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Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is [#permalink]
28 Mar 2013, 10:20

1

This post received KUDOS

Expert's post

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. _________________

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

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

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: If m and n are integers and 36/3^4 = 1/3^m + 1/3^n , what is [#permalink]
28 Feb 2015, 05:50

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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