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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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23 Aug 2012, 02:11

4

This post was BOOKMARKED

Am not srue as to how to substitute values and slove this. Am writing the algebric solution to this problems

If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n a^(n-1) - a^n > 0 a^(n-1) (1 -a) >0 Either both a^(n-1) and (1-a) are > 0 or both are < 0 taking the case in which both are > 0 a^(n-1) >0 1-a> 0 => a<1 and a^(n-1) > 0 so, only possible solution to this is a < 0 and n-1 = even => n = odd

taking the case in which both are < 0 a^(n-1) < 0 and 1-a < 0 => a > 1 There is no solution to thsi equation as a^(n-1) for a > 1 will always be > 0

So, SUFFICIENT!

(2) a^n > a^(3n) a^n - a^3n >0 a^n(1-a^2n) > 0 Either both a^n and (1-a^2n) are > 0 or both are < 0 taking the case when both are > 0 a^n > 0 , 1 -a^2n > 0 => a^2n < 1 and a^n > 0 This equation is true for negative a and negative even values of n, But n > 1 so, NO solution!

taking the case when both are < 0 a^n < 0 , 1 -a^2n < 0 => a^2n > 1 and a^n < 0 This equation is true for negative values of a and odd values of n, So n is odd, SUFFICIENT

If you factor, as in the post above, you can avoid cases if you arrange things differently. Rewriting Statement 1, noticing that a cannot be 1, we can factor out the smaller power, so can factor out a^(n-1) :

0 > a^n - a^(n-1) 0 > a^(n-1) (a - 1)

Now one term on the right side must be positive, and the other must be negative. But if a is a positive integer, that can't possibly be true -- both terms will be positive. So a must be negative, and (a-1) must be the negative term, and a^(n-1) must be the positive term. Since a is negative, that means the exponent must be even, so n-1 is even and n is odd.

Similarly for Statement 2, notice a cannot be -1, 0 or 1, and we have

0 > a^(3n) - a^n 0 > a^n [a^(2n) - 1]

and again, one term must be positive, the other negative. But a^(2n) has an even exponent, so it must be positive and so a^(2n) - 1 is the positive term (using the fact that a cannot be 0, 1 or -1). That makes a^n the negative term. But if a^n is negative, a must be negative, and n must be odd.

I don't think you need to do the problem algebraically, however. You can just imagine how different numbers would 'behave' in the inequalities given. So if you know that n is a positive integer greater than 1, and you know

a^(n-1) > a^n

then you might be able to see that a must be negative, since this inequality won't ever be true if a > 1. And if we know the base is negative, and we can see that one exponent is even and the other is odd (since the exponents are just consecutive integers), we know that one side of the inequality will become positive, and the other will remain negative. Clearly the larger side needs to be positive, which makes n-1 even, and n odd.

Similarly for the other Statement, the inequality

a^n > a^(3n)

simply cannot be true if a > 1. Since 1, 0 or -1 don't work in this inequality, a must be -2 or smaller. But then if n is even, it wouldn't matter if a is negative or positive, since even exponents just 'erase' our negative signs. And we know it must matter, since the inequality won't work if we have positive bases on either side. So the inequality will only work if n is odd so that the left and right sides both remain negative.
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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04 Oct 2012, 03:06

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conty911 wrote:

If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n (2) a^n > a^(3n)

Please provide me pointers to any similar questions, also if possible

Can some one explain steps to tackle such problems with least amount of time.For eg. how to select the values for plugging etc. THanks

From the conditions in both statements, we can deduce that \(a\) cannot be neither \(0\) nor \(1.\) A little algebraic manipulation can help in understanding the inequalities and also in choosing values for \(a\). Usually it is easier to compare a product of numbers to \(0\), so it is worth rearranging the inequalities such that there is a comparison to \(0\) involved.

(1) \(a^{n-1}>a^n\) can be rewritten as \(a^{n-1}-a^n>0\) or \(a^{n-1}(1-a)>0\). If \(a>1\), doesn't matter if \(n\) is even or odd, the given inequality cannot hold. If \(a<0,\) then necessarily \(n\) must be odd. Sufficient.

(2)\(a^n>a^{3n}\) can be rewritten as \(a^n-a^{3n}\) or \(a^n(1-a^{2n})>0\). Since \(a\) cannot be neither \(0\) nor \(1,\) \(\,\,1-a^{2n}\) is certainly negative. (Now \(a\) cannot be \(-1\) either.) Then \(a^n\) is negative if \(a\) is negative and \(n\) is odd, and this is the only way the given inequality can hold. If \(a\) is positive, doesn't matter the value of \(n\), the inequality cannot hold. Sufficient.

Answer D.
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Last edited by EvaJager on 05 Oct 2012, 01:08, edited 3 times in total.

Re: If a and n are integers and n > 1, is n odd? [#permalink]

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26 Apr 2013, 22:01

IanStewart wrote:

conty911 wrote:

If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n (2) a^n > a^(3n)

If you factor, as in the post above, you can avoid cases if you arrange things differently. Rewriting Statement 1, noticing that a cannot be 1, we can factor out the smaller power, so can factor out a^(n-1) :

0 > a^n - a^(n-1) 0 > a^(n-1) (a - 1)

Now one term on the right side must be positive, and the other must be negative. But if a is a positive integer, that can't possibly be true -- both terms will be positive. So a must be negative, and (a-1) must be the negative term, and a^(n-1) must be the positive term. Since a is negative, that means the exponent must be even, so n-1 is even and n is odd.

