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.

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:

The answer is E.
The statement I doesnt make sense with "Z" and non mention of T.
And for Statement II, there are both yes and no answers.

like 9 = 3^2 where 2 is not a factor of 9.
also 27 = 3^3 where 3 is a factor of 27.

On combining statements I and II,
n = 3^(n-z)
n=3^n/3^z
n= t/3^z (t= 3^n, from statement II)
For all postive integers of Z (including Zero), n is a factor of T
But for negative integers, we can only say that n is multiple of T, but not the factor.

Re: DS- Positive integers and factors - Sounds easy [#permalink]
20 Jan 2011, 16:20

Sorry for pulling up this old thread, but I googled it because it is a question in the OG12 and I would never come up with an approach like that in the OG12. The initial post states the question incorrectly. Instead of a z, that is subtracted in the exponent, it is actually a 2. You can find the question on p. 310 #66 in the OG12.

Well, it makes sense to me that the answer is C, but not by dividing statement (1) by (2)?! I see that each of the statements are not sufficient considered solely, but instead of that weird approach, I would find out that n in the first statement has to be 3, which goes along with statement (2), since 3 is a factor of 27.

Isn't it true that the questions in the OG are arranged from easy to tough in the particular sections? Sometimes I need five secs for one of the lower questions and then others like these take, if you can solve it at all, a decent amount of time.

Re: DS- Positive integers and factors - Sounds easy [#permalink]
21 Jan 2011, 09:13

I think the answer should be C. From (1) we can find it that n will always be multiple of 3, or n can be 1. if put this in (2), it clearly means that n will be a factor of t. n can be 1,3,9,27 etc... and 3(exp)1,3,9,27 will always be multiple of n and 3.

Re: DS- Positive integers and factors - Sounds easy [#permalink]
21 Jan 2011, 11:12

2

This post received KUDOS

Expert's post

selines wrote:

Sorry for pulling up this old thread, but I googled it because it is a question in the OG12 and I would never come up with an approach like that in the OG12. The initial post states the question incorrectly. Instead of a z, that is subtracted in the exponent, it is actually a 2. You can find the question on p. 310 #66 in the OG12.

Well, it makes sense to me that the answer is C, but not by dividing statement (1) by (2)?! I see that each of the statements are not sufficient considered solely, but instead of that weird approach, I would find out that n in the first statement has to be 3, which goes along with statement (2), since 3 is a factor of 27.

Isn't it true that the questions in the OG are arranged from easy to tough in the particular sections? Sometimes I need five secs for one of the lower questions and then others like these take, if you can solve it at all, a decent amount of time.

Yes the question is from OG and it should be:

If n and t are positive integers, is n a factor of t ?

(1) n = 3^(n-2) --> n=3 (only integer solution for this equation), but we know nothing about t, so this statement is not sufficient.

(2) t = 3^n --> if n=1 then the answer will be YES but if n=2 then t=9 and the answer will be NO. Not sufficient.

(1)+(2) As n=3 then t=3^n=27 and the answer to the question will be YES as 3 is a factor of 27. Sufficient.

Re: If n and t are positive integers, is n a factor of t? (1) n [#permalink]
28 Jul 2012, 01:07

2

This post received KUDOS

Expert's post

alchemist009 wrote:

hello Bunuel, How'd you figure it out in statement 1 that n=3. i do comprehend that n must be 3 but i cant figure it out by doing algebra.

please help

You can find that by trial and error: n=1 and n=2 does not satisfy n = 3^(n-2), but n=3 does. Now, if n>3 (4, 5, 6, ...), then RHS is always greater than LHS, so n=3 is the only solution.

You can solve this problem without finding the value of n in (1):

n = 3^{n-2} --> n=\frac{3^n}{9} --> 9n=3^n

For (1)+(2): since from (2) t = 3^n, then t=9n, hence n is a factor of t.

Re: If n and t are positive integers, is n a factor of t ? [#permalink]
23 Sep 2014, 04:42

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. _________________

Re: If n and t are positive integers, is n a factor of t ? [#permalink]
23 Sep 2014, 06:01

is n a factor of t? is t/n= integer ?

statement 1 : n= 3^(n-2) nothing is given about t... statement is insufficient

statement 2 : t=3^n let n=2, then t=9 n is not a factor of t .... false let n=3, then t=27 n is a factor of t .... true statement is insufficient

both statements combined n= 3^(n-2)... given n=3^n/3^2 n=t/3^2 ..... (replacing 3^n by t as given in statement 2) t/n= 3^2 t/n= integer Therefore n is a factor of t.

Ans - C

gmatclubot

Re: If n and t are positive integers, is n a factor of t ?
[#permalink]
23 Sep 2014, 06:01

It’s been a long time, since I posted. A busy schedule at office and the GMAT preparation, fully tied up with all my free hours. Anyways, now I’m back...

Burritos. Great, engaging session about how to network properly. How better can it get? Hosted jointly by Human Capital Club and Engineers in Management, we had a chance to...