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# If m and n are positive integers, is root(m)^n an integer?

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Math Expert
Joined: 02 Sep 2009
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If m and n are positive integers, is root(m)^n an integer? [#permalink]

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09 Jul 2012, 04:56
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Difficulty:

15% (low)

Question Stats:

68% (00:39) correct 32% (00:50) wrong based on 963 sessions

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If m and n are positive integers, is $$(\sqrt{m})^n$$ an integer?

(1) $$\sqrt{m}$$ is an integer.
(2) $$\sqrt{n}$$ is an integer.

Diagnostic Test
Question: 44
Page: 26
Difficulty: 600
[Reveal] Spoiler: OA

_________________

Kudos [?]: 129002 [0], given: 12187

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129002 [0], given: 12187

Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]

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09 Jul 2012, 04:56
Expert's post
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SOLUTION

If m and n are positive integers, is $$(\sqrt{m})^n$$ an integer?

(1) $$\sqrt{m}$$ is an integer --> $$(\sqrt{m})^n=(integer)^{positive \ integer}=integer$$. Sufficient.

(2) $$\sqrt{n}$$ is an integer. If $$m=n=1$$, then the answer is YES but if $$m=2$$ and $$n=1$$, then $$(\sqrt{m})^n=\sqrt{2}\neq{integer}$$, so in this case the answer is NO. Not sufficient.

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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]

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09 Jul 2012, 05:27
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Bunuel wrote:
If m and n are positive integers, is $$(\sqrt{m})^n$$ an integer?

(1) $$\sqrt{m}$$ is an integer.
(2) $$\sqrt{n}$$ is an integer.

Hi,

Difficulty level: 600

We have to find whether $$(\sqrt{m})^n$$ is an integer,
or $$m^{\frac n2}$$ is an integer.

Given, m & n are integers,

Using (1),
$$\sqrt{m}$$ is an integer
then, $$(integer)^{(integer)}$$ is Integer. Sufficient.

Using (2),
$$\sqrt{n}$$ is an integer, or n is an integer.
lets say, m=3, n=9,
$$(\sqrt{m})^n$$=$$(\sqrt{3})^9$$ which is not an integer. Insufficient.

Regards,

Kudos [?]: 515 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129002 [0], given: 12187

Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]

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13 Jul 2012, 04:09
SOLUTION

If m and n are positive integers, is $$(\sqrt{m})^n$$ an integer?

(1) $$\sqrt{m}$$ is an integer --> $$(\sqrt{m})^n=(integer)^{positive \ integer}=integer$$. Sufficient.

(2) $$\sqrt{n}$$ is an integer. If $$m=n=1$$, then the answer is YES but if $$m=2$$ and $$n=1$$, then $$(\sqrt{m})^n=\sqrt{2}\neq{integer}$$, so in this case the answer is NO. Not sufficient.

_________________

Kudos [?]: 129002 [0], given: 12187

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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]

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17 Jul 2014, 20:49
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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]

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21 Dec 2015, 10:45
Hello from the GMAT Club BumpBot!

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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]

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26 Mar 2017, 07:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If m and n are positive integers, is root(m)^n an integer?   [#permalink] 26 Mar 2017, 07:05
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