Data Sufficiency

If x, y and z are positive integers such that x^4*y^3 = z^2, is x^9-y^6 odd?

1.(x^4*y^3)/(x^2+y^2) can be written in the form 4k + 3, where k is a positive integer.
2. z = x + y

OA: D

@ Experts - It took me more than 3 mins(nearly 4 mins) to solve. Could you please share any smarter approach for fast solution ? Thanks in advance!

RBBmba@2014 wrote:If x, y and z are positive integers such that x^4*y^3 = z^2, is x^9-y^6 odd?

1.(x^4*y^3)/(x^2+y^2) can be written in the form 4k + 3, where k is a positive integer.
2. z = x + y

Every test-taker should know the following:
EVEN ± EVEN = EVEN.
ODD ± ODD = EVEN.
EVEN ± ODD = ODD.
ODD ± EVEN = ODD.

Also:
(EVEN)(EVEN) = EVEN.
(EVEN)(ODD) = EVEN.
(ODD)(EVEN) = EVEN.
(ODD)(ODD) = ODD.

By extension:
An even integer raised to a positive power stays EVEN.
An odd integer raised to a positive power stays ODD.

Whether an integer is even or odd is known as the PARITY of the integer.

Is x� - y� odd?
The answer to the question stem is YES if x and y are different parities (one is even, while the other is odd).
Question stem, rephrased:
Are x and y different parities?

Statement 1:
(x�y³)/(x²+y²) = 4k + 3.
(x�y³)/(x²+y²) = EVEN + ODD.
(x�y³)/(x²+y²) = ODD.
x�y³ = (ODD)(x²+y²).

Test whether it's possible for x and y to be different parities.

Case 1: x is odd and y is even
The red equation becomes:
(odd)(even) = (odd)(odd + even)
even = (odd)(odd).
even = odd.
Doesn't work.
Thus, it is not possible that x is odd and y is even.

Case 2: x is even and y is odd
The red equation becomes:
(even)(odd) = (odd)(even + odd)
even = (odd)(odd).
even = odd.
Doesn't work.
Thus, it is not possible that x is even and y is odd.

Since x and y cannot be different parities, the answer to the question stem is NO.
SUFFICIENT.

Statement 2:
Given z = x+y and x�y³ = z², test whether it's possible for x and y to be different parities.

Case 1: x is odd and y is even
The blue equation becomes:
z = odd + even
z = odd.

Since z=odd, the green equation becomes:
(odd)(even) = odd
even = odd.
Not possible.
Thus, it is not possible that x is odd and y is even.

Case 2: x is even and y is odd
The blue equation becomes:
z = even + odd
z = odd.

Since z=odd, the green equation becomes:
(even)(odd) = odd
even = odd.
Not possible.
Thus, it is not possible that x is even and y is odd.

Since x and y cannot be different parities, the answer to the question stem is NO.
SUFFICIENT.

The correct answer is D.

Did it in the nearly same way but it took me more than 3.5 mins, so thought whether there could be any faster approach!

GMATGuru,

Thanks for your incredibly thorough explanation. I'm also wondering the same thing. Is there a way to get through a question like this any faster? If it really would take 4 minutes. Should I guess on something like this? Thanks!!

RBBmba@2014 wrote:Did it in the nearly same way but it took me more than 3.5 mins, so thought whether there could be any faster approach!

For clarity purposes, I tested two cases in my post above:
Case 1: x is odd and y is even
Case 2: x is even and y is odd
But it is necessary to test only ONE of these two cases.

The problem includes 4 expressions:
x�y³
x� - y�
x²+y²
x+y.

The first expression will be EVEN whether x is odd and y is even (Case 1) or x is even and y is odd (Case 2).
The remaining expressions will be ODD whether x is odd and y is even (Case 1) or x is even and y is odd (Case 2).
Since both cases yield the SAME RESULTS, we can evaluate the two statements by testing only Case 1.