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"

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Please let me know if I missed any case.

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~fluke

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