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If a^n ≠ 0 and n is a positive integer, is n odd?
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Updated on: 20 Oct 2013, 04:43
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If a^n ≠ 0 and n is a positive integer, is n odd? (1) a^n + a^(n+1) < 0 (2) a is an integer.
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Originally posted by jellybean23 on 19 Oct 2013, 12:14.
Last edited by Bunuel on 20 Oct 2013, 04:43, edited 1 time in total.
Edited the question and added the OA.



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Re: If a^n ≠ 0 and n is a positive integer, is n odd?
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19 Oct 2013, 13:42
Statement 1:
This essentially says \(a\) is negative and thereby one of the terms \(a^n\) or \(a^n^+^1\) is negative (both can't be negative as either \(n\) is even or \(n+1\) is even)
Let's assign \(a=2\). So \(a^n + a^n^+^1 < 0\) only when \(n\) is even. But what happens if \(a\) is not an integer? Assign \(a=0.1\), then for the equation to be less than zero, \(n\) has to be odd. INSUFFICIENT
Statement 2:
This tells us that \(a\) is an integer, but does not say anything about \(n\). INSUFFICIENT
Statement 1 & 2: as discussed under Statement 1, when \(a\) is a negative integer, \(n\) has to be even for the equation to be less than zero. So we know \(n\) is not odd. SUFFICIENT
Answer is C.



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Re: If a^n ≠ 0 and n is a positive integer, is n odd?
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19 Oct 2013, 20:11
n >0 Statement 1 : a^n + a^(n+1) < 0 lets put n =1 (odd) , a+a^2 <0 .................. , that means 0 > a > 1, lets put n=2 (even) , a^2 + a^3 < 0 ........... , that means a < 0 but no info about a so no conclusion about n.. Insufficient..
Statement 2 : a is an integer , but no conclusion about n... insufficient...
Taking both statements together.... only n=2 valid .... which gives n is not an odd no.. Answer C



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Re: If a^n ≠ 0 and n is a positive integer, is n odd?
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19 Oct 2013, 21:50
+1 Kudos !!!! great explanation by both of you. Thanks
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Re: If a^n ≠ 0 and n is a positive integer, is n odd?
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20 Oct 2013, 04:54
If a^n ≠ 0 and n is a positive integer, is n odd?a^n ≠ 0 and n is a positive integer implies that a ≠ 0. (1) a^n + a^(n+1) < 0 > \((a+1)a^n<0\). Two cases: \(a+1>0\) and \(a^n<0\). From first inequality we have that \(a>1\). If a is a negative number (say 1/2), then \((negative)^n\) to be negative n must be odd. \(a+1<0\) and \(a^n>0\). From first inequality we have that \(a<1\), so a is a negative number. Now, \((negative)^n\) to be positive n must be even. Not sufficient. (2) a is an integer. Clearly insufficient. (1)+(2) Consider the same two cases but now take into account that a is an integer: \(a+1>0\) and \(a^n<0\). From first inequality we have that \(a>1\). Since we know that \(a\neq{0}\), then a must be a positive integer (1, 2, 3, ...). Next, \((positive)^n\) cannot be negative, thus this case is out. \(a+1<0\) and \(a^n>0\). From first inequality we have that \(a<1\), so a is a negative integer. Now, \((negative)^n\) to be positive n must be even. Sufficient. Answer: C. Hope it's clear.
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Re: If a^n ≠ 0 and n is a positive integer, is n odd?
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30 Mar 2018, 23:37
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