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If a^n ≠ 0 and n is a positive integer, is n odd?

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If a^n ≠ 0 and n is a positive integer, is n odd?  [#permalink]

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New post 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?  [#permalink]

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New post 19 Oct 2013, 13:42
1
1
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?  [#permalink]

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New post 19 Oct 2013, 20:11
2
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?  [#permalink]

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New post 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?  [#permalink]

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New post 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?  [#permalink]

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Re: If a^n ≠ 0 and n is a positive integer, is n odd?   [#permalink] 30 Mar 2018, 23:37
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