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# Odds and Evens..

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VP
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Odds and Evens.. [#permalink]  03 Apr 2006, 21:07
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If x,y and z are integers and xy + z is odd, is x even ?

1. xy + xz is even
2. y + xz is odd
Manager
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E ??.. let me know if this is correct. Will post explanation.
Manager
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I think A .

Keeping in mind xy +z is odd -- eqn 1

From first xy +xz is even if x is even irrespective of combination of y and z that satisfies the eqn 1

from second y+xz is odd irrespective of x being odd or even .

I hope my logic is correct but i wont be surprise if i messed up somewhere. Very good question .
Manager
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I picked numbers and both A and B are not correct so,

E
Director
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If we substract second equation from first we get x*(y-1)=odd, since even-odd=odd then both X and (y-1) should be odd
so think C) is correct
GMAT Club Legend
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1) x(y+z) = even

x can be even or odd. Insufficient.

2) y+xz = odd
x can be even or odd. Insufficient.

Using 1) and 2)

We have y+xz = odd.

If y = odd, we know xz = even. If x = even, z = odd or even since x(y+z) will always be even if x is even.

If y = even, xz = odd. z cannot be even because y+z would be be odd, and if x is odd, then st1 cannot be satisfied. so x must be even.

Ans C
VP
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Re: Odds and Evens.. [#permalink]  04 Apr 2006, 13:22
lhotseface wrote:
If x,y and z are integers and xy + z is odd, is x even ?

1. xy + xz is even
2. y + xz is odd

I think it is A.

Given
xy+z = Odd -(I)
xy+xz = Even -(II)

Subtract (I) from (II)

z(x-1) = Even-Odd = Odd
=> z(x-1) must be Odd
=> both z and x-1 are odd.

If x-1 is Odd => x is Even

A it is!
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Manager
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i think its C ... ywilfred is right from the stem z could be anything even or odd i f thats the case and x(y+z) is even ... x could be even or odd ... you need II) also ... good job
VP
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Re: Odds and Evens.. [#permalink]  04 Apr 2006, 16:39
OA is A. Giddi also gave the OE.

giddi77 wrote:
lhotseface wrote:
If x,y and z are integers and xy + z is odd, is x even ?

1. xy + xz is even
2. y + xz is odd

I think it is A.

Given
xy+z = Odd -(I)
xy+xz = Even -(II)

Subtract (I) from (II)

z(x-1) = Even-Odd = Odd
=> z(x-1) must be Odd
=> both z and x-1 are odd.

If x-1 is Odd => x is Even

A it is!
Senior Manager
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I posted answer C. I have read giddi's reply but I do not understand OAs answer. I don't understand if we use both equations to reach a solution, how come answer is A. Any help is appreciated.
VP
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gmatacer wrote:
I posted answer C. I have read giddi's reply but I do not understand OAs answer. I don't understand if we use both equations to reach a solution, how come answer is A. Any help is appreciated.

May be you could put some values and check the equations.
Remember:

O+/-E = O
O+/-O = E
E+/-E = E

and

O*O = O
O*E = E
E*E = E

if X is even and we are given that X-Y is odd, then Y has be odd.
Eg: If 10 - Y = ODD (say 3), then Y has to be ODD (= 7 in this eg).. HTH
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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