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If x, y, and z are positive integers, is x+y divisible by 2?

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Math Revolution GMAT Instructor
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If x, y, and z are positive integers, is x+y divisible by 2?  [#permalink]

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New post 26 Jan 2018, 01:17
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A
B
C
D
E

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  35% (medium)

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[GMAT math practice question]

If x, y, and z are positive integers, is x+y divisible by 2?

1) x+z is divisible by 2
2) y+z is divisible by 2

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Re: If x, y, and z are positive integers, is x+y divisible by 2?  [#permalink]

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New post 26 Jan 2018, 08:02
MathRevolution wrote:
[GMAT math practice question]

If x, y, and z are positive integers, is x+y divisible by 2?

1) x+z is divisible by 2
2) y+z is divisible by 2


(1) x+z is even. This will happen if either both x/z are odd or both x/z are even. But nothing given about y. So not sufficient

(2) y+z is even. This will happen if either both y/z are odd or both y/z are even. But nothing given about x. So not sufficient

Combining the two statements, we could have two cases:
If z is odd, then by first statement x also has to be odd, and by second statement, y also has to be odd. In this case x+y = odd+odd = even, thus divisible by 2.
If z is even, then by first statement x also has to be even, and by second statement, y also has to be even. In this case x+y = even+even = even, thus divisible by 2.

So in any case, x+y is divisible by 2. This gives us a definite YES to the question asked. Sufficient.

Hence C answer
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Re: If x, y, and z are positive integers, is x+y divisible by 2?  [#permalink]

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New post 28 Jan 2018, 18:44
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):
There are two cases to consider.
Case 1: If x, y, and z are all even, then the answer is ‘yes’
Case 2: If x, y, and z are all odd, then the answer is also ‘yes’
Since no other cases are possible, both conditions are sufficient, when taken together.

Since this is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since it tells us nothing about y, condition 1) is not sufficient.

Condition 2)
Since it tells us nothing about y, condition 2) is not sufficient.


Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Answer: C
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Re: If x, y, and z are positive integers, is x+y divisible by 2?  [#permalink]

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New post 01 May 2018, 10:57
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

If x, y, and z are positive integers, is x+y divisible by 2?

1) x+z is divisible by 2
2) y+z is divisible by 2


Target question: Is x+y divisible by 2?

Statement 1: x+z is divisible by 2
No information about y.
Statement 1 is NOT SUFFICIENT

Statement 2: y+z is divisible by 2
No information about x.
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: If x+z is divisible by 2, then we can say that x+z = 2k for some integer k
Statement 2: If y+z is divisible by 2, then we can say that y+z = 2j for some integer j

Now take the two red equations add ADD them to get: x + y + 2z = 2k + 2j
Subtract 2z from both sides to get: x + y = 2k + 2j - 2z
Factor right side: x + y = 2(k + j - z)
This tells us that x+y is DEFINITELY divisible by 2
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Re: If x, y, and z are positive integers, is x+y divisible by 2? &nbs [#permalink] 01 May 2018, 10:57
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