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If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is

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If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post Updated on: 27 Mar 2013, 03:37
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If m and n are integers and \(\frac{36}{3^4}=\frac{1}{3^m}+\frac{1}{3^n}\), what is the value of m+n?

A. -2
B. 0
C. 2
D. 3
E. 5

Originally posted by DoItRight on 26 Mar 2013, 13:32.
Last edited by Bunuel on 27 Mar 2013, 03:37, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Exponent Question  [#permalink]

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New post 26 Mar 2013, 20:43
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9
DoItRight wrote:
If m and n are integers and \(\frac{36}{3^4}=\frac{1}{3^m}+\frac{1}{3^n}\), what is the value of m+n?

A. -2
B. 0
C. 2
D. 3
E. 5

Please explain your work.


We have \(\frac{36}{3^4}\) = \(2^2/3^2\) = 4/9.

Thus, 4/9 = \(\frac{1}{3^m}+\frac{1}{3^n}\)

or (3+1)/9 = \(\frac{1}{3^m}+\frac{1}{3^n}\)

or 1/3 + 1/9 = \(\frac{1}{3^m}+\frac{1}{3^n}\)

Thus, m = 1, n =2, m+n = 3.

D.
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Re: Exponent Question  [#permalink]

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New post 26 Mar 2013, 15:15
Not sure why I get this as m+n=2

My workings: 36 / 9 * 9 = 1 / 3^m + 1 / 3^n
4 / 9 = {3^n + 3^m} / 3^(n+m)
since 9 = 3^2, we get 3^2=3^(m+n)
hence m+n = 2

Ans C
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Re: Exponent Question  [#permalink]

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New post 26 Mar 2013, 20:37
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\(\frac{36}{3^4}\) can be rewritten as \(\frac{36}{81}\) -> \(\frac{12}{27}\). \(3^3=27\). Therefore, m+n=3.

To check \(\frac{1}{3^2}\) -> \(\frac{3}{27}\) & \(\frac{1}{3^1}\) -> \(\frac{9}{27}\)

\(\frac{3}{27}\) + \(\frac{9}{27}\) = \(\frac{12}{27}\)
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 28 Mar 2013, 10:03
Hey vinaymimani,

How did you know when to break 4/9 into 1/9+3/9? I tried using alegbra like srcc25anu did and got nowhere.
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 28 Mar 2013, 10:20
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DoItRight wrote:
Hey vinaymimani,

How did you know when to break 4/9 into 1/9+3/9? I tried using alegbra like srcc25anu did and got nowhere.


Firstly, solving it algebraically, we have
\(4/9 = 1/3^m + 1/3^n\)

or 4 = 3^(2-m)+3^(2-n)

Now we know that m,n are integers. Moreover, two expressions, both being powers of 3 add upto 4. Thus, it must be of the form : 3+1. Thus, m=2,n=1 or m=1,n=2.

In either case, m+n = 3.

Now as to how it struck me to split 4/9, when you observe \(4/9 = 1/3^m+1/3^n\), one might notice that after any cancelling of common factors on the rhs, we should end up with a 9 in the denominator. Also, seeing that 4 = 3+1, one can think that maybe there is a (1/3) and a (1/9) on the rhs, where the lcm is 9, which leads to the (3+1) part in the numerator. I hope it was of some help.
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 28 Mar 2013, 11:47
Quote:
Firstly, solving it algebraically, we have
or 4 = 3^(2-m)+3^(2-n)

When I tried the alegbra way, I got
4 / 9 = 3^n + 3^m / 3^(n+m) I did this but i gave me n+m together but it made the numerator very messy, and hit a dead end.

Quote:
Now as to how it struck me to split 4/9, when you observe , one might notice that after any cancelling of common factors on the rhs, we should end up with a 9 in the denominator. Also, seeing that 4 = 3+1, one can think that maybe there is a (1/3) and a (1/9) on the rhs, where the lcm is 9, which leads to the (3+1) part in the numerator. I hope it was of some help.

Kudos on recognizing the split. Thanks for breaking it down.



Does anyone where I can practice these types of questions? Questions where you have to recognize certain splits?
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 03 May 2014, 12:24
I tried cross multiplying the right side.
got: 36/3^4 = (3^m)*(3^n)/(3^mn)

from this i wasn't able to find an answer.

