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

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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.
[Reveal] Spoiler: OA
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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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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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10 Jun 2011, 09:24
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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

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10 Jun 2011, 09:39
My very first kudo! This is special... thank you fluke.
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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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24 Dec 2011, 18:52
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?????????

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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
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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
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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.

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07 Jun 2012, 21:40
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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07 Jun 2012, 21:59
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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07 Jun 2012, 22:55
yep i got now. i am just missing the x^y= 16 part.
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07 Jun 2012, 23:37
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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Re: If x and y are integers, is x^y<y^x ? [#permalink]

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