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16 Sep 2014, 01:17
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If \(x\) and \(y\) are even integers, which of the following must also be an even integer?
A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)
A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)
16 Sep 2014, 01:17
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Official Solution:
If \(x\) and \(y\) are even integers, which of the following must also be an even integer?
A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)
Let's evaluate each option.
A. \(x^y\). If \(y = 0\) and \(x\) is a nonzero even number, then \(x^y\) will be 1, which is odd.
B. \(2*\frac{x}{y}\). If \(x = 2\) and \(y = 4\), then the expression equals 1, which is odd.
C. \(\frac{x - y}{x + y}\). This option might not even be an integer. For example, if \(x=2\) and \(y = 4\), the result is not an integer.
D. \(\frac{x^2 - y^2}{2} = \frac{(x - y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\). Therefore, this option is always even.
E. \((x + 1)(y - 1)\). Since \(x\) and \(y\) are even integers, \((x + 1)(y - 1)=odd*odd=odd\).
Thus, only option D must always be an even integer.
Answer: D
If \(x\) and \(y\) are even integers, which of the following must also be an even integer?
A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)
Let's evaluate each option.
A. \(x^y\). If \(y = 0\) and \(x\) is a nonzero even number, then \(x^y\) will be 1, which is odd.
B. \(2*\frac{x}{y}\). If \(x = 2\) and \(y = 4\), then the expression equals 1, which is odd.
C. \(\frac{x - y}{x + y}\). This option might not even be an integer. For example, if \(x=2\) and \(y = 4\), the result is not an integer.
D. \(\frac{x^2 - y^2}{2} = \frac{(x - y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\). Therefore, this option is always even.
E. \((x + 1)(y - 1)\). Since \(x\) and \(y\) are even integers, \((x + 1)(y - 1)=odd*odd=odd\).
Thus, only option D must always be an even integer.
Answer: D
18 Apr 2015, 16:16
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Here, if x=y (Question didn't say that x and y are different, so we can consider this use case as well), then option (D) becomes 0/2 =0, which is not an even integer. Can you please let me know your thoughts on this?
