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!
If x, y and z are positive integers such that x^4y^3 = z^2
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Every test-taker should know the following: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
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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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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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!!
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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For clarity purposes, I tested two cases in my post above: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!
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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