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

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Director
Joined: 14 Sep 2005
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Location: South Korea
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m and n are integers. Is m^n an integer? 1) n^m is positive [#permalink]  11 Nov 2005, 08:09
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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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Auge um Auge, Zahn um Zahn !

SVP
Joined: 24 Sep 2005
Posts: 1893
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Kudos [?]: 138 [0], given: 0

Re: DS - Integer [#permalink]  11 Nov 2005, 08:20
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.
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 180 [0], given: 2

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
VP
Joined: 22 Aug 2005
Posts: 1123
Location: CA
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Kudos [?]: 42 [0], given: 0

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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Whether you think you can or think you can't. You're right! - Henry Ford (1863 - 1947)

Senior Manager
Joined: 07 Jul 2005
Posts: 405
Followers: 3

Kudos [?]: 19 [0], given: 0

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.
Senior Manager
Joined: 07 Jul 2005
Posts: 405
Followers: 3

Kudos [?]: 19 [0], given: 0

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

getting slow. =(

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