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Exponents - If n and t are positive integers, is n a factor

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Exponents - If n and t are positive integers, is n a factor

by II » Sun Aug 17, 2008 12:10 am
Hi, Interested in seeing the various approachs people have in answering this question ... please illustrate your logic/thinking when answering this. Thanks.

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

(1) n = 3^(n-2)

(2) t = 3^n
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Re: Exponents - If n and t are positive integers, is n a fac

by sudhir3127 » Sun Aug 17, 2008 12:14 am
II wrote:Hi, Interested in seeing the various approachs people have in answering this question ... please illustrate your logic/thinking when answering this. Thanks.

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

(1) n = 3^(n-2)

(2) t = 3^n
are u sure abt the statement 1. it doesnt say anything abt "T"?
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Re: Exponents - If n and t are positive integers, is n a fac

by sudhir3127 » Sun Aug 17, 2008 12:22 am
II wrote:Hi, Interested in seeing the various approachs people have in answering this question ... please illustrate your logic/thinking when answering this. Thanks.

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

(1) n = 3^(n-2)

(2) t = 3^n

are u this is not the question.. because i have seen such a question on some forum..

If n and t are positive integers, is n a factor of t?
(1) n = 3^(n-z)
(2) t = 3^n
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by II » Sun Aug 17, 2008 12:39 am
Hi Sudhir,
I have posted the question correctly. Everything you see there is correct.
Also see attachment, which is screen shot of the question.
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If n and t are positive integers.jpg
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by sudhir3127 » Sun Aug 17, 2008 1:28 am
i go with C.

here the explanation.

Statement 1. it doesnt say anything abt t .. hence in suffcient.

statement 2.

assume if n= 1 then t= 3 , n is a factor of t but if n= 2 t=9 but if n= 2 then t = 9 .. then n is not a factor of t. hence insufficient..

when we take both together..

statement 1 :
n has to greater than or equal to 2 to be an interger.. thus the possible values of n are 1,3,9,27.....

and statement 2. the t can be 3, 27......

thus we are clear that every for all the values of n and t... n is a factor of t.

Hope it helps..

do let me know if u have any doubts..
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by pepeprepa » Sun Aug 17, 2008 1:37 am
If n and t are positive integers, is n a factor of t ?

(1) n = 3^(n-2)
Alone, this has no sens, the logic is just that there is no "t" in the equation so it is useless.

(2) t = 3^n
If n=2 then t=9, n is not a a factor of t
But if n=1 and t=3, n is a factor of t (3=3*1)
So it is insufficient, given we have example and counter-example.

(1) & (2)
n = 3^(n-2) can be written like that: n=(3^n)/(3^2)
Then, 9*n=3^n

Thanks to the 2) we have,
t=9*n

So it is clear n is a factor of t given they are both integers.
My answer is C
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by II » Sun Aug 17, 2008 1:42 am
pepeprepa wrote:If n and t are positive integers, is n a factor of t ?

(1) n = 3^(n-2)
Alone, this has no sens, the logic is just that there is no "t" in the equation so it is useless.

(2) t = 3^n
If n=2 then t=9, n is not a a factor of t
But if n=1 and t=3, n is a factor of t (3=3*1)
So it is insufficient, given we have example and counter-example.

(1) & (2)
n = 3^(n-2) can be written like that: n=(3^n)/(3^2)
Then, 9*n=3^n

Thanks to the 2) we have,
t=9*n

So it is clear n is a factor of t given they are both integers.
My answer is C
Excellent explanation ... thanks !
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by II » Sun Aug 17, 2008 4:16 am
did anyone else have other approaches to this one ?
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by sharad » Mon Aug 18, 2008 8:02 am
IMO a

this gives n as 1

n = 3^(n-2) or 3 = 3^(1)

1 is factor for all the numbers...

B is insufficient
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by pepeprepa » Mon Aug 18, 2008 8:37 am
yep sharad you seem to be right the only solution for n=3^(n-2) is 1 ...
Need to check more the proposals and not skip them so fast

II can you give OA
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by kiran.raze » Mon Aug 18, 2008 9:30 am
Hi Pepeprepa,

The solution to n=3^(n-2) is not 1 but n=3, when we have 3=3;

which only means n=3 and so we cannot be sure whether it can be a factor of
t,

and option 2 alone is not sufficient i.e t=3^n because t/n = 3^n/n , which for n=1 is divisible and n=2 is not divisible , therefore insufficient.

Together, of course we have t/n= 3^(n-2)/3^n which equals 9

therefore sufficient ...

Thanks,
Kiran
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by pepeprepa » Mon Aug 18, 2008 9:36 am
Thanks for claryfing this post man, I don't know why I bugged between 1 and 3.
Kind of small things which make you doubt, I was right indeed :D
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by II » Mon Aug 18, 2008 1:34 pm
yes ... official answer is C.
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by arorag » Mon Aug 18, 2008 6:56 pm
There is no need to put the values.

key point:
Is t/n= integer

Combining

3^n/3^(n-2)= 9
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by drabblejhu » Wed Sep 16, 2009 9:20 pm
Yes, combining by dividing expressions for t and n is great. Don't you need to be careful to test B with numbers, though? To see that you can get yes and no with varying n values?

Thanks,
J
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