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Re: DS - Is x even? [#permalink]
28 Jul 2009, 22:14
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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
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Expert's post
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. _________________
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Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
27 Sep 2009, 08:58
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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
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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
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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
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