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# If m and n are nonzero integers, is m^n an integer? (1) n^m

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If m and n are nonzero integers, is m^n an integer? (1) n^m [#permalink]  21 Feb 2011, 11:26
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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]  21 Feb 2011, 11:35
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]  21 Feb 2011, 11:58
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]  21 Feb 2011, 18:53
We cannot know whether n is positive, so E.
Re: QR DS 79   [#permalink] 21 Feb 2011, 18:53
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# If m and n are nonzero integers, is m^n an integer? (1) n^m

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