m and n are integers. Is m^n an integer? 1) n^m is positive

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m and n are integers. Is m^n an integer? 1) n^m is positive [#permalink]

11 Nov 2005, 09:09

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m and n are integers. Is m^n an integer?

1) n^m is positive
2) n^m is an integer
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Re: DS - Integer [#permalink]

11 Nov 2005, 09:20

gamjatang wrote:

m and n are integers. Is m^n an integer?

1) n^m is positive
2) n^m is an integer

to make m^n an integer ---> n>= 0
but from 1 and 2, we can't conclude whether it is
take example : n= -1, m =2 , 2^-1 = 1/2

E it is.

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11 Nov 2005, 13:02

this is E...

we need to know if N is a postive integer...we dont know that from (1) and (2)

cause -2^0=1..which is positive. and is an integer...

2^0

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11 Nov 2005, 13:17

agree.

for m^n to be integer,
(a) m = 1, n = -1
or
(b) m any integer, n = 0

or
(c)m any integer(except 0) and n positive integer

together (1) and (2) cancels (a) and (b) but still (c) is not implied.
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11 Nov 2005, 21:31

m^n will always be an integer except the following case

m = 0 and n <0 -> Remember we cannot have a zero in the denominator

(1) Sufficient

If n^m is positive implies that n > 0.

(2) Insufficient

If m = 0, n^m is still an integer.

I pick A.

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11 Nov 2005, 21:35

opps... forgot about the negative cases....

getting slow. =(

The answer should be E.

[#permalink] 11 Nov 2005, 21:35

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m and n are integers. Is m^n an integer? 1) n^m is positive

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m and n are integers. Is m^n an integer? 1) n^m is positive

m and n are integers. Is m^n an integer? 1) n^m is positive

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