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If x and y are positive integers, is the following cube root

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solarzj
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If x and y are positive integers, is the following cube root [#permalink] New post   09 Jun 2012, 07:22
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If x and y are positive integers, is the following cube root an integer?

\((x + y^2)^{(\frac{1}{3})}\)

(1) \(x = y^2(y-1)\)
(2) \(x = 2\)

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Re: If x and y are positive integers, is the following cube root [#permalink] New post   09 Jun 2012, 07:45
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If x and y are positive integers, is the following cube root an integer?

\(\sqrt[3]{x + y^2}\)

(1) x = y^2(y-1) --> \(x=y^3-y^2\) --> \(\sqrt[3]{x+y^2}=\sqrt[3]{y^3-y^2+y^2}=\sqrt[3]{y^3}=y=integer\). Sufficient.

(2) x = 2 --> if \(y=1\) then the answer is NO but if \(y=5\) then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.
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Re: If x and y are positive integers, is the following cube root [#permalink] New post   18 Dec 2016, 15:46
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Hello Friends,
please correct the question:
1) x = y^2 * (y - 1) AND NOT x = y^2(y - 1).
Needless to say that i assumed the latter form :(
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Re: If x and y are positive integers, is the following cube root [#permalink] New post   09 Jun 2012, 07:23
[quote="solarzj"]If x and y are positive integers, is the following cube root an integer?

(x + y^2)^(1/3)

1. x = y^2(y-1)
2. x = 2

So if I understand correctly.

In 1.) if I expand x = y^2 * (y-1) to x = y^3 - y ^2 and substitute in

cubeRoot(x + y^2) => i get cubeRoot(y^3 - y ^2 + y^2)
which leads to cubeRoot(y^3)
because both y^2 cancel out.

After this I can say, based on the information of the question (x and Y are positive integers) that y is a positive INTEGER

and because cubeRoot(y^3) => y ... the answer is yes, cubeRoot(x + y^2) = y = a positive INTEGER
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Re: If x and y are positive integers, is the following cube root [#permalink] New post   17 Aug 2020, 12:40
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solarzj wrote:
If x and y are positive integers, is the following cube root an integer?

(x + y^2)^(1/3)

(1) x = y^2(y-1)
(2) x = 2


This is a badly designed question, because if you try to use both statements together, there is no positive integer value possible for y, so the two statements are not consistent. That can never happen in a real GMAT question, and it means answer C is not much of a 'trap' answer here, because you end up with an equation you can't even solve. If this were a real GMAT question, you'd find it easy to solve for y using both Statements, and thus might be tempted to pick C (even though A is the right answer, as you can see immediately by substituting for x).
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Re: If x and y are positive integers, is the following cube root [#permalink] New post   24 Dec 2020, 00:09
I interpreted it as y^2(y-1) instead of (y^2) * (y-1). Hence got it wrong.

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Re: If x and y are positive integers, is the following cube root [#permalink]
24 Dec 2020, 00:09
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