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

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Joined: 13 Mar 2018
Posts: 14
Re: If n and t are positive integers, is n a factor of t ?  [#permalink]

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30 Dec 2018, 08:25
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.

Hi Bunnel

May you please explain, for what reasons n=3 is the only integer solution to eqn of st.1
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Re: If n and t are positive integers, is n a factor of t ?  [#permalink]

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05 Jan 2019, 13:46
SWAPNILP wrote:
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.

Hi Bunnel

May you please explain, for what reasons n=3 is the only integer solution to eqn of st.1

Okay I get it now. I struggled to understand why the answer wasn't E. Took me probably 20 minutes to figure it out!

Statement (1) tells us one thing. n = 3^(n-2). Okay so we know for sure that n has to be equal to 3.

How else could we satisfy n = 3^(n-2) without making n=3?

Let's try plugging in n=2. We get 2 = 3^(2-2) -----> 2 = 3^0 -----> 2 = 1. Impossible.

The key take away from Statement (1) is that n is equal to 3.

So when we combine Statement (1) and (2) we only have to plug in n=3.

t = 3^3^(3-2) ----> 3^3^1 ----> 3^3=27. -----> T= 27 and n=3. Sufficient.
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Re: If n and t are positive integers, is n a factor of t ?   [#permalink] 05 Jan 2019, 13:46

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