If m and n are nonzero integers, is m^n 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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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"

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:

GMAT assassins aren't born, they're made,
Rich

Hi VeritasPrepKarishma
Can you add your two cents? I am not able to understand both statements.
WR,
Arpit
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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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Given: m, n - integers, m, n are not equal to zero

Statement 1: n^m is positive
n = 1, m = 1, n^m = 1, hence m^n = 1...we get YES
n = -1, m = 2, n^m = 1, hence m^n = 1/2...we get NO

Statement 1 is not Sufficient.

Statement 2: n^m is an integer
n = 1, m = 1, n^m = 1, hence m^n = 1...we get YES
n = -1, m = 2, n^m = 1, hence m^n = 1/2...we get NO

Statement 2 is not Sufficient.

Combining both statements, we can use same examples

Combing is also not Sufficient.


Answer E.



Thanks,
GyM
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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

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

(1) n^m is positve
(2) n^m is an integer.

m,n=......-2,-1,1,2.... any integer

Statement 1: n^m is positive

................................n^m.............................m^n
If m=2, n=2 then 2^2=4 (positive) 2^2=4 (Integer)
If m=2, n=-2 then (-2)^2=4 (positive) 2^-2=1/4 (Not Integer)

Hence, Not Sufficient

Statement 2: n^m is an integer.

........................................ n^m....................m^n
If m=2, n=2 then 2^2=4 (integer) 2^2=4 (Integer)
If m=2, n=-2 then (-2)^2=4 (integer) 2^-2=1/4 (Not Integer)

Hence, Not Sufficient

Statement 1 and 2 combined

..................................n^m....................................m^n
If m=2, n=2 then 2^2=4 (positive+ Integer) 2^2=4 (Integer)
If m=2, n=-2 then (-2)^2=4 (positive+ Integer) 2^-2=1/4 (Not Integer)

Not Sufficient

Ans E

(1) Let n= -2, m= 2, \((-2)^2=4\) is positive number, 2^-2=\(\frac{1}{4}\) is not an integer. Yes
Again, n= 2, m= 2, \((2)^2=4\) is positive number \(2^2=4\) is an integer, No
INSUFFICIENT.

(2) Let n= -2, m= 2, \((-2)^2=4\) is an integer, 2^-2=\(\frac{1}{4}\) but is not an integer. Yes
Again, n= 2, m= 2, \((2)^2=4\) is an integer, \(2^2=4\) is an integer, No

INSUFFICIENT

Considering both
Let n= -2, m= 2, \((-2)^2=4\) is a positive number and an integer, but 2^-2= \(\frac{1}{4}\) is not an integer. Yes
Again, n= 2, m= 2, \((2)^2=4\) is a positive and an integer, \(2^2=4\) is an integer, No

INSUFFICIENT

THE ANSWER IS E
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(1) n^m > 0

two possible cases:

(a) if n < 0 then m should be even
(b) if n > 0 then m can be any number

From these 2 cases we can see that n can take up negative values therefore m^n may not be an integer.

INSUFFICIENT

(2) n^m is an integer

According to this statement m>= 0 but n can take up any value.

Again since n can be (+) or (-) therefore INSUFFICIENT

(1)(2) combining the statements wont make a difference, n can still take up negative values.

INSUFFICIENT

The answer is E.

Solution:

We need to determine if m^n is an integer, given that m and n are nonzero integers. Notice that m^n is an integer if n is positive OR if n is negative and m is either 1 or -1.

Statement One Alone:

If n = 2 and m = 4, then m^n = 4^2 = 16 is an integer. However, if n = -2 and m = 4, then m^n = 4^(-2) = 1/16 is not an integer. Statement one alone is not sufficient.

Statement Two Alone:

If n = 2 and m = 4, then m^n = 4^2 = 16 is an integer. However, if n = -2 and m = 4, then m^n = 4^(-2) = 1/16 is not an integer. Statement two alone is not sufficient.

Statements One and Two Together:

With the two statements, n^m is a positive integer. We can use the same examples for statement one and statement two to see that both statements together are not sufficient.

Answer: E
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