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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