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If an ≠ 0 and n is a positive integer, is n odd?
(1) a^n + a^(n + 1) < 0
(2) a is an integer.
(1) a^n + a^(n + 1) < 0
(2) a is an integer.
12 Oct 2014, 13:35
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If an ≠ 0 and n is a positive integer, is n odd?
(1) a^n + a^(n + 1) < 0.
First of all, notice that this statement implies that a must be a negative number (if it's not, then a^n + a^(n + 1) = positive + positive > 0). Next, work on the given inequality: \(a^n + a^{(n + 1)} < 0\) --> \(a^n(1+a)< 0\).
If \(-1<a<0\), then \(1+a=positive\) and that would mean that \(a^n=(negative)^n\) must be negative, which can happen if n is odd.
If \(a<-1\), then \(1+a=negative\) and that would mean that \(a^n=(negative)^n\) must be positive, which can happen if n is even.
Not sufficient.
(2) a is an integer. Clearly insufficient.
(1)+(2) From above we have that a is a negative integer less than -1 (it cannot be -1 because in this case inequality from the first statement won't hold true), which means that n must be even. Sufficient.
Answer: C.
Hope it's clear.
(1) a^n + a^(n + 1) < 0.
First of all, notice that this statement implies that a must be a negative number (if it's not, then a^n + a^(n + 1) = positive + positive > 0). Next, work on the given inequality: \(a^n + a^{(n + 1)} < 0\) --> \(a^n(1+a)< 0\).
If \(-1<a<0\), then \(1+a=positive\) and that would mean that \(a^n=(negative)^n\) must be negative, which can happen if n is odd.
If \(a<-1\), then \(1+a=negative\) and that would mean that \(a^n=(negative)^n\) must be positive, which can happen if n is even.
Not sufficient.
(2) a is an integer. Clearly insufficient.
(1)+(2) From above we have that a is a negative integer less than -1 (it cannot be -1 because in this case inequality from the first statement won't hold true), which means that n must be even. Sufficient.
Answer: C.
Hope it's clear.
P.S. Please name topics properly. Rule 3 here: rules-for-posting-please-read-this-before-posting-133935.html
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