Bunuel wrote:

Official Solution:

(1) \(x^y=\frac{1}{81}\). As both \(x\) and \(y\) are negative integers then \(x^y=\frac{1}{81}=(-9)^{-2}=(-3)^{-4}\) hence \(xy=18\) or \(xy=12\). Note that as negative integer (\(x\)) in negative integer power (\(y\)) gives positive number (\(\frac{1}{81}\)), then the power must be negative even number. Not sufficient.

(2) \(y^x=-\frac{1}{64}\). As the result is negative then \(x\) must be negative odd number, so: \(y^x=-\frac{1}{64}=(-4)^{-3}=(-64)^{-1}\) hence \(xy=12\) or \(xy=64\). Not sufficient.

(1)+(2) Only one pair of negative integers \(x\) and \(y\) satisfies both statements \(x=-3\) and \(y=-4\). Therefore \(xy=12\). Sufficient.

Answer: C

Very good question.

In statement 2, I forget the SPECIAL CASE in which x = -1.

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