If x is an integer, is x even?

[VJesus12]

If x is an integer, is x even?

(1) x^2 - y^2 = 0
(2) x^2 + y^2 = 18

The OA is the option C.

I don't know how using both statements I can get an answer here. Experts, can you help me? Thanks in advanced.

[Vincen]

Hello.

I solve it as follows:

If $$x^2-y^2=0\ \Rightarrow\ x^2=y^2\ \Rightarrow\ x=\pm\ y.$$ Since I don't know the value of y, hence I don't know if x is or not even. (1) is not sufficient.

If $$x^2+y^2=18$$ hence: $$x=3,\ y=3\ \Rightarrow\ \ 3^2+3^2=18.$$ but $$x=4,\ y=\sqrt{2}\ \Rightarrow\ \ 4^2+\left(\sqrt{2}\right)^2=18.$$ Hence, (2) is not sufficient.

Now, using both, I added the expressions and get $$2x^2=18\ \Rightarrow\ x^2=9\ \Rightarrow\ \ x=\pm3$$ Now, no matter the sign of x, I have that x is NOT even (is odd). This answer is sufficient.

The correct option is C

[Jay@ManhattanReview]

If x is an integer, is x even?

(1) x^2 - y^2 = 0
(2) x^2 + y^2 = 18

The OA is the option C.

I don't know how using both statements I can get an answer here. Experts, can you help me? Thanks in advanced.

We are given that x is an integer.

We have to determine whether x is even.

Let's take each statement one by one.

(1) x^2 - y^2 = 0
=> x^2 = y^2
=> |x| = |y|

Case 1: Say x = y = 1. x is NOT even; the answer is No.
Case 2: Say x = y = 2. x is even; the answer is Yes. Insufficient.

(2) x^2 + y^2 = 18

Case 1: Say x = 1. then y = 17^(1/2); x is NOT even; the answer is No.
Note that y is not necessarily an integer.

Case 2: Say x = 2. then y = 14^(1/2); x is even; the answer is Yes. Insufficient.

(1) and (2) together

From (1), we know that x^2 = y^2 and from (2), we know that x^2 + y^2 = 18

Thus, 2x^2 = 18 => x = ±3; Thus x is not even. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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