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Manager  G
Joined: 27 May 2010
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1 00:00

Difficulty:   35% (medium)

Question Stats: 66% (01:36) correct 34% (01:28) wrong based on 56 sessions

### HideShow timer Statistics Is $$\sqrt{\sqrt[n]{m}}$$ an integer?

(1) m = n + 14
(2) m = 5n + 6

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Intern  B
Joined: 17 Jan 2014
Posts: 28
Re: Is (m^(1/n))^(1/2) an integer?  [#permalink]

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Pls explain answer...how both r sufficient together? And what is concept for this question to answer

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Manhattan Prep Instructor G
Joined: 04 Dec 2015
Posts: 756
GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: Is (m^(1/n))^(1/2) an integer?  [#permalink]

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prashanths wrote:
Is $$\sqrt{\sqrt[n]{m}}$$ an integer?

(1) m = n + 14
(2) m = 5n + 6

Interesting problem! Here's how I approached it:

First, I noticed that both statements together would definitely allow me to answer the question. That's because both statements together will give you the values of m and n, and that would let you calculate the exact answer, if you really wanted to. Here's what that made me think: uh-oh, this might be a C trap. I'm going to try very carefully to prove that each statement alone is insufficient before I assume the answer is C.

Then, I rewrote the question to look a little clearer:

Is $$m^\frac{1}{2n}$$ an integer?

Statement 1: Let's start writing out cases and see if we can find some integers and non-integers.

First case: n = 1, m = 15. Is $$15^\frac{1}{2}$$ an integer? NO.

Second case: n = 2, m = 16. Is $$16^\frac{1}{4}$$ an integer? YES, and its value is 2.

The answer can be either YES or NO, so this statement is insufficient. Eliminate A and D.

Statement 2: Let's write out some more cases.

First case: If n = 1, m = 5(1) + 6 = 11. Is Is $$11^\frac{1}{2}$$ an integer? NO.

Second case: n = 2, m = 16. Is $$16^\frac{1}{4}$$ an integer? YES, and its value is 2.

The answer can be either YES or NO, so this statement is insufficient. Eliminate B.

Statements together: We can solve and find that m = 16, n = 2, and the answer is definitely YES. That's why the answer is C.
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Manager  B
Joined: 23 Jul 2015
Posts: 57
Re: Is (m^(1/n))^(1/2) an integer?  [#permalink]

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ccooley wrote:
prashanths wrote:
Is $$\sqrt{\sqrt[n]{m}}$$ an integer?

(1) m = n + 14
(2) m = 5n + 6

Interesting problem! Here's how I approached it:

First, I noticed that both statements together would definitely allow me to answer the question. That's because both statements together will give you the values of m and n, and that would let you calculate the exact answer, if you really wanted to. Here's what that made me think: uh-oh, this might be a C trap. I'm going to try very carefully to prove that each statement alone is insufficient before I assume the answer is C.

Then, I rewrote the question to look a little clearer:

Is $$m^\frac{1}{2n}$$ an integer?

Statement 1: Let's start writing out cases and see if we can find some integers and non-integers.

First case: n = 1, m = 15. Is $$15^\frac{1}{2}$$ an integer? NO.

Second case: n = 2, m = 16. Is $$16^\frac{1}{4}$$ an integer? YES, and its value is 2.

The answer can be either YES or NO, so this statement is insufficient. Eliminate A and D.

Statement 2: Let's write out some more cases.

First case: If n = 1, m = 5(1) + 6 = 11. Is Is $$11^\frac{1}{2}$$ an integer? NO.

Second case: n = 2, m = 16. Is $$16^\frac{1}{4}$$ an integer? YES, and its value is 2.

The answer can be either YES or NO, so this statement is insufficient. Eliminate B.

Statements together: We can solve and find that m = 16, n = 2, and the answer is definitely YES. That's why the answer is C.

How did you figure that m &n are integers to begin with?
Intern  B
Joined: 17 Jan 2014
Posts: 28
Re: Is (m^(1/n))^(1/2) an integer?  [#permalink]

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Mentioned in question itself, m&n as integer, hope I am right

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Manager  G
Joined: 27 May 2010
Posts: 170
Re: Is (m^(1/n))^(1/2) an integer?  [#permalink]

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Well, actually I don't think it needs to be mentioned that m and n are integers since if we solve the two equations the only value we get are n = 2 and m = 16.
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Please give Kudos if you like the post Re: Is (m^(1/n))^(1/2) an integer?   [#permalink] 26 Jun 2019, 03:15
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