M26-37

Official Solution:


If \(x\) and \(y\) are non-negative integers and \(x+y \gt 1\), is \((x+y)^{xy}\) an even integer?

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

(2) \(2^x+3^y=\sqrt[(x+y)]{25}\).

Since \(2^x+3^y\) must be an integer, \(\sqrt[(x+y)]{25}\) must be an integer as well. The only possible value for \(\sqrt[(x+y)]{25}\) is \(5\), which implies that \(x+y=2\). Thus, \(2^x+3^y=5\). From this equation, we have two possible scenarios:

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: \((x+y)^{xy}=2^0=1\), which is 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: \((x+y)^{xy}=2^1=2\), which is even.

Since the two scenarios give different answers, statement (2) is not sufficient.


Answer: A
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Great question, but quite difficult to solve it under 2 mins

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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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