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Target question: Is x an even integer?If x,y and z are integers and xy+z is an odd number, is x an even integer?
1) xy + xz is an even integer.
2) y + xz is an odd integer.
Given: xy+z is an odd number
Statement 1: xy+xz is an even integer.
Notice that statement 1 has an xy term, and the given information also has an xy term. We can use this to our advantage.
We know the property: Even - Odd = Odd
So, we can conclude that (xy + xz) - (xy + z) is odd
Simplify to get: xz - z is odd
Factor: z(x - 1) is odd
IMPORTANT: if the product of two integers is odd, then both of those integers must be odd as well.
So, z must be odd
And (x-1) must be odd, which means x must be even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: y + xz is an odd integer.
There are several conflicting sets of values that meet this condition. Here are two:
Case a: x = 0, y = 1 and z = 1, in which case x is even
Case b: x = 1, y = 2 and z = 1, in which case x is odd
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer = A
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Statement 1: xy + xz is evenIf 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
(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.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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Student Review #3