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16 Sep 2014, 01:20
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If \(a\) and \(b\) are negative integers, is \(a^b\) negative?
(1) \(ab\) is even.
(2) \(a + b\) is even.
(1) \(ab\) is even.
(2) \(a + b\) is even.
16 Sep 2014, 01:20
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Official Solution:
If \(a\) and \(b\) are negative integers, is \(a^b\) negative?
(1) \(ab\) is even.
If \(a=b=-2\), then \(a^b=\frac{1}{4} \gt 0\), but if \(a=-2\) and \(b=-1\), then \(a^b=-\frac{1}{2} \lt 0\). Not sufficient.
(2) \(a+b\) is even.
If \(a=b=-2\), then \(a^b=\frac{1}{4} \gt 0\), but if \(a=-1\) and \(b=-1\), then \(a^b=-1 \lt 0\). Not sufficient.
(1)+(2) From \(a+b\) being even, it follows that \(a\) and \(b\) are either both odd or both even. Now, if both are odd, then \(ab\) being even won't hold true, so both \(a\) and \(b\) must be even. Hence, \(a^b\) will be the result of a negative number raised to an even power, \(a^b=negative^{even}\), which is positive. Sufficient.
Answer: C
If \(a\) and \(b\) are negative integers, is \(a^b\) negative?
(1) \(ab\) is even.
If \(a=b=-2\), then \(a^b=\frac{1}{4} \gt 0\), but if \(a=-2\) and \(b=-1\), then \(a^b=-\frac{1}{2} \lt 0\). Not sufficient.
(2) \(a+b\) is even.
If \(a=b=-2\), then \(a^b=\frac{1}{4} \gt 0\), but if \(a=-1\) and \(b=-1\), then \(a^b=-1 \lt 0\). Not sufficient.
(1)+(2) From \(a+b\) being even, it follows that \(a\) and \(b\) are either both odd or both even. Now, if both are odd, then \(ab\) being even won't hold true, so both \(a\) and \(b\) must be even. Hence, \(a^b\) will be the result of a negative number raised to an even power, \(a^b=negative^{even}\), which is positive. Sufficient.
Answer: C
