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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] New post 26 Mar 2013, 13:32
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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
[Reveal] Spoiler: OA

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] 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] 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: Exponent Question [#permalink] New post 26 Mar 2013, 20:43
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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

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: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is [#permalink] 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] 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] 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] New post 03 May 2014, 00:56
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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is [#permalink] 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] 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] 04 May 2014, 19:02
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