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If x and y are integers, is x^y<y^x ? [#permalink]
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10 Jun 2011, 07:27
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If x and y are integers, is x^y<y^x ? (1) x^y= 16 (2) x and y are consecutive even integers.
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Re: Another exponent question [#permalink]
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10 Jun 2011, 07:42
I would say that the asnwer is A.
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?



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Re: Another exponent question [#permalink]
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10 Jun 2011, 09:09
jaizen wrote: I would say that the asnwer is A.
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
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Re: Another exponent question [#permalink]
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10 Jun 2011, 09:24
fluke wrote: jaizen wrote: I would say that the asnwer is A.
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 The answer still holds.



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Re: Another exponent question [#permalink]
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10 Jun 2011, 09:39
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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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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.
please let me know if i'm missing something.



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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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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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Updated on: 03 Jan 2012, 19:36
Guys initially marked but now confused..... 1. it will satisfy for both x=4, y=2 and x=2 y=4 ....the value is not greater But for x=16 y=1 X^Y>Y^X so we are getting two answers....so then how can the answer be A????????? Can anyone please explain????
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Originally posted by mydreammba on 24 Dec 2011, 18:52.
Last edited by mydreammba on 03 Jan 2012, 19:36, edited 1 time in total.



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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, 11:26
I feel it is c. Because different values of X and Y are giving different relations.
Lets se, X=2, Y=4 then x^y=y^x
if we have x=16, y=1, then x^y > y^x
So, I feel C is correct answer.



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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
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.
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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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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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Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]
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18 Jan 2012, 12:37
If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x and y are consecutive even integers.
(1) x^y = 16 x y can be 2 4 4 2 16 1
not sufficient.
(2) x and y are consecutive even integers. we don't know whether x > y or y > x.
not sufficient
1 + 2
x y can be 2 4 => x^y = 16 and y^x = 16 4 2 => x^y = 16 and y^x = 16 0 2 => x^y = 0 and y^x = 1 2 4 => x^y = 1/16 and y^x = 1/16
not sufficient.



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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
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. Answer: A.
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If x and y are integers, is x^y < y^x ? (1) x^y = 16 (2) x and y are consecutive even integers. according to statement 1 the possible pair of x and y would be (16, 1), (1, 16), (2,4), (4,2) 2^4=4^2 in this case the answer is no if x= 1 then 1^16<16^1 in this case the answer is yes if x=16 then 16^1>1^16 in this case the answer is No so statement 1 is insufficient . am i right or am i missing something?
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alchemist009 wrote:
according to statement 1 the possible pair of x and y would be (16, 1), (1, 16), (2,4), (4,2)
You are wrong: For the statement 1, the possible is (2,4) or (4,2) only. 1^16=1 not 16.



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yep i got now. i am just missing the x^y= 16 part.
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Hi,
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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Re: If x and y are integers, is x^y<y^x ? [#permalink]
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02 Nov 2017, 00:59
The answer must be C because the two of the conditions derived from the first equation is sufficient when x=2 or 4. But what about the value of x as 16. Please correct the answer.



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Re: If x and y are integers, is x^y<y^x ? [#permalink]
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02 Nov 2017, 01:23




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