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If x y, and z are integers and xy + z is an odd integer, is

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If x y, and z are integers and xy + z is an odd integer, is [#permalink] New post 22 May 2008, 05:26
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If x y, and z are integers and xy + z is an odd integer, is x an even integer?

1) xy + xz is an even integer.

2) y + xz is an odd integer.

** Edited **
My bad, just made a correction.

Last edited by mbawaters on 22 May 2008, 07:30, edited 1 time in total.
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Re: DS - interesting prep [#permalink] New post 22 May 2008, 05:33
question is missing is x or z or what is an integer??
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Re: DS - interesting prep [#permalink] New post 22 May 2008, 08:54
mbawaters wrote:
If x y, and z are integers and xy + z is an odd integer, is x an even integer?

1) xy + xz is an even integer.

2) y + xz is an odd integer.

** Edited **
My bad, just made a correction.



I am 0 for 2 when answering this question. 7-t60544
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Re: DS - interesting prep [#permalink] New post 22 May 2008, 09:21
mbawaters wrote:
If x y, and z are integers and xy + z is an odd integer, is x an even integer?

1) xy + xz is an even integer.

2) y + xz is an odd integer.

** Edited **
My bad, just made a correction.


I believe the answer is D.

I tried several numbers. Although im not proud of how long it took me.
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Re: DS - interesting prep [#permalink] New post 22 May 2008, 12:06
Answer is A

you can represent any odd number by 2n+1 and even number by 2n , I have realized using this property helps me a lot in solving odd/even questions

First piece of infomation xy + z is odd so xy+z = 2n+1

1) xy + xz is an even integer.

re- write it as (xy+z)-z+xz = 2n+1 + z(x-1) = even ....since 2n+1 is odd ...the term z(x-1) must be odd and so (x-1) must be odd , this tells us that x must be even .

2) 2) y + xz is an odd integer.

re-write it as

y+(xy+z)-(xy+z)+xz = 2n+1 +(x-1)(z-y) = odd ...for this to be true (x-1)(z-y) must be even and so either (x-1) is even or (z-y) even or both are even ....insuff

I did this in less than a minute time ....
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Re: DS - interesting prep [#permalink] New post 23 May 2008, 04:05
rpmodi wrote:
Answer is A

you can represent any odd number by 2n+1 and even number by 2n , I have realized using this property helps me a lot in solving odd/even questions

First piece of infomation xy + z is odd so xy+z = 2n+1

1) xy + xz is an even integer.

re- write it as (xy+z)-z+xz = 2n+1 + z(x-1) = even ....since 2n+1 is odd ...the term z(x-1) must be odd and so (x-1) must be odd , this tells us that x must be even .

2) 2) y + xz is an odd integer.

re-write it as

y+(xy+z)-(xy+z)+xz = 2n+1 +(x-1)(z-y) = odd ...for this to be true (x-1)(z-y) must be even and so either (x-1) is even or (z-y) even or both are even ....insuff

I did this in less than a minute time ....

Hello,

I tried also to use the same pattern but what i'm not sure about is this : so (x-1) must be odd in your explaining from stmt 1
Because if z(x-1) is odd, it doesn't necessarily lead to (x-1) odd ?
Can you please explain? Thx!
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Re: DS - interesting prep [#permalink] New post 25 May 2008, 07:41
Poullo , if product of two terms is odd , both the terms are odd .

try any combination ; if one of the number is even product will be even . HTH
Re: DS - interesting prep   [#permalink] 25 May 2008, 07:41
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If x y, and z are integers and xy + z is an odd integer, is

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