Similarly for Statement 2, notice a cannot be -1, 0 or 1, and we have

0 > a^(3n) - a^n 0 > a^n [a^(2n) - 1]

and again, one term must be positive, the other negative. But a^(2n) has an even exponent, so it must be positive and so a^(2n) - 1 is the positive term (using the fact that a cannot be 0, 1 or -1). That makes a^n the negative term. But if a^n is negative, a must be negative, and n must be odd.

I don't think you need to do the problem algebraically, however. You can just imagine how different numbers would 'behave' in the inequalities given. So if you know that n is a positive integer greater than 1, and you know

a^(n-1) > a^n

then you might be able to see that a must be negative, since this inequality won't ever be true if a > 1. And if we know the base is negative, and we can see that one exponent is even and the other is odd (since the exponents are just consecutive integers), we know that one side of the inequality will become positive, and the other will remain negative. Clearly the larger side needs to be positive, which makes n-1 even, and n odd.

Similarly for the other Statement, the inequality

a^n > a^(3n)

simply cannot be true if a > 1. Since 1, 0 or -1 don't work in this inequality, a must be -2 or smaller. But then if n is even, it wouldn't matter if a is negative or positive, since even exponents just 'erase' our negative signs. And we know it must matter, since the inequality won't work if we have positive bases on either side. So the inequality will only work if n is odd so that the left and right sides both remain negative.

-------------

Hi Ian, I like the way you solved this Question, but I really can't understand the 2nd stmt.. explanation. Pls explain it to me. Thanks !!
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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26 Apr 2013, 22:04

Bunuel wrote:

Archit143 wrote:

If a and n are integers and n > 1, is n odd?

(1) (a)^n–1 > a^n

(2) a^n > a^(3n)

-----------------------------------------

Hi Bunuel....

I have read the explanation provided by Ian ,... but I'm unable to understand the explanation for 2nd statement ,, & I think Ian is no longer a regular visitor to the GMAT Club ..... Can You pls... Make me understand this question ....& solution Pls,,,,,, Thanks !!
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S1) a^(n–1) > a^n This can happen in two cases. If a is negative (e.g. -2^2 > -2^3) of if a is fraction (e.g. (1/2)^2 > (1/2)^3. Since we are told that both a and n are integers, later case is not possible. So we can conclude that a is negative. Now to hold the inequality a^(n–1) > a^n true when a being negative, n-1 must be even and hence n must be odd. Sufficient

S2) In the same way as S1 this choice is also sufficient

Well , i feel a bit intimidated by equation questions so it took me 2:25 to solve this one. But if you keep the initial hesitation aside , this can be solved in less than 2 min.

From 1) a^n-1 > a ^n. From given conditions ie n> 1 and a and n being integers , this is only possible if a is negative and n-1 is even integer. so n is an odd integer. From 2) a^n > a^3n. Again only possible if a is negative and n has to be odd.

So D is the answer.

Last edited by Bluelagoon on 29 Apr 2013, 17:09, edited 1 time in total.

a^(n–1) > a^n thus , a^(n-1) - a^n > 0 , thus a^n (a^-1 - 1) > 0 i.e. a^n [ (1/a) - 1] >0 and n>1 and 1/a is a fraction therefore a^n has to be -ve , this is only possible when a is -ve integer and n is odd .....suff

from 2

a^n > a^(3n), thus , a^n - a^ 3n > 0 thus a^n ( 1- a^2n) > 0 and since a,n are integers and n>1 and since (1-a^2n) is definitely -ve therefore a^n is -ve and n is odd ....suff

Re: If a and n are integers and n > 1, is n odd? [#permalink]

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17 Jun 2015, 01:42

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If a and n are integers and n > 1, is n odd? [#permalink]

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18 Dec 2015, 08:33

conty911 wrote:

If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n (2) a^n > a^(3n)

Please provide me pointers to any similar questions, also if possible

Can some one explain steps to tackle such problems with least amount of time.For eg. how to select the values for plugging etc. THanks

In the MGMAT CATs there are lots of such caliber (time consuming ). Could not solve this one on the CAT because of the time issue, had to guess.

(1) \(a^(n–1) > a^n\) here we have to check 2 cases n=ODD and n=EVEN ODD: In order this expression to be true a must be either \(-ve or a fraction\), a is an integer, so a is -ve EVEN: The only possibility this expression to hold true - a must be a positive fraction, BUT it's not a fraction, but an integer Thus, it holds true ONLY if n is ODD, Sufficient

(2) Same game as above, and you get here that n must be an ODD integer Answer D
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n (2) a^n > a^(3n)

In the original condition, there are 2 variables(a,n) and 1 equation(n>1, when it comes to inequality questions, an inequality also can be an equation.), which should match with the number equation. So you need 1 more equation. For 1) 1 equation, for 2 1equation, which is likely to make D the answer. In 1), from a^(n-1)>a^n, if a>1, it is impossible because n-1>n impossible as well as a=0,1. So, it has to be a<0. In this case, n=odd and n-1=even. Therefore, a^(n-1)>0, a^n<0 is yes, which is sufficient. In 2), from a^n>a^(3n), if a>1, it is impossible as well because n>3n is impossible so as a=0,1,-1. So, it has to be a<-1. In order to satisfy a^n>a^(3n) from 3n>n, only can n=odd be possible, which is yes and sufficient. Therefore, the answer is D.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Both the terms can either be positive or negative Taking the first case i.e. positive Say ((1-a)/a) = 1 => a=1/2 Substituting the value of a in a^n For the second term i.e. a^n to be positive (1/2)^n can take the value of n as either positive or negative Why is then A sufficient?

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