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16 Sep 2014, 01:26
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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) \(2^x+3^y=\sqrt[(x+y)]{25}\)
(1) \(2^{x-y}=\sqrt[(x+y)]{16}\)
(2) \(2^x+3^y=\sqrt[(x+y)]{25}\)
16 Sep 2014, 01:26
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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
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
General Discussion
07 Aug 2016, 02:52
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I think this is a high-quality question and I agree with explanation. Fantabulous question!
