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

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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]

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25 Feb 2016, 23:46

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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08 Mar 2016, 10:20

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Chiragjordan wrote:

Hey MIKE Can you help with this one..

attempted twice.. got it wrong both times.. I choose A both the times...

Whenever you are quoting a user, use "@"before the correct username. I believe you want to ask inputs from mikemcgarry from Magoosh.

As for this question, it hinges on the observation that \(A^B\) will be <0 when A < 0 for B=odd. Thus for \(A^B\) to be an integer ---> A =\(\pm\) 1 and B can be any odd integer (\(\neq\) 0). Analyse the given statements in light of this information.

Per statement 1, \(B^A\)< 0 ---> The only possible case is B < 0 and A= odd. If B = -1, A = any power, you get a yes to the question asked but if B = -3 and A = 1, you get -1/3 = no for the question asked. Not sufficient.

Per statement 2, \(A^B\) < 0 ---> The only possible case is A<0 and B = odd. Same logic as that for statement 1. Not sufficient.

Combining, you get that A = B = odd negative integer and as such you get a yes if A=B=-1 but you get a NO for A=-3 and B = -1.

Re: If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]

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13 Mar 2016, 21:46

Engr2012 wrote:

Chiragjordan wrote:

Hey MIKE Can you help with this one..

attempted twice.. got it wrong both times.. I choose A both the times...

Whenever you are quoting a user, use "@"before the correct username. I believe you want to ask inputs from mikemcgarry from Magoosh.

As for this question, it hinges on the observation that \(A^B\) will be <0 when A < 0 for whatever value of B. Thus for \(A^B\) to be an integer ---> A =\(\pm\) 1 and B can be any integer (\(\neq\) 0). Analyse the given statements in light of this information.

Per statement 1, \(B^A\)< 0 ---> The only possible case is B < 0 and A= odd. If B = -1, A = any power, you get a yes to the question asked but if B = -3 and A = 1, you get -1/3 = no for the question asked. Not sufficient.

Per statement 2, \(A^B\) < 0 ---> The only possible case is A<0 and B = odd. Same logic as that for statement 1. Not sufficient.

Combining, you get that A = B = odd negative integer and as such you get a yes if A=B=-1 but you get a NO for A=-3 and B = -1.

Hence E is thus the correct answer.

Hope this helps.

Thank you so much for the explanation here is what i think i made the mistake=> In the first case i neglected A being 1 or -1 So combining the two statements => A can be -1 B=-21=> integer and A= anything but -1 ,B=anything => non integer.. Is this understanding correct? regards Also whats the point of tagging the name when they only respond when they want else they DON'T.. Regards Stone Cold Steve Austin
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If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]

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08 Apr 2016, 05:10

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If \(a\) and \(b\) are nonzero integers, is \(a^b\) an integer?

(1) \(b^a\) is negative

This can be true when \(b\) is negative integer(odd or even) and \(a\) is odd(negative or postive) If \(a=1\), then all cases of \(a^b\) is an integer If \(a=3\), None of the cases give an integer for \(a^b\)

Not sufficient

(2) \(a^b\) is negative

Negative can be an integer or decimal or real number as well.

for \(b=1\) & \(a=-3,-2\) we have some values of \(a^b\) as integer and some are not integer(decimal) values.

Thus insufficient.

Combining 1 and 2 we get both a and b as odd and negative integers Try the intended expression \(a^b\) with values (a,b) as (-1,-3) and (-3,-1). we get both integer and non integer values -3 and -0.333. Thus combining both the statements is also insufficient.

Ans E _________________

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Re: If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]

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02 May 2017, 15:31

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.

The KEY here is to pick numbers. Let's go:

1) \(b^{a}\) is negative - We know from this equation that a=odd AND b=negative -- TEST VALUES: > \(-3^{3}\) = -27. INTEGER. > \(-3^{-3}\) = \(\frac{-1}{27}\). NOT AN INTEGER. * KEY: b is negative, but a is just "odd"...this means a can be positive or negative, and this differentiator makes or breaks the problem

ELIMINATE A&D

\(a^{b}\) is negative - We know from this equation that a=negative AND b=odd -- TEST VALUES: > \(-5^{3}\) = -125. INTEGER. > \(-5^{-3}\) = \(\frac{-1}{125}\). NOT AN INTEGER. * KEY: a is negative, but b is just "odd"...this means b can be positive or negative, and this differentiator makes or breaks the problem

ELIMINATE B

Between C&E - We know from BOTH equations that BOTH a&b need to be ODD AND NEGATIVE. -- TEST VALUES: > \(-3^{-3}\) = \(\frac{-1}{27}\). NOT AN INTEGER. > \(-1^{-3}\) = 1. INTEGER.