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Re: DS - Is x even? [#permalink]
28 Jul 2009, 22:14

26

This post received KUDOS

7

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IMO A

1. xy + xz is an even integer - SUFFICIENT Given: xy + z is odd ...(i) xy + xz is even ...(ii)

subtracting (ii) from (i) we get xz - z, which should be odd (* since odd - even = odd) => z(x-1) is odd => both z and (x-1) is odd => since (x-1) is odd, x must be even.

2. y + xz is an odd integer -INSUFFICIENT Given: xy + z is odd ...(i) y + xz is odd ...(ii)

subtracting (ii) from (i) we get xy + z - y - xz = (x-1)(y-z) , which should be even => either (x-1) is even or (y-z) is even ....insufficient to determine _________________

Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
26 Sep 2009, 21:52

You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
27 Sep 2009, 08:58

IanStewart wrote:

You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.

Fr St2, y+xz odd xy+z odd => y+zx+xy+z even => y(x+1)+z(x+1) even => (y+z)(x+1) event x+1 can be odd or even means that x can be even or odd, insuff

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
19 Mar 2014, 18:41

Given condition: xy + z = odd implies either xy = odd (x =odd and y = odd) and z = even or xy = even (x or y can be odd and even respectively and vice versa) and z = odd

condition 1:

xy + xz = even; Implies x(y+z) = even which again implies the following:

i) x even and y+z = odd - where again y or z can be odd and even respectively and vice versa ii) x odd and y +z = even - where again y and z has to be both odd or both even

inconclusive

condition 2:

y + xz = odd

again inconclusive 1 + 2: Add xy + z + y + xz = odd + odd implies: (x + 1)(y+z) = even and x (y+z) is also even according to 2.. so y + z = even <y and z both even or y + z both odd>, x can be odd or even but by 1 xy + z = odd which means y and z both odd, so x is even.

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
20 Mar 2014, 05:09

3

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Expert's post

1

This post was BOOKMARKED

Mountain14 wrote:

jlgdr wrote:

Aleehsgonji 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

Odd/Even questions can be usually solved quite easily if one tries some operations with the statements

We want to know if x is even integer

We are given that xy+z is odd

Statement 1

xq + xz is even

Subtracting

z(x+1) is odd

Therefore, x+1 should be odd and x should be even

Sufficient

Statement 2

Not sufficient

Answer is A

Just my 2c

Cheers J

I am not clear with the red part.

When you subtract xy + z=odd from xy+xz=even you'll get: xz-z=even-odd=odd --> z(x-1)=odd. For the product of two integers to be odd, both of them must be odd --> z and x-1 are odd. If x-1=odd, then x must be even: x-1=x-odd=odd --> x=odd+odd=even.

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
21 Mar 2014, 19:46

1

This post received KUDOS

Expert's post

Odd(O) Even (E) given: x,y,z integers xy+z=O so only the following scenarios can fulfill the constraints a) EO+O b) EE+O c) OE+O d) OO+E

question: x=E?

1) x(y+z)=E i. (E)(O+O) --> fits scenario a -->yes, x can be even ii. (O)(E+E) --> n/a - doesn't fit any scenarios iii. (O)(O+O) --> n/a - doesn't fit any scenarios

stop testing, x can't be odd, sufficient

2) y+xz = O i. E+(O)(O) --> fits scenario a -->yes, x can be even ii. O+(E)(E) --> n/a - doesn't fit any scenarios iii. O+(O)(E) --> fits scenario d -->no, x can be odd

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...