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Is x an even integer? (GMAT PREP 1)

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by alex.gellatly » Fri Aug 03, 2012 10:51 pm
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

Thanks!
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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Source: — Data Sufficiency |

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by niketdoshi123 » Sat Aug 04, 2012 3:05 am
alex.gellatly 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

Thanks!
From statement 1 we can answer that x is even

If statement 2 is true x can be odd or even..
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by alex.gellatly » Wed Aug 15, 2012 7:28 pm
niketdoshi123 wrote:
alex.gellatly 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

Thanks!
From statement 1 we can answer that x is even

If statement 2 is true x can be odd or even..
So... the correct answer is [spoiler]A?[/spoiler]
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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by GMATGuruNY » Wed Aug 15, 2012 8:45 pm
alex.gellatly 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

Thanks!
Statement 1: xy + xz is even
(xy + xz) - (xy + z) = even - odd
xz - z = odd
z(x-1) = odd.
Since odd*odd = odd, x-1 must be odd, implying that x itself is even.
SUFFICIENT.

Statement 2: y + xz is odd
(y + xz) + (xy + z) = odd + odd
y + x(z+y) + z = even
x(y+z) + (y+z) = even
(y+z)(x+1) = even.
Since it's possible that x+1 is even or that x+1 is odd, no way to determine whether x is even.
INSUFFICIENT.

The correct answer is A.
Last edited by GMATGuruNY on Wed Aug 15, 2012 9:02 pm, edited 1 time in total.
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by crisro » Wed Aug 15, 2012 8:53 pm
We have xy+z odd

1) xy+xz even
if we add/subtract odd and even we are getting odd, so
xy+xz-(xy+z)=xz-z=z(x-1) odd
for z(x-1) to be odd both z and (x-1) have to be odd,
so x has to be even.
Sufficient

2) y+xz odd
if we add/subtract two odd numbers we are getting even
xy+z+y+xz=x(y+z)+y+z=(y+z)(x+1)even
if y+z=even x+1 or x can be either way odd or even
xy+z-(y+xz)=xy-y-xz+z=y(x-1)-z(x-1)=(x-1)(y-z) even
if y-z=even x-1 or x can be either way odd or even
Insufficient
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