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Odds and Evens..

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Odds and Evens.. [#permalink] New post 03 Apr 2006, 21:07
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A
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E

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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
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 [#permalink] New post 03 Apr 2006, 21:33
E ??.. let me know if this is correct. Will post explanation.
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 [#permalink] New post 04 Apr 2006, 02:35
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 .
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 [#permalink] New post 04 Apr 2006, 02:46
I picked numbers and both A and B are not correct so,

E
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 [#permalink] New post 04 Apr 2006, 05:49
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
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 [#permalink] New post 04 Apr 2006, 06:59
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
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Re: Odds and Evens.. [#permalink] New post 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!
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 [#permalink] New post 04 Apr 2006, 14:56
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
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Re: Odds and Evens.. [#permalink] New post 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!
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 [#permalink] New post 04 Apr 2006, 19:29
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.
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 [#permalink] New post 04 Apr 2006, 19:48
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
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  [#permalink] 04 Apr 2006, 19:48
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