Bunuel wrote:

Official Solution:

(1) \(2^{x-y}=\sqrt[(x+y)]{16}\)

\(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\);

Equate the powers: \(x-y=\frac{4}{x+y}\);

\((x-y)(x+y)=4\).

Now, since both \(x\) and \(y\) are integers (and \(x+y \gt 0\)) then \(x-y=2\) and \(x+y=2\) so, \(x=2\) and \(y=0\). Therefore, \((x+y)^{xy}=2^0=1=odd\), so the answer to the question is No. Sufficient. (Note that \(x-y=1\) and \(x+y=4\) is not a valid scenario (solution), since in this case we would have that \(x=2.5\) and \(y=1.5\), which is not possible as given that both unknowns must be integers)

(2) \(2^x+3^y=\sqrt[(x+y)]{25}\). Obviously \(\sqrt[(x+y)]{25}\) must be an integer (since \(2^x+3^y=integer\)) and as \(x+y=integer\) then the only solution is \(\sqrt[(x+y)]{25}=\sqrt[2]{25}=5\) giving \(x+y=2\). So, \(2^x+3^y=5\). From that, two scenarios are possible:

A. \(x=2\) and \(y=0\) (notice that \(x+y=2\) holds true): \(2^x+3^y=2^2+3^0=5\), and in this case: \((x+y)^{xy}=2^0=1=odd\);

B. \(x=1\) and \(y=1\) (notice that \(x+y=2\) holds true): \(2^x+3^y=2^1+3^1=5\), and in this case: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A

Hi

BunuelCan you please explain how we arrive at the last fraction here? Thanks.

\(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\);

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