If x, y and z are positive integers, is xz even?

even - even = even

odd + odd = even.

Statement 1:

2xy = even. So x has to be even. Thus xz must be even. Sufficient.

Statement 2:

case 1:

let x be 2 and

\(x^2 = even\)

xz = even.

case 2:

x = 3 and z = 3

\(3^2 + 3*3 = 18\)

but 3*3 = 9 , which is odd.

Thus not sufficient .

The best answer is A.

Asked: If x, y and z are positive integers, is xz even?

(1) (2xy − x) is even
2xy is even ; x is even
xz will also be even since x is even
SUFFICIENT

(2) (x^2 + xz) is even
If x is even, x^2 is even and xz is also even ; x^2 + xz is even
But if x is odd; x^2 is odd ; xz is also odd
NOT SUFFICIENT

IMO A

Hi Bunuel
eve should be replaced with even in highlighted red portion.

This looks like a "GMAT code" sort of question, since they're asking me (and telling me) fairly complicated things about even/odd. Usually with these questions, they're actually communicating something much simpler, but they've dressed it up a bit to make the problem look harder than it really is.

Question stem: Is xz even? xz will be even if either x or z is even, or both. So really, the answer would be "yes" if x or z is even, and "no" if they're both odd.

(1) (2xy − x) is even

What is this actually saying? First, 2xy will always be an even number no matter what the values of x and y are. So, it doesn't really factor into the statement here. It's basically saying "Even - x = even".

When we subtract x from some even value, we get another even value. The only way this will happen is if x itself is even.

Therefore, this statement really says "x is even." So, the answer to the question is "yes," and the statement is sufficient.

(2) (x^2 + xz) is even

I don't see anything here that I already know is even or odd, unlike the previous statement. Instead of approaching it that way, I'll see if I can factor it to simplify things. x^2 + xz factors to x(x + z). In other words, x(x+z) is even. That means either x is even, or x+z is even, or both.

Does this give us a definite answer to the question?

One possible way to read this statement is that x is even. In that case, the answer to the question is "yes."

But another possible way to read it is that x+z is even. That can happen if x and z are both odd. So, it's possible that x and z are both odd, in which case, the answer is "no."

Since we can get both a "yes" answer and a "no" answer, the statement is insufficient.

The correct answer is A.

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.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Condition 1) tells \(2xy-x = x(2y-1)\) is an even integer and \(2y-1\) is an odd integer.
Thus, condition 1) tells \(x\) is an even integer.
Then \(xz\) is an even integer.
Since condition 1) yields a unique solution, it is sufficient.

Condition 2) tells \(x^2+xz = x(x+z)\) is even.
It means both \(x\) and \(z\) are odd integers or \(x\) is an even integer.
If both \(x\) and \(z\) are odd integers, then \(xz\) is an odd integer and the answer is 'yes'.
If \(x\) is an even integer, then \(xz\) is an even integer and the answer is 'no'.

Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.

The product of x and z will be even if at least one of them is even.

From statement I alone, 2xy -x = even.
2xy will definitely be even. LHS can be even only if x is also even. If x is even, xz will definitely be even.
Statement I alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, \(x^2\) +xz = even.
This can happen in two ways.
Case 1: Both \(x^2\) and xz are odd. Is xz even? We answer with a NO.
Case 2: Both \(x^2\) and xz are even. Is xz even? We answer with a YES.
Statement II alone is insufficient to answer the question with a definite YES or NO. Answer option D can be eliminated.

The correct answer option is A.

Hope that helps!