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# If x, n, and y are all positive integers, is x^n divisible

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If x, n, and y are all positive integers, is x^n divisible [#permalink]

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25 Jan 2011, 21:42
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If x, n, and y are all positive integers, is x^n divisible by y?

(1) x is divisible by y^n.
(2) x^y is divisible by y.
[Reveal] Spoiler: OA

Last edited by MichelleSavina on 26 Jan 2011, 05:21, edited 1 time in total.

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Re: Is X^n divisible by y? [#permalink]

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25 Jan 2011, 22:07
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MichelleSavina wrote:
If x, n, and y are all positive integers, is X^n divisible by y?
(1) x is divisible by y^n.
(2) x^n is divisible by y.

Hi!

There must be a mistake in your post, since (2) directly answers the question, which will never happen on the actual GMAT (and which would also make the correct answer D, not A).

Ignoring (2) for now, let's start by breaking down the question stem: we need to determine whether y is a factor of x^n. What do we need? Information about the relationship among the variables.

(1) tells us that y^n is a factor of x. Well, since x, y and n are positive integers, we know that y^n is a multiple of y and x is a factor of x^n. Accordingly, (1) tells us that a multiple of y goes into a factor of x^n. If this is true, then y must go into x^n - sufficient.

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Re: Is X^n divisible by y? [#permalink]

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26 Jan 2011, 03:02
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MichelleSavina wrote:
If x, n, and y are all positive integers, is X^n divisible by y?
(1) x is divisible by y^n.
(2) x^n is divisible by y.

I think (2) should be: x^y is divisible by y

If x, n, and y are all positive integers, is X^n divisible by y?

(1) x is divisible by y^n --> since $$x$$, $$n$$, and $$y$$ are positive integers then $$x$$ is divisible by $$y$$ too (note that if we were not told that $$n$$ is positive then we couldn't say that: $$x=1$$, $$y=2$$ and $$n=0$$ --> $$x$$ is divisible by $$y^n$$ but not by $$y$$) --> as $$x$$ is divisible by $$y$$ then $$x^n$$ is divisible by $$y$$ (again as $$n$$ is a positive integers). Sufficient.

(2) x^y is divisible by y --> if $$x=y=n=1$$ then the answer will be YES but if $$x=2$$, $$y=4$$ and $$n=1$$ then $$x^y=2^4=16$$ and 16 is divisible by $$y=4$$ but $$x^n=2$$ is not divisible by $$y=4$$, so the answer in this case will be NO. Not sufficient.

Answer: A.
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Re: Is X^n divisible by y? [#permalink]

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26 Jan 2011, 05:22
sorry about that... its edited...
thankx for the clarification..

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Re: If x, n, and y are all positive integers, is x^n divisible [#permalink]

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25 Jul 2014, 02:12
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Re: If x, n, and y are all positive integers, is x^n divisible [#permalink]

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31 Jul 2015, 02:18
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Re: If x, n, and y are all positive integers, is x^n divisible [#permalink]

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06 Apr 2017, 19:18
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Re: If x, n, and y are all positive integers, is x^n divisible [#permalink]

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06 Apr 2017, 21:53
(I) Sufficient.
x^n = (y^n*k)^n.

Clearly x^n is divisible by y.

(II) Not true for diff. cases.

Answer A.
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Re: If x, n, and y are all positive integers, is x^n divisible   [#permalink] 06 Apr 2017, 21:53
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# If x, n, and y are all positive integers, is x^n divisible

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