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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.