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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
_________________

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]

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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(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]

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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.

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]

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21 Mar 2014, 20:46

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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

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 1

xy+xz is even xy+z is odd

Subtract the two. xy+xz-xy-z = xz-z

xz-z will be odd (even - odd will always result in odd)

z(x-1) will be odd

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

Sufficient.

Statement 2

y+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 (x-1) are odd.

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

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(1) Since xy + z is odd and xy +xz is even, subtracting we can deduce that z-xz = z(1-x) 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

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(x-1) = 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

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(x-1) = 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.
_________________

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

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26 May 2017, 09:25

Solved this problem by plugging numbers: Given: xy + z = odd integer

Now, plug numbers that satisfy the above condition: [ I plugged numbers by considering the scenarios : (i) xy = even and z = odd (ii) xy = odd and z = even]

X Y Z 2 1 3 1 2 3 2 4 3 3 1 2 1 3 2 3 5 2

Use the above values in given conditions to check whether X can be odd or even:

Statement 1: xy+xz = even => x(y+z) = even substitute values from above table and identify the values that satisfy statement 1 You will get : x = 2 and corresponding (y,z) as (1,3) and (4,3). Rest of cases does not satisfy statement 1. Hence we can say that X is even and cannot be odd SUFFICIENT

Statement 2: y+xz = odd substitue values in statement 2 to identify values that satisfy it 1 + 2*3 = 7 (x=2) 2 + 1*3 = 5 (x=1) This means x can be both odd and even. Hence INSUFFICIENT