Responding to a pm:

Question: If x and y are integers, does (x)^y * (y)^-x = 1?

(1) (x)^x > y

(2) x > (y)^y

Let's re-arrange the question first:

Is

(x)^y * (y)^{-x} = 1?

Is

(x)^y = (y)^x?

Check this post for a detailed discussion on this:

try-this-one-700-level-number-properties-103461.html#p805817So,

(x)^y = (y)^x when x = y or x and y take values 2,4 or -2,-4

Look at the statements now:

(1)

(x)^x > yWe know this relation is true for many random values of x and y e.g. x = 4, y = 5 etc. So the answer to the question is NO in this case.

(x)^y is not equal to

(y)^x.

But does it hold for any values which will make

(x)^y = (y)^x?

Yes it does! If x = y, x^x > y is true for say, x = y = 3. 3^3 is greater than 3. So x and y can take values which will give the answer YES.

Not sufficient.

(2)

x > (y)^yAgain, it holds for many random values of x and y e.g. x = 10, y = 2 etc. So the answer to the question is NO in this case.

But does it hold for any values which will make

(x)^y = (y)^x?

Let's see. If x = y, x cannot be greater than

y^y. Check for a few values to figure out the pattern.

If x = 4 and y = 2, x is not greater than

y^y.

Similarly, it doesn't work for x = -2, y = -4 and x = -4 and y = -2 since x will be negative while y^y will be positive.

Therefore, if

x > (y)^y,

(x)^y = (y)^x cannot hold for any values of x and y. Hence answer to the question stays NO.

Sufficient.

Answer (B).

good Q.. thanks for the explanation...