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If A and B are nonzero integers, is A^B an integer? (1) B^A [#permalink]
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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]
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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]
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28 Apr 2011, 19:48
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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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28 Apr 2011, 22:21
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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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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.



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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. 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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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 negativeThis 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 negativeNegative 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 2we 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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08 Apr 2016, 13:37
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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08 Apr 2016, 15:38
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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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 problemELIMINATE 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 problemELIMINATE BBetween 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. HENCE, CORRECT ANSWER = EKudos please if you find this helpful




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