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

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

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

\(\frac{1}{9}+\frac{3}{9}=\frac{1}{3^2}+\frac{1}{3^1}=\frac{1}{3^m}+\frac{1}{3^n}\)

\(m+n=1+2=3\)


If you are not able to simplify the LHS 4/9, then you can proceed further in the following manner.

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

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

Now, m and n have to be both different when it comes down to being even or odd..
\(3^{odd}+3^{odd}\) is even but not a multiple of 4.
Similarly, \(3^{even}+3^{even}\) is even but not a multiple of 4.

Thus, the RHS has to be \(3^{odd}+3^{even}\), as it is a multiple of 4.
so, m+n=even+odd=odd

Only D and E possible.

If any one of the m and n are negative, the answer becomes fraction, and we will not get two sides equal.

D. 3....m+n=1+2=3.
Substitute and we get the answer.

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

\(\frac{36}{3^4}\) can be rewritten as \(\frac{36}{81}\) -> \(\frac{12}{27}\). \(3^3=27\). Therefore, m+n=3.

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

\(\frac{3}{27}\) + \(\frac{9}{27}\) = \(\frac{12}{27}\)

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.

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.


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?

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

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...

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.

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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.

Posted from my mobile device

so my question is - what is wrong in the approach above?

using the numerator/denominator scales can't we say that 9=3^(m+n)? then would give m+n=2. i know that on top we have 4=3^m+3^n - so it that we have formed two equations that don't have a valid solution that this approach work?

Hi yahya729,

The reason why that approach is not correct is because it completely ignores the value of the numerator. If you want to reduce the fraction 36/81 to 4/9, then that's fine, but you still need the calculation to equal 4/9 - and while that calculation does manage to make the denominator equal 9, it doesn't make the numerator equal 4.

GMAT assassins aren't born, they're made,
Rich

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