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If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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28 Jul 2009, 22:25
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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
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Last edited by Bunuel on 07 Mar 2013, 02:10, edited 1 time in total.
Edited the question and added the OA.



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Re: DS  Is x even? [#permalink]
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28 Jul 2009, 23:14
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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(x1) is odd => both z and (x1) is odd => since (x1) 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 = (x1)(yz) , which should be even => either (x1) is even or (yz) is even ....insufficient to determine
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Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
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26 Sep 2009, 22:52
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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(x1) is odd. Since this is a product, z must be odd, and x1 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]
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27 Sep 2009, 09:58
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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(x1) is odd. Since this is a product, z must be odd, and x1 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



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Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
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28 Sep 2009, 05:34
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It took me 15 seconds.
1) sufficient (x is even) 2) insufficient (try even y, odd x, odd z; even x, odd y, z)
A



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Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
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29 Sep 2009, 19:26
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Took me less than a min
if XY+Z is odd these are the possiblities
X Y Z
E E O O E O E O O O O E
so substituting these values in Option A gives the solution write away as Even. in B we have 2 diff results so
Answer is A



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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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31 Jan 2014, 09:31
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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



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



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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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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 evenSufficient 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: \(xzz=evenodd=odd\) > \(z(x1)=odd\). For the product of two integers to be odd, both of them must be odd > \(z\) and \(x1\) are odd. If \(x1=odd\), then x must be even: \(x1=xodd=odd\) > \(x=odd+odd=even\). Hope it's clear.
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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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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 stop testing, x can be either even or odd insufficient A
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Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]
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nikhilpoddar wrote: Please solve... took me 4 seconds, but I got it wrong



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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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23 Feb 2015, 06:56
Amazing explanation as usual Bunuel. May I know how did you validate statement (2) !
[quote="Bunuel"][quote="Mountain14"][quote="jlgdr"][quote="Aleehsgonji"]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



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



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Re: DS ODD EVEN [#permalink]
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rahsin wrote: If x, y, 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 Statement 1xy+xz is even xy+z is odd Subtract the two. xy+xzxyz = xzz xzz will be odd (even  odd will always result in odd) z(x1) will be odd This is only possible when both z and (x1) are odd. x1 is odd. This means that x is even. Sufficient. Statement 2y+xz is odd xy+z is odd Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z). (x+1)(y+z) will be even (odd+odd is even) Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result) or x+1 can be odd and y+z can be even (Even * odd will result in an even result) or both (x+1) and (y+z) can be even (Even * even will result in an even result) So x+1 can be both odd and even. In other words, x can be both even and odd. This is only possible when both z and (x1) are odd. x1 is odd. This means that x is even. NOT Sufficient. A is the answer.



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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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31 Jan 2017, 19:03
(1) Since xy + z is odd and xy +xz is even, subtracting we can deduce that zxz = z(1x) is odd. Thus 1  x is odd, so x must be even. SUFF (2) SUbtracting, we see that xy  y + z  xz = y(x  1)  z(x  1) =(y  z)(x  1) is even. If y  z is odd, x  1 is even and thus x is odd. On the other hand, if y  z is even, x is even. NOT SUFF Hence A.
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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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15 Mar 2017, 00:52
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 >z(x1) = odd so X= even
(2) y + xz is an odd integer >(Y+Z)(X+1)= even,{given xy + z=odd,either XY =odd and Z =even,satisfies in statement 2 so X can be odd} { Now XY= even and Z=odd, either X/Y =even or X&Y =even,As both wont satisfy in statement 2 so X cannot be even} Hence i got answer D



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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]
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15 Mar 2017, 01:07
sreenu7464 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 >z(x1) = odd so X= even
(2) y + xz is an odd integer >(Y+Z)(X+1)= even,{given xy + z=odd,either XY =odd and Z =even,satisfies in statement 2 so X can be odd} { Now XY= even and Z=odd, either X/Y =even or X&Y =even,As both wont satisfy in statement 2 so X cannot be even} Hence i got answer D In (2), I see no problem if x is even, y is odd and z is odd so xy+z is odd and y+xz is odd.
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Re: If x, y and z are integers and xy + z is an odd integer, is
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