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If m and n are nonzero integers, is m^n an integer?

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If m and n are nonzero integers, is m^n an integer? [#permalink] New post 21 Feb 2011, 10:26
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If m and n are nonzero integers, is m^n an integer?

(1) n^m is positve
(2) n^m is an integer.
[Reveal] Spoiler: OA

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Re: QR DS 79 [#permalink] New post 21 Feb 2011, 10:35
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If m and n are nonzero integers, is m^n an integer?
(1) n^m is positve
(2) n^m is an integer.

(1) n=-2, m=2; n^m=(-2)^2=4. +ve;
m^n=(2)^-2=1/4=0.25. Not an integer

n=1; m=1; n^m=1^1=1; +ve
m^n=1^1=1; Integer.

Not Sufficient.

(2) n=-2, m=2; n^m=(-2)^2=4. integer;
m^n=(2)^-2=1/4=0.25. Not an integer

n=1; m=1; n^m=1^1=1; integer
m^n=1^1=1; Integer.

Not Sufficient.

Combining both;
n=-2, m=2; n^m=(-2)^2=4. integer and +ve;
m^n=(2)^-2=1/4=0.25. Not an integer.

n=1; m=1; n^m=1^1=1; integer and +ve
m^n=1^1=1; Integer.

Ans: "E"
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Re: QR DS 79 [#permalink] New post 21 Feb 2011, 10:58
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Baten80 wrote:
If m and n are nonzero integers, is m^n an integer?
(1) n^m is positve
(2) n^m is an integer.


If m and n are nonzero integers, is m^n an integer?

If n is a positive integer then m^n will be an integer for any value of m (taking into account that both are nonzero integers).
If n is negative then m^n will be an integer if and only m=1 or m=-1, for example: (-1)^(-2)=1/(-1)^2=1

So basically we are asked: is n positive or m=|1|?

(1) n^m is positive --> either m=even (and in this case n can take any value) or n=positive (and in this case m can take any value). Not sufficient.

(2) n^m is an integer --> either m=positive (and in this case n can take any value) or m=negative and in this case n=1 or -1. Not sufficient.

(1)+(2) If n^m=(-1)^2=positive integer, then the answer will be NO as m^n=2^(-1)=1/2 but if n^m=1^2=positive integer, then the answer will be YES as m^n=2^1=2. Not Sufficient.

Answer: E.
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Re: QR DS 79 [#permalink] New post 21 Feb 2011, 17:53
We cannot know whether n is positive, so E.
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Re: If m and n are nonzero integers, is m^n an integer? [#permalink] New post 26 Mar 2015, 02:29
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Re: If m and n are nonzero integers, is m^n an integer? [#permalink] New post 26 Mar 2015, 11:03
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Hi All,

This DS question is built around a few Number Property rules. You can take advantage of these rules by TESTing VALUES and keeping your TESTs simple....

We're told that M and N are NON-ZERO INTEGERS. We're asked if M^N is an integer. This is a YES/NO question.

Fact 1: N^M is POSITIVE

IF....
N = 1
M = 1
1^1 is positive
1^1 is an integer and the answer to the question is YES

IF....
N = -2
M = 2
(-2)^2 is positive
2^(-2) is NOT an integer and the answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: N^M is an integer

The same two TESTs that we used in Fact 1 also 'fit' Fact 2....

IF....
N = 1
M = 1
1^1 is an integer
1^1 is an integer and the answer to the question is YES

IF....
N = -2
M = 2
(-2)^2 is an integer
2^(-2) is NOT an integer and the answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we have the SAME TESTs for both Facts which give us a YES and a NO answer.
Combined, INSUFFICIENT

Final Answer:
[Reveal] Spoiler:
E


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Re: If m and n are nonzero integers, is m^n an integer?   [#permalink] 26 Mar 2015, 11:03
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