Can someone help with this?

How did you guys find the way of breaking it down?
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 04 May 2014, 19:02
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ronr34 wrote:
I tried cross multiplying the right side.
got: 36/3^4 = (3^m)*(3^n)/(3^mn)
from this i wasn't able to find an answer.

Can someone help with this?

How did you guys find the way of breaking it down?


You don't cross multiply but take an LCM

For RHS, If you take the LCM, we get

\((3^n+ 3^m)/ 3^{(m+n)}\)

So, we have 36/81 or 4/9 = \((3^n+ 3^m)/ 3^{(m+n)}\)

It is better to reduce the LHS to 4/9 and then take 9 to the RHS.

The expression becomes 4= 9/3^m + 9/3^n

or \(4= 3^{(2-m)}+3^{(2-n)}\)

Now when you see a 4 on LHS this should tell you that one of the terms on RHS is of the type 3^0 =1 so either m=2 or n=2,plugging in these values and we can get value for n and m+n=3 in any case

The best way forward is given above by mau5 or you end up pluggin in multiple values(If you don't simplify the algebra) and loose the plot as there will be too many iterations you may have to try before getting the answer.

mau5's method is pretty neat.

Hope it helps
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 28 Feb 2015, 05:50
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cherryli2015 wrote:
If m and n are integers and 36/(3^4) = 1/(3^m) + 1/(3^n), what is the value of m+n?

Thanks very much ...

hi cherry,
right side = 36/3^4= 4/9..
left side = 1/(3^m) + 1/(3^n)=(3^m+3^n)/3^(m+n) =4/9...
so one can solve for m as 1 when n=2 or m as 2 when n=1.. m+n =3, in each case...
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 24 Jun 2015, 08:31
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DoItRight wrote:
If m and n are integers and \(\frac{36}{3^4}=\frac{1}{3^m}+\frac{1}{3^n}\), what is the value of m+n?

A. -2
B. 0
C. 2
D. 3
E. 5



My first thought was to try to write 36 in the form of \(3^x + 3^y\).

\(3^0=1, 3^1=3, 3^2=9, 3^3=27.\)

We can write 36 = 9 + 27 = \(3^2 + 3^3\)

Using the above in LHS,

\(\frac{36}{3^4}\) = \(\frac{3^2 + 3^3}{3^4}\) = \(\frac{3^2}{3^4}\) + \(\frac{3^3}{3^4}\) = \(\frac{1}{3^2}\) + \(\frac{1}{3^1}\)

So m = 2, n = 1 => m + n=3

So D
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 05 Nov 2016, 17:03
Easiest way to solve is to simplify the LHS and then test out values for m and n on RHS.

36 = (3^2)(2^2)

36/(3^4) = [(3^2)(2^2)]/(3^4) = (2^2)/(3^2) --> 4/9

m and n combination that will give this is m+n=3 (i.e. 1+2)

D is the correct answer.
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 04 Feb 2018, 11:59
Hi All,

Quant questions on the GMAT are often built around number 'patterns' of some kind, so your ability to 'play around' with a question and find the pattern(s) behind it can help you to get past complex-looking questions without too much trouble.

Here, we're told that M and N are INTEGERS and that 36/(3^4) = 1/(3^M) + 1/(3^N). We're asked for the value of M+N.

To start, I'm going to do a quick comparison assuming that there were no variables at all...

36/(3^4) = 36/81
1/3 = 27/81

So 1/3 + 1/3 = 54/81 > 36/81. This proves that at least one of the two variables has to be greater than 1 (we have to make at least one of the denominators BIGGER so that we can shrink that fraction and get the sum to equal 36/81.

Let's try making one of the variables 1 and one of the variables 2...

1/3 = 27/81
1/(3^2) = 1/9 = 9/81

27/81 + 9/81 = 36/81
This is an EXACT MATCH for the other side of the equation, so this MUST be the solution. M+N = 1+2 = 3

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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is  [#permalink]

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New post 25 Nov 2018, 10:15
The way I solved this was to first minimize the LHS to 4/9. (It makes the numbers easier to play with in your head). Then, I just plugged and played until I found something that works. Which is D. Also note that answer choice A and B have to have at least one negative number somewhere.

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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is &nbs [#permalink] 25 Nov 2018, 10:15
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