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If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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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.
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Originally posted by p2bhokie on 12 Oct 2014, 13:14.
Last edited by Bunuel on 12 Oct 2014, 13:16, edited 1 time in total.
Renamed the topic and edited the question.



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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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12 Oct 2014, 13:15
Bunuel, can you please explain why isn't the correct ans 'A'? Thanks.



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Re: Negative Power. [#permalink]
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12 Oct 2014, 13:16
p2bhokie wrote: Bunuel, can you please explain why isn't the correct ans 'A'? Thanks. Sorry forgot to copy you VeritasPrepKarishma



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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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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. P.S. Please name topics properly. Rule 3 here: rulesforpostingpleasereadthisbeforeposting133935.html
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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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12 Oct 2014, 13:42
If an ≠ 0 and n is a positive integer, is n odd? (1) a^n + a^(n + 1) < 0 (2) a is an integer. Dear p2bhokie, Please find the reasoning below (1) If A lies between 1 and 0, then the mentioned equation will be negative for all odd values of N and positive for even values of N. We'll concentrate on negative values. Lets directly assign values to a and nA = 0.1 N = 1 = (.1)^1 + (.1)^2 = .1 + .01 =  .09 If A is lesser than 1, then the equation will be negative only for Even values of NA = 2 N = 2 = (2)^2 + (2)^3 = 4  8 = 4 Thus (1) alone will not be able to answer whether N is odd. Considering (2) If A is an integer, there will be various combinations of A and N to produce a negative value Combining 1 & 2 will eliminate the possibility of A being a number between 0 and 1. Thus this combination will be suffice to answer. Hope it helps you.
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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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02 May 2016, 09:05
p2bhokie wrote: If an ≠ 0 and n is a positive integer, is n odd? (1) a^n + a^(n + 1) < 0 (2) a is an integer. The solution is attached.
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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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02 May 2016, 22:49
p2bhokie wrote: If an ≠ 0 and n is a positive integer, is n odd? (1) a^n + a^(n + 1) < 0 (2) a is an integer. Here is my thoughtprocess on this question: a and n both are not 0. n is a positive integer. Is n odd? (1) a^n + a^(n + 1) < 0 The sum of powers of a is negative. So a has to be negative. The powers are n and (n+1). One of them will be odd, the other even. The even power will make the term positive so the term with the odd power should be the greater absolute value term (so that the sum is negative). So (n+1) should be odd and hence n should be even. But wait, stmnt 2 makes you wonder whether you should consider fractions. If a is a fraction, then the term with the smaller power will have greater absolute value so that should be negative. In that case, n should be odd and n+1 should be even. Hence this statement alone is not sufficient. (2) a is an integer. This statement alone is definitely not sufficient. Using both, \(a^{n+1}\) should be the greater absolute value term and hence should be negative. So n+1 should be odd which makes n even. Sufficient. Answer (C)
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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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04 May 2016, 07:38
There are 2 variables (a and n) in the original condition. In order to match the number of variables to the number of equations, we need 2 equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer choice. Using both the condition 1) and the condition 2), we get n=even and n+1=odd. The answer becomes no and the conditions are sufficient. Thus, the correct answer is C.  For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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23 Dec 2016, 22:31
Bunuel wrote: 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. P.S. Please name topics properly. Rule 3 here: rulesforpostingpleasereadthisbeforeposting133935.html Thanks a lot for this explaination. I had a query that when a^n(1+a)<0 this means either a^n<0 or 1+a<0 hence a<1 doesn't this eliminate the possibility of a lying in the range 1<n<0????? Please help.



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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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24 Dec 2016, 00:50
AmritaSarkar89 wrote: Bunuel wrote: 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. P.S. Please name topics properly. Rule 3 here: rulesforpostingpleasereadthisbeforeposting133935.html Thanks a lot for this explaination. I had a query that when a^n(1+a)<0 this means either a^n<0 or 1+a<0 hence a<1 doesn't this eliminate the possibility of a lying in the range 1<n<0????? Please help. \(a^n(1+a)< 0\) means that there can be two cases: \(a^n > 0\) and \(1+a<0\) OR \(a^n < 0\) and \(1+a>0\)
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Re: If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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24 Dec 2016, 01:31
Quality Question.
Here is my solution to this one =>
Given data > a*n≠0> Both are non zero numbers n=> positive integer a=> integer or non integer
We need to see if n is odd or not.
Statement 1>
a^n+a^n+1<0
Hence a must be negative
If a=> (∞,1) => n must be even so that n+1 will be odd If a=>(1,0)=> n must be odd so that n+1 will be even
Hence not sufficient
Statement 2> a is an integer No clue of n Hence not sufficient
Combining the two statements > a<0 and an integer > a can never be 1 Hence a<1 Thus n must be even
Hence sufficient
Hence C
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If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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24 Dec 2016, 05:26
VeritasPrepKarishma wrote: p2bhokie wrote: If an ≠ 0 and n is a positive integer, is n odd? (1) a^n + a^(n + 1) < 0 (2) a is an integer. Here is my thoughtprocess on this question: a and n both are not 0. n is a positive integer. Is n odd? (1) a^n + a^(n + 1) < 0 The sum of powers of a is negative. So a has to be negative. The powers are n and (n+1). One of them will be odd, the other even. The even power will make the term positive so the term with the odd power should be the greater absolute value term (so that the sum is negative). So (n+1) should be odd and hence n should be even. But wait, stmnt 2 makes you wonder whether you should consider fractions. If a is a fraction, then the term with the smaller power will have greater absolute value so that should be negative. In that case, n should be odd and n+1 should be even. Hence this statement alone is not sufficient. (2) a is an integer. This statement alone is definitely not sufficient. Using both, \(a^{n+1}\) should be the greater absolute value term and hence should be negative. So n+1 should be odd which makes n even. Sufficient. Answer (C) Hello Karishma mam, If statement1 changes to a^n + a^(n + 1) > 0?
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If an ≠ 0 and n is a positive integer, is n odd? [#permalink]
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29 Dec 2016, 01:15
understandZERO wrote: VeritasPrepKarishma wrote: p2bhokie wrote: If an ≠ 0 and n is a positive integer, is n odd? (1) a^n + a^(n + 1) < 0 (2) a is an integer. Here is my thoughtprocess on this question: a and n both are not 0. n is a positive integer. Is n odd? (1) a^n + a^(n + 1) < 0 The sum of powers of a is negative. So a has to be negative. The powers are n and (n+1). One of them will be odd, the other even. The even power will make the term positive so the term with the odd power should be the greater absolute value term (so that the sum is negative). So (n+1) should be odd and hence n should be even. But wait, stmnt 2 makes you wonder whether you should consider fractions. If a is a fraction, then the term with the smaller power will have greater absolute value so that should be negative. In that case, n should be odd and n+1 should be even. Hence this statement alone is not sufficient. (2) a is an integer. This statement alone is definitely not sufficient. Using both, \(a^{n+1}\) should be the greater absolute value term and hence should be negative. So n+1 should be odd which makes n even. Sufficient. Answer (C) Hello Karishma mam, If statement1 changes to a^n + a^(n + 1) > 0? Hi. See if You change the statements to a^n + a^(n + 1) > 0 > N can be either even or odd for a = positive integer. But beware that a^n + a^(n + 1) > 0 does not mean that a must be positive. a can be negative too E.g a=10 and n=9
The answer to your Question would be E as a can be positive or negative and n can be even/odd.
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