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# If x and y are both integers, which is larger, x^x or y^y?

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Senior Manager
Joined: 08 Nov 2010
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If x and y are both integers, which is larger, x^x or y^y? [#permalink]

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26 Aug 2011, 15:09
00:00

Difficulty:

55% (hard)

Question Stats:

58% (00:56) correct 42% (01:03) wrong based on 36 sessions

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If x and y are both integers, which is larger, x^x or y^y?

1. x = y + 1
2. x^y > x and x is positive.
[Reveal] Spoiler: OA

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Kudos [?]: 133 [0], given: 161

Manager
Joined: 14 Mar 2011
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27 Aug 2011, 10:12
1
KUDOS
If x and y are both integers, which is larger, x^x or y^y?

1. x = y + 1 is not sufficient as it can not determined which one is larger x^x or y^y
2. x^y > x and x is positive. we know x is +ve but we not sure about value of y it -ve or +ve

if we combine both we get x ^y > x & x is + positive integer we imply x > 1 so x^x is alwys greater y^y

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Senior Manager
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26 Aug 2011, 19:21
22 views and no reply? Can someone help me out here. I can solve it by trials and error, but I am looking for a better way. Thanks.
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Kudos [?]: 133 [0], given: 161

Manager
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26 Aug 2011, 19:44
which values did you compute...wont x be always greater in statement 1..

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Senior Manager
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27 Aug 2011, 03:34
Take for example the case of y=-2, x=-1

(-1)^-1 = -1
And
(-2)^-2 = 1/4

So Y^y is bigger...
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27 Aug 2011, 04:53
144144 wrote:
If x and y are both integers, which is larger, x^x or y^y?

1. x = y + 1
2. x^y > x and x is positive.

Difficult. I can't solve it algebraically. I did it by substitution as well:

It is apparent that if both x and y were greater than 1, 1 would be sufficient. However, this is not so; thus we'll have to consider both +ves and -ves.

1. x=y+1
x=2, y=1; x^x>y^y
x=-1, y=-2; y^y>x^x
Not Sufficient.

2. x^y > x for x>0
Now, for y<=1, the above statement won't hold good. Also, x must be greater than 1 to make the expression valid, because 1^(anything>=0)=1 and y>1.

x=9,y=10, y^y>x^x
x=10,y=9, x^x>y^y
Not Sufficient.

Combining both statements, we know
x>1 and y>1; and x>y
Sufficient.

Ans: "C"
***********************************

Please let me know if I missed any case.
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27 Aug 2011, 20:35
stmt 1 - clearly insufficient

stmt 2 can be solved to be

x^(y-1) Multiplied by (x-1) > 0 indicating both x-1 and x^(y-1) are the same sign.

Now both will be positive as x>0 is given - indicates x>1

but insuff as nothing is known about y

combining it with stmt 1 = we get x>y>0 and x>1 which is sufficent for the question.
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Re: X and Y   [#permalink] 27 Aug 2011, 20:35
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