficient.

a+b x and y have to be either both even or odd integers.

C it is.
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Re: Is (x-y)*(x+y) an even integer? [#permalink]

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11 Jun 2011, 14:25

Statement 1 is not sufficient by itself therefore neither A nor D could be the answer.

Statement 2 is not sufficient if we don't know if they are integers or not.

If used together, (x+y) = even implies that either x and y are both even or both odd.

Therefore, x-y = even as well. Which implies their product has to be even. Hence option c.
Re: Is (x-y)*(x+y) an even integer?   [#permalink] 11 Jun 2011, 14:25
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# Is (x-y)*(x+y) an even integer?

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