Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

23 Aug 2012, 02:11

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

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

Show Tags

23 Aug 2012, 18:35

1

This post received KUDOS

Expert's post

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. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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

Show Tags

04 Oct 2012, 03:06

4

This post received KUDOS

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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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]

Show Tags

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 !! _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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

Show Tags

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 !! _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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]

Show Tags

17 Jun 2015, 01:42

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

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 _________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

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

Show Tags

19 Dec 2015, 05:28

1

This post received KUDOS

Expert's post

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. _________________

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?

gmatclubot

Re: If a and n are integers and n > 1, is n odd?
[#permalink]
21 May 2016, 04:27

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...