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Re: If x y, and z are integers and xy + z is an odd integer, is [#permalink]
22 May 2008, 12:06

3

This post received KUDOS

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

Re: If x y, and z are integers and xy + z is an odd integer, is [#permalink]
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!

Re: If x y, and z are integers and xy + z is an odd integer, is [#permalink]
28 Aug 2014, 01:41

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