Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: DS - interesting prep [#permalink]
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

Re: DS - interesting prep [#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!