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From 1, we can say have the following combinations. x=2, y=4. Ans: Not greater x=4, y=2. Ans: Not greater x=16, y=1. Ans: Not greater

All three possibilities of combination gives us a consistent answer as 'No'. Therefore A is sufficient.

What is the OA?

What if: x=-2, y=4 x=-4, y=2

I forgot to include signs. Well, since (1) says that the x^y = 16, this means that y has to be a positive, even number. So we can add two more combinations. x=-4, y=2 x=-2, y=4

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

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24 Dec 2011, 10:45

guys, i was looking at this again.. and the OA seems debatable.

the question does not say if x<y or not.

so for 1, if x=16 and y = 1 then \(x^y\) = 16 and \(y^x\) = 1. then \(x^y\) > \(y^x\) but if y=16 and x =1 then \(x^y\) = 1 and \(y^x\) = 16. and \(x^y\) < \(y^x\).

so A is not sufficient.

combining the 2 statements, the only possibility will be (x,y) = (4,2) and (2,4). in both cases however, \(x^y\) = \(y^x\) and the ans to the question will be a No. So i think the ans to this question should be C.

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

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24 Dec 2011, 12:44

\(\)

mmm..even i was confused, a bit.. X^Y can be -4^2, -2^4, 2^4 or 16^1 but in all cases above y^X is smaller than 16...so i guess its A In DS problem any concrete answer YES or NO can be an answer so i guess we do have an answer here.. _________________

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Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

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01 Jan 2012, 12:32

1

This post received KUDOS

dreambeliever wrote:

guys, i was looking at this again.. and the OA seems debatable.

the question does not say if x<y or not.

so for 1, if x=16 and y = 1 then \(x^y\) = 16 and \(y^x\) = 1. then \(x^y\) > \(y^x\) but if y=16 and x =1 then \(x^y\) = 1 and \(y^x\) = 16. and \(x^y\) < \(y^x\).

so A is not sufficient.

combining the 2 statements, the only possibility will be (x,y) = (4,2) and (2,4). in both cases however, \(x^y\) = \(y^x\) and the ans to the question will be a No. So i think the ans to this question should be C.

please let me know if i'm missing something.

you dont need to know whether x>y.

there are 3 possibilities for statement 1. x=2 y =4 x=4 y=2 x=16 y=1 for all three the answer to the question is NO. so it is sufficient. _________________

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

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07 Jan 2012, 16:21

1. x^y = 16. Only possible relationship is x=4,y=2. In this case x^y=y^x. Suff. 2. Say x=2, y=3, we get 2^3<3^2. Say x=3, y=4, we get 3^4>4^3. Insuff.

A _________________

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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If x and y are integers, is x^y<y^x? (1) x^y=16 [#permalink]

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18 Jan 2012, 13:54

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If x and y are integers, is x^y<y^x?

We have an YES/NO data sufficiency question. In a Yes/No Data Sufficiency question, each statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".

(1) \(x^y=16\), since \(x\) and \(y\) are integers then following cases are possible: \(x=-4\) and \(y=2\) --> \(x^y=16>\frac{1}{16}=y^x\) --> the answer to the question is NO; \(x=-2\) and \(y=4\) --> \(x^y=16>\frac{1}{16}=y^x\) --> the answer to the question is NO; \(x=2\) and \(y=4\) --> \(x^y=16=y^x\) --> the answer to the question is NO; \(x=4\) and \(y=2\) --> \(x^y=16=y^x\) --> the answer to the question is NO; \(x=16\) and \(y=1\) --> \(x^y=16>1=y^x\) --> the answer to the question is NO.

As you can see in ALL 5 possible cases the answer to the question "is \(x^y<y^x\)?" is NO. Thus this statement is sufficient.

(2) x and y are consecutive even integers --> if \(x=2\) and \(y=4\) the answer will be NO but if \(x=0\) and \(y=2\) the answer will be YES. Not sufficient.

Since it is mentioned that x and y are integers. \(x^y = 16\) gives following solution, (x,y) = (-4,2), (-2,4), (2,4), (4,2) & (16,1)

to check \(x^y < y^x\) using above values of (x,y) \(16 = (-4)^2\) \(16 = (-2)^4\) \(16 = (2)^4\) \(16 = (4)^2\) \(16 = (16)^1\)

Although the answer is same (i.e, (A)), but I want to emphasis on the fact that all the cases should be considered. This would help in other DS questions.

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