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26 May 2017, 04:43
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When m, n, and p are positive integers, is \(m^{2}+n^{3}+p^{5 }\) divisible by \(2\)?
(1) \(m = n + p\)
(2) \(m\), \(n\), and \(p\) are 3 lengths of a certain right triangle and \(m < n < p\).
(1) \(m = n + p\)
(2) \(m\), \(n\), and \(p\) are 3 lengths of a certain right triangle and \(m < n < p\).
26 May 2017, 04:43
Kudos
Bookmarks
Official Solution:
You can take the 1st step of the variable approach and modify the original condition and the question. Then, "is \(m^{2}+n^{3}+p^{5 }\) divisible by 2?" is asking "is \(m^{2}+n^{3}+p^{5}=even\)?"
In the original condition, there are 3 variables \((m, n, p)\) and in order to match the number of variables to the number of equations, there must be 3 equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2),
In the case of con 1), you get \((m, n, p)=(even, odd, odd)\), \((even, even, even)\), or \((odd, even, odd)\). Thus, you get \(m^{2}+n^{3}+p^{5} = even\) in any cases, hence yes, it is sufficient.
In the case of con 2), according to the Pythagorean theorem, from \(m^{2}+n^{2}=p^{2}\), you get \((m, n, p) = (even, odd, odd), (even, even, even),\) or \((odd, even, odd)\). Thus, you get \(m^{2}+n^{3}+p^{5} = even\) in any cases, hence yes, it is sufficient. Therefore, you get con 1) = con 2). D is the answer.
Answer: D
You can take the 1st step of the variable approach and modify the original condition and the question. Then, "is \(m^{2}+n^{3}+p^{5 }\) divisible by 2?" is asking "is \(m^{2}+n^{3}+p^{5}=even\)?"
In the original condition, there are 3 variables \((m, n, p)\) and in order to match the number of variables to the number of equations, there must be 3 equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2),
In the case of con 1), you get \((m, n, p)=(even, odd, odd)\), \((even, even, even)\), or \((odd, even, odd)\). Thus, you get \(m^{2}+n^{3}+p^{5} = even\) in any cases, hence yes, it is sufficient.
In the case of con 2), according to the Pythagorean theorem, from \(m^{2}+n^{2}=p^{2}\), you get \((m, n, p) = (even, odd, odd), (even, even, even),\) or \((odd, even, odd)\). Thus, you get \(m^{2}+n^{3}+p^{5} = even\) in any cases, hence yes, it is sufficient. Therefore, you get con 1) = con 2). D is the answer.
Answer: D
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