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# If A and B are nonzero integers, is A^B an integer? (1) B^A

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If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]  28 Apr 2011, 17:15
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If A and B are nonzero integers, is $$A^B$$ an integer?

(1) $$B^A$$ is negative
(2) $$A^B$$ is negative

Please explain the most efficient way to attack this question.
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Re: Is a^b an integer [#permalink]  28 Apr 2011, 19:45
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a B^A <0 => B is essentially negative and A is Odd number. For B=A = -1 and for B=-1 and A=3 the values are different.
b A^B <0 => A is essentially negative and B is an Odd number. For similar values the equation gives different outcomes.

for a+b, A=B= -1 and for A= -3 and B= -1 the values are different.Hence IMO E.

Under such condition we have to always check for A=B values and A> or <B values.
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Re: Is a^b an integer [#permalink]  28 Apr 2011, 19:48
The answer is E i guess

A. B^A is <0
that means that B<0 and A is odd
case 1 :
So consider A = -1 B is 3

(-1)^3 = -1 YES

Consider A=-3 and B=-2

(-3)^-2 = -1/9 NO Insufficient

B. A^B <0 A < 0 B is odd

so this is also insufficient as we can use the same values as above

combining both A,B <0 and A,B Odd numbers

so the case fails whenever A is -1 since -1 is also an odd number

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Re: Is a^b an integer [#permalink]  28 Apr 2011, 22:21
1
KUDOS
(1)
B^A = -ve

So B is -ve and A is odd (-ve or +ve)

(1) is insufficient, because if A is -ve, B^A may not be an integer.

(2)
A^B is -ve

So A is -ve and B is odd ( +ve or -ve)

(2) is insufficient as well

(1) and (2) say :

A and B are -ve and odd

So A^B may/may not be an integer

(-1)^-1 = -1

(-3)^-3 = not an integer

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Re: If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]  23 Dec 2014, 15:14
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Re: If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]  19 Jan 2015, 04:40
I guess positivity and negativity doesn't say much about a number being integer or a decimal. So you should go for E.
Re: If A and B are nonzero integers, is A^B an integer? (1) B^A   [#permalink] 19 Jan 2015, 04:40
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