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If x, y and z are positive integers such that x^4y^3 = z^2

Expert replies
by RBBmba@2014 » Tue May 05, 2015 2:23 am
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!
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Source: — Data Sufficiency |

by GMATGuruNY » Tue May 05, 2015 5:26 am
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.
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by RBBmba@2014 » Tue May 05, 2015 9:14 pm
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!
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by andymal » Wed May 06, 2015 12:14 pm
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!!
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by GMATGuruNY » Thu May 07, 2015 3:06 am
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.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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I unlock the best way for YOU to solve problems.

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