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If x and y are positive integers

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Source: — Data Sufficiency |

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by [email protected] » Tue Aug 01, 2017 12:44 pm
Hi jjjinapinch,

We're told that X and Y are POSITIVE INTEGERS. We're asked if (X)(Y) is EVEN. This is a YES/NO question. This question deals with Number Properties, so you can solve it using those rules and/or by TESTing VALUES.

In this question, it's helpful to consider the possible answers to the given question. The ONLY way for the answer to be "NO" is if BOTH X and Y are ODD. In all other circumstances, the answer to the question will be "YES." You can actually use this knowledge "against" each of the two Facts and eliminate possibilities....

1) X^2 + Y^2 - 1 is divisible by 4

For Fact 1, we can use a Number Property to eliminate a possibility.....
IF.... X and Y are BOTH ODD, then we would have....
(Odd)^2 + (Odd)^2 - 1....
Odd + Odd - Odd = ALWAYS ODD.... but an Odd number CANNOT be divisible by 4, so there is no way that both X and Y can be Odd. This eliminates the possibility of a "No" answer... so the only option that is left is "Yes" (thus, the answer is ALWAYS YES).
Fact 1 is SUFFICIENT

2) X + Y is ODD

For Fact 2, the only way to end up with an ODD sum is if one variable is EVEN and the OTHER is ODD. Since (EVEN)(ODD) = EVEN, the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer: D

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by Jay@ManhattanReview » Tue Aug 01, 2017 9:52 pm
jjjinapinch wrote:If x and y are positive integers, is xy even?

(1) x^2 + y^2 - 1 is divisible by 4
(1) x + y is odd

Official Guide question
Answer: D
Given that x and y are positive integers. We have to determine whether xy is even.

For xy to be even at least one of x and y must be even. xy is not even if both x and y are odd.

Statement 1: x^2 + y^2 - 1 is divisible by 4.

Since (x^2 + y^2 - 1) is divisible by 4, an even number, (x^2 + y^2 - 1) must be even.

(x^2 + y^2 - 1) --> even
=> x^2 + y^2 --> even + 1 --> even + odd --> odd

If the sum of two positive integers is odd, then each of the positive integers is even.

Thus, x^2 --> even and y^2 ---> even. This follows that x --> even and y --> even.

Thus, xy --> even. Sufficient.

Statement 2: (x + y) is odd.

If the sum of two positive integers is odd, then each of the positive integers is even.

This follows that x --> even and y --> even.

Thus, xy --> even. Sufficient.

The correct answer: D

Hope this helps!

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by GMATGuruNY » Wed Aug 02, 2017 6:06 am
jjjinapinch wrote:If x and y are positive integers, is xy even?
(1) x^2 + y^2 - 1 is divisible by 4
(1) x + y is odd
xy will not be even -- in other words, xy will be ODD -- only if x and y are both odd.
Question stem, rephrased:
Are x and y both odd?

Statement 1: x² + y² - 1 is divisible by 4
x² + y² - 1 = (multiple of 4)
x² + y² = (multiple of 4) + 1
x² + y² = EVEN + ODD
x² + y² = ODD.
The equation in blue implies that x and y CANNOT both be odd, since ODD² + ODD ² = EVEN.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.

Statement 2: x + y is odd
The statement above implies that x and y CANNOT both be odd, since ODD + ODD = EVEN.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.

The correct answer is D.
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by Jeff@TargetTestPrep » Tue Dec 19, 2017 9:18 am
jjjinapinch wrote:If x and y are positive integers, is xy even?
(1) x^2 + y^2 - 1 is divisible by 4
(1) x + y is odd
We need to determine whether the product of x and y is even.

Statement One Alone:

x^2 + y^2 − 1 is divisible by 4.

Since 4 is an even number, we need x^2 + y^2 − 1 to be even. In order for x^2 + y^2 − 1 to be even, we need x^2 + y^2 to be odd. If the sum of two squares is odd, one of them must be odd and the other must be even. This means that either x = odd and y = even OR x = even and y = odd. In either case, the product of x and y will be even. Statement one alone is sufficient to answer the question.

Statement Two Alone:

x + y is odd.

Since x + y = odd, either x = odd and y = even OR x = even and y = odd. In either case, the product of x and y will be even. Statement two alone is sufficient to answer the question.

Answer: D

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