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

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

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Updated on: 28 Jul 2012, 01:52
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If n and t are positive integers, is n a factor of t ?

(1) n = 3^(n-2)
(2) t = 3^n
[Reveal] Spoiler: OA

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Originally posted by TOUGH GUY on 12 Dec 2005, 06:26.
Last edited by Bunuel on 28 Jul 2012, 01:52, edited 1 time in total.
Edited the question and added the OA.
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12 Dec 2005, 09:54
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.

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13 Dec 2005, 06:25
Yes, exp means power of exponents.

A alone makes no sense.

B alone is confusing, if we do pick up numbers, we realise than that

if

n=1; t=3 yes n is a factor of t;
n=2; t=9 no n is not a factor of t;

OA is E
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Re: DS- Positive integers and factors - Sounds easy [#permalink]

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20 Jan 2011, 17:20
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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.
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Re: DS- Positive integers and factors - Sounds easy [#permalink]

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21 Jan 2011, 10:13
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.

Is my assumption right?
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Re: DS- Positive integers and factors - Sounds easy [#permalink]

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21 Jan 2011, 12:12
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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.

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

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27 Jul 2012, 18:50
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.

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

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28 Jul 2012, 02:07
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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.

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.

Hope it' helps.
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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, 07:01
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
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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19 Jan 2015, 00:56
Ya original question has z in place of '2'in the exponent....ans for tat is E

But for the given question
Ans is C

just observe t/n =9...tats enough
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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24 Jan 2015, 03:07
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.

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

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24 Jan 2015, 04:26
miriampirooz 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.

----
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.

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

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26 Jan 2015, 02:22
Ya original question has z in place of '2'in the exponent....ans for tat is E

But for the given question
Ans is C

just observe t/n =9...tats enough

Yes i did this way too. (3^n)/((3^n)/(3^2))=9
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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09 Jul 2015, 00:12
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?
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If n and t are positive integers, is n a factor of t ? [#permalink]

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09 Jul 2015, 00:28
noTh1ng wrote:
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 $$3^{3n} * 1/3^6$$
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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09 Jul 2015, 00:34
GMATinsight wrote:
noTh1ng wrote:
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?
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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09 Jul 2015, 00:42
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noTh1ng wrote:
GMATinsight wrote:
noTh1ng wrote:
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

Third seems the Best to me

I hope it Helps!
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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

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06 Sep 2016, 23:59
Statement 1: n=3^(n-2). No correlation with 't' is mentioned. Hence, insufficient.

Statement 2: t=3^n. Take n=1, then n is a factor of t. However, n=2 is not a factor of t. Hence, insufficient.

Combined: St:1 can be re-written as, n=3^n.3^-2, with information from statement 2, n=t.3^-2, n=t/9. Given that n&t are positive integers, the information is sufficient.
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Re: If n and t are positive integers, is n a factor of t ? [#permalink]

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05 Dec 2017, 19:43
TOUGH GUY wrote:
If n and t are positive integers, is n a factor of t ?

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

We need to determine whether t/n = integer

Statement One Alone:

n = 3^(n - 2)

Since we do not have any information regarding t, statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

t = 3^n

We can substitute some numbers for n. For example, if n = 1, then t = 3^1 = 3 and 1 is a factor of 3. However, if n = 2, then t = 3^2 = 9 but 2 is not a factor of 9. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using the information from statements one and two, we can substitute 3^(n - 2) for n and 3^n for t in our question: t/n = integer ?

(3^n)/3^(n - 2) = integer ?

Since we are dividing similar bases, we can subtract the exponents and keep the base. Then we have:

3^(n - n + 2) = integer ?

3^2 = integer ?

9 = integer ?

Since 9 IS an integer. We have answered “yes” to the question.

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Re: If n and t are positive integers, is n a factor of t ?   [#permalink] 05 Dec 2017, 19:43
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