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If \(n\) is a positive integer, is \(n\) even?
(1) \(n^{n + 1}\) is even.
(2) \((n + 1)^n\) is odd.
(1) \(n^{n + 1}\) is even.
(2) \((n + 1)^n\) is odd.
16 Sep 2014, 00:46
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Official Solution:
If \(n\) is a positive integer, is \(n\) even?
Keep in mind that since \(n\) is a positive integer, it must be either an even or odd positive integer.
(1) \(n^{n + 1}\) is even.
If \(n\) were a positive odd integer, then \(n^{n + 1}\) would be in the form \(odd^{even}\), which will be odd, not even. Therefore, \(n\) must be even. Sufficient.
(2) \((n + 1)^n\) is odd.
Similarly, here, if \(n\) were a positive odd integer, then \((n + 1)^n\) would be in the form \(even^{odd}\), which will be even, not odd. Therefore, \(n\) must be even. Sufficient.
Answer: D
If \(n\) is a positive integer, is \(n\) even?
Keep in mind that since \(n\) is a positive integer, it must be either an even or odd positive integer.
(1) \(n^{n + 1}\) is even.
If \(n\) were a positive odd integer, then \(n^{n + 1}\) would be in the form \(odd^{even}\), which will be odd, not even. Therefore, \(n\) must be even. Sufficient.
(2) \((n + 1)^n\) is odd.
Similarly, here, if \(n\) were a positive odd integer, then \((n + 1)^n\) would be in the form \(even^{odd}\), which will be even, not odd. Therefore, \(n\) must be even. Sufficient.
Answer: D
31 Dec 2021, 06:43
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I think this is a high-quality question and I agree with explanation. NICE ONE BUT I THINK ITS 600 LEVEL.
