DS - Odd even

[karthikpandian19]

If x and y are integers and (x^2)y + 5x is odd, which of the following must be even?


(A) x

(B) y

(C) xy + 5

(D) xy - 5

(E) y + 5

[Anurag@Gurome]

If x and y are integers and x2y + 5x is odd, which of the following must be even?

I'm assuming the underlined term is x²y

Algebraic Approach:
(x²y + 5x) = x(xy + 5) is odd
Hence, both x and (xy + 5) is odd.

As (xy + 5) is odd, xy must be even.
Now, as x is odd, y must be even

The correct answer is B.

[GMATGuruNY]

If x and y are integers and x2y + 5x is odd, which of the following must be even?


(A) x

(B) y

(C) xy + 5

(D) xy - 5

(E) y + 5

Try to show that the answers DON'T have to be even.

x²y + 5x = x(xy + 5).
For this product to be odd, factors x and xy+5 must both be odd.
Let x=1 and xy+5 = 7.
Solving the second factor for y, we get:
1*y + 5 = 7
y=2.

Now we plug x=1 and y=2 into the answers and eliminate any answer choice that does not yield an even result.

A: x=1. Eliminate A.
B: y=2. Hold onto B.
C: xy+5 = 1*2 + 5 = 7. Eliminate C.
D: xy-5 = 1*2 - 5 = -3. Eliminate D.
E: y+5 = 2+5 = 7. Eliminate E.

The correct answer is B.

[sams_online]

If x and y are integers and x2y + 5x is odd, which of the following must be even?

I'm assuming the underlined term is x²y

Algebraic Approach:
(x²y + 5x) = x(xy + 5) is odd
Hence, both x and (xy + 5) is odd.

As (xy + 5) is odd, xy must be even.
Now, as x is even, y must be even

The correct answer is B.

Correction: as x is odd, y must be even

[Anurag@Gurome]

Correction: as x is odd, y must be even

Thanks for pointing the typo.
Edited my reply.

[karthikpandian19]

OA is B