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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]

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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]

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21 Feb 2011, 11:35
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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]

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21 Feb 2011, 11:58
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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]

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21 Feb 2011, 18:53
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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]

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26 Mar 2015, 03:29
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Re: If m and n are nonzero integers, is m^n an integer? [#permalink]

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26 Mar 2015, 12:03
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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]

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03 Apr 2016, 09:55
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Re: If m and n are nonzero integers, is m^n an integer? [#permalink]

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26 Apr 2017, 11:02
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Re: If m and n are nonzero integers, is m^n an integer? [#permalink]

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07 Aug 2017, 09:33
Hi VeritasPrepKarishma
Can you add your two cents? I am not able to understand both statements.
WR,
Arpit
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If m and n are nonzero integers, is m^n an integer? [#permalink]

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06 Sep 2017, 09:23
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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.

m and n are non zero integers. So m and n could be positive or negative.
Is m^n an integer?

When will m^n be an integer? When m is an integer (which we are given it is) and n is a positive integer. Also when m = 1/-1 and n is any integer.
So (2)^5, (-5)^3, 1^(-2) etc

Given that m and n are non-zero,

(1) n^m is positve

n^m will be positive when n is positive and m can be any integer.
e.g. 3^(-2), 7^4 etc.
In this case m^n is an integer.

n^m will be positive when n is negative and m is even.
e.g. (-3)^2, (-4)^(-6) etc.
In this case m^n may not be an integer.

Not sufficient.

(2) n^m is an integer.

n^m is an integer when n is any integer and m is positive.
e.g. 4^2, (-2)^3 etc.
In this case, m^n may be an integer (2^4) or may not be an integer (3^(-2)).
We already have two cases so this alone is not sufficient

Using both, n^m is a positive integer.
So n could be positive integer and m may be a positive integer e.g. 3^2, 7^3 etc. Here m^n will be an integer.
or if n is 1, m could be any integer e.g. 1^7. Here m^n will be an integer
or if n is -1, m should be an even integer e.g. (-1)^2, (-1)^-4. Here m^n may not be an integer e.g. (-4)^(-1)

Both statements together are not sufficient.

Answer (E)
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If m and n are nonzero integers, is m^n an integer?   [#permalink] 06 Sep 2017, 09:23
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