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

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Bunuel
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If m and n are positive integers, is root(m)^n an integer? [#permalink] New post   09 Jul 2012, 04:56
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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.
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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink] New post   09 Jul 2012, 04:56
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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.

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

Thus, Answer is (A)

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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink] New post   03 Nov 2020, 19:28
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[quote="Bunuel"]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.

1) √m = Ineteger
i.e. \((\sqrt{m})^n = Integer^{Integer} \)
But \(Integer^{Integer} = Integer\)
SUFFICIENT

2)√n = integer does not confirm whether n is even (Yes) or not (No) hence
NOT SUFFICIENT

Answer: Option A
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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink] New post   05 Nov 2020, 04:54
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If m and n are positive integers, is \((\sqrt{m})^n\) an integer?

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

\(Integer^{integer}\) = integer. SUFFICIENT.

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

We can get a yes and no answer here because we don't know if \(\sqrt{m}\) is an integer. INSUFFICIENT.

Answer is A.
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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink] New post   21 Jan 2023, 18:37
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.

Diagnostic Test
Question: 44
Page: 26
Difficulty: 600



here we know both m and n are positive integers


we need to find if this is an integer
\((\sqrt{m})^n\)



Lets look at statement 1

root m is an integer

so we know (int)^(int) is an integer

So this is sufficient as (root m)^n is an integer according to this


Looking at st 2 we know that root n is integer but this statement doesnt help as we know nothing about root m'

so this st is insufficient


Answer (A)
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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink] New post   16 Mar 2024, 03:40
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Re: If m and n are positive integers, is root(m)^n an integer? [#permalink]
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