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16 Sep 2014, 01:01
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If \(m\) and \(n\) are positive even integers, which of the following must also be an even integer?
A. \((m + 1)^{|n|}\)
B. \((mn)^{|m - n|}\)
C. \(n^{|m|} * n\)
D. \((m-3)^n * (m-1)^2\)
E. \((m + 1)(m - 1)\)
A. \((m + 1)^{|n|}\)
B. \((mn)^{|m - n|}\)
C. \(n^{|m|} * n\)
D. \((m-3)^n * (m-1)^2\)
E. \((m + 1)(m - 1)\)
16 Sep 2014, 01:01
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Official Solution:
If \(m\) and \(n\) are positive even integers, which of the following must also be an even integer?
A. \((m + 1)^{|n|}\)
B. \((mn)^{|m - n|}\)
C. \(n^{|m|} * n\)
D. \((m-3)^n * (m-1)^2\)
E. \((m + 1)(m - 1)\)
Let's eliminate options A, D, and E right away. Since \(m\) and \(n\) are positive even integers, then:
\((m + 1)^{|n|} = odd^{positive \ even \ number} = odd\);
\((m - 3)^n * (m - 1)^2 = odd * odd = odd\);
\((m + 1)(m - 1) = odd * odd = odd\).
Option B, \((mn)^{|m - n|}\), will not be an even integer if \(m = n\). In this case, \((mn)^{|m - n|} = (positive \ even \ number)^0 = 1\), which is odd.
Thus, option C is the only one remaining: \(n^{|m|} * m = (positive \ even \ number)^{|positive \ even \ number|} * even = even * even = even.\)
Answer: C
If \(m\) and \(n\) are positive even integers, which of the following must also be an even integer?
A. \((m + 1)^{|n|}\)
B. \((mn)^{|m - n|}\)
C. \(n^{|m|} * n\)
D. \((m-3)^n * (m-1)^2\)
E. \((m + 1)(m - 1)\)
Let's eliminate options A, D, and E right away. Since \(m\) and \(n\) are positive even integers, then:
\((m + 1)^{|n|} = odd^{positive \ even \ number} = odd\);
\((m - 3)^n * (m - 1)^2 = odd * odd = odd\);
\((m + 1)(m - 1) = odd * odd = odd\).
Option B, \((mn)^{|m - n|}\), will not be an even integer if \(m = n\). In this case, \((mn)^{|m - n|} = (positive \ even \ number)^0 = 1\), which is odd.
Thus, option C is the only one remaining: \(n^{|m|} * m = (positive \ even \ number)^{|positive \ even \ number|} * even = even * even = even.\)
Answer: C
06 May 2023, 16:06
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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.
