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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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New post 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.

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Re: QR DS 79  [#permalink]

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New post 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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New post 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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New post 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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New post 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:

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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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

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New post 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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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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New post 02 Jul 2019, 05:57
Try my approach:
here m and n

There are 4 combinations possible:

m and n positive
m and n negative
m positive n neg
n pos and n neg

take m and n as 2 and 3

change the signs of 2 and 3 respectively

Compare with each option whether you can prove that wrong.

If you can' t prove wrong, then that is the answer
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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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New post 16 Jul 2019, 19:06
ShivamoggaGaganhs
Quote:
Try my approach:
here m and n

There are 4 combinations possible:

m and n positive
m and n negative
m positive n neg
n pos and n neg

take m and n as 2 and 3

change the signs of 2 and 3 respectively

Compare with each option whether you can prove that wrong.

If you can' t prove wrong, then that is the answer


I believe this approach can land you into trouble because you are missing a few important cases in case of exponents
1. What if one of them has a value 1
2. What if one of them is even or odd

It was just fluke that you could get the answer for this question with your method but it might not work every time as 1 behaves different from other integers in case of exponents and so does even and odd exponents.

Just my 2 cents :)
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Re: If m and n are nonzero integers, is m^n an integer?   [#permalink] 16 Jul 2019, 19:06
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