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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]

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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.
[Reveal] Spoiler: OA
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Re: Is a^b an integer [#permalink]

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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]

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New post 28 Apr 2011, 20:48
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

E is the answer.
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Re: Is a^b an integer [#permalink]

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New post 28 Apr 2011, 23:21
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(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

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

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New post 19 Jan 2015, 05:40
I guess positivity and negativity doesn't say much about a number being integer or a decimal. So you should go for 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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Re: If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]

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New post 08 Mar 2016, 11:03
Hey MIKE Can you help with this one..

attempted twice.. got it wrong both times..
I choose A both the times...
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If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]

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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.

Hence E is thus the correct answer.

Hope this helps.
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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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New post 13 Mar 2016, 22:46
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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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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New post 08 Apr 2016, 14:37
I cant understand why people are assuming that both premise A or B(eg B^A i negative) talking about integers

Option A says B^A is negative, it doesnt say that B^A is neg interger so why all are assuming it to be.

Answer should be A since we can find out the sign of B will be neg so A^B will never be an integer.

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

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New post 08 Apr 2016, 16:38
varundixitmro2512 wrote:
I cant understand why people are assuming that both premise A or B(eg B^A i negative) talking about integers

Option A says B^A is negative, it doesnt say that B^A is neg interger so why all are assuming it to be.

Answer should be A since we can find out the sign of B will be neg so A^B will never be an integer.

Statement 1 is suff


No one is ASSUMING anything. Refer to the solution mcp.php?i=main&mode=post_details&f=141&p=1656135 that clearly uses 2 distinct cases. Alternately, look at the following 2 cases:

Case 1: B=-1 and A = 3 ---> B^A < 0 and A^B \(\neq\) integer but

Case 2: B=-2 and A = -1 ---> B^A < 0 and A^B = integer

Thus you clearly get 2 different answers for the question asked --> Is A^B an integer ---> Thus this statement is NOT sufficient.

Hope this helps.
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If A and B are nonzero integers, is A^B an integer? (1) B^A   [#permalink] 08 Apr 2016, 16:38
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If A and B are nonzero integers, is A^B an integer? (1) B^A

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