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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.

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):

Quick Algebra question for Statement 1&2 combined:

If I plug in n = 3^(n-2) into t = 3^n I get: n = 3^(3^(n-2))

when I rewrite it I eventually come to 3^3n * 1/3^6 = t

However this does not help me in any way... Where am i going wrong?

The highlighted steps are out of Sink

\(a^{(b^c)}\) is NOT equal to \(a^b*a^c\)

Whereas, \((a^b)^c\) = \(a^b*a^c\)

i.e. \(3^{(3^{(n-2)})}\) is NOT same as \(3^{3n} * 1/3^6\)

Thank you, so the only way would be to plug in values for n for \(3^{(3^{(n-2)})}\) ?

Or is there any way to rewrite this?

There are three ways

1) Plug-in the Values from Options 2) Take Logarithm on both sides and then solve further 3) Change the method and follow the methods given in other explanations

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]

Show Tags

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]

Show Tags

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: If n and t are positive integers, is n a factor of t ? [#permalink]

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23 Sep 2014, 04:42

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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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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.

Re: If n and t are positive integers, is n a factor of t ? [#permalink]

Show Tags

24 Jan 2015, 02:07

Bunuel wrote:

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.

Answer: C.

---- I think this question is mathematically a wrong question. The equation n=3^(n-2) does not qualified for any existing integer. Can you find any integer that can put this equation for n and get n=3^(n-2)????? This question does not make sense for me.

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.

Answer: C.

---- I think this question is mathematically a wrong question. The equation n=3^(n-2) does not qualified for any existing integer. Can you find any integer that can put this equation for n and get n=3^(n-2)????? This question does not make sense for me.

Have you read the very post you are quoting??? Please re-read!
_________________

Re: If n and t are positive integers, is n a factor of t ? [#permalink]

Show Tags

09 Jul 2015, 00:01

GMATinsight wrote:

There are three ways

1) Plug-in the Values from Options 2) Take Logarithm on both sides and then solve further 3) Change the method and follow the methods given in other explanations

Third seems the Best to me

I hope it Helps!

It does Option 3) should indeed be the way to go, however when I first solved the problem I just did not see it happens...

gmatclubot

Re: If n and t are positive integers, is n a factor of t ?
[#permalink]
09 Jul 2015, 00:01

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