m+n=?

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m+n=?

by Max@Math Revolution » Fri Jan 25, 2019 1:04 am

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[Math Revolution GMAT math practice question]

m+n=?

1) (4^m)(2^n)=16
2) (2^{2m})(4^n)=64

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by fskilnik@GMATH » Fri Jan 25, 2019 6:24 am

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Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

m+n=?

1) (4^m)(2^n)=16
2) (2^{2m})(4^n)=64
$$? = m + n$$

$$\left( 1 \right)\,\,\,{2^{2m + n}} = {4^m} \cdot {2^n} = {2^4}\,\,\,\,\,\,\mathop \Rightarrow \limits^{2\,\, \notin \,\left\{ { - 1,0,1} \right\}} \,\,\,\,\,2m + n = 4$$
$$\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2,0} \right)\,\,\,\, \Rightarrow \,\,\,? = 2 \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {0,4} \right)\,\,\,\, \Rightarrow \,\,\,? = 4 \hfill \cr} \right.$$

$$\left( 2 \right)\,\,\,{2^{2\left( {m + n} \right)}} = {2^{2m}} \cdot {4^n} = {2^6}\,\,\,\,\,\,\mathop \Rightarrow \limits^{2\,\, \notin \,\left\{ { - 1,0,1} \right\}} \,\,\,\,\,2\left( {m + n} \right) = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 3$$


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by Max@Math Revolution » Sun Jan 27, 2019 5:07 pm

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Condition 2) is equivalent to m + n = 3 as shown below:
(2^{2m})(4^n)=64
=> (2^{2m})(2^2n})=2^6
=> 2^{2m+2n}=2^6
=> 2m+2n = 6
=> m + n = 3
Condition 2) is sufficient.

Condition 1)
(4^m)(2^n)=16
=> (2^{2m})(2^n)=2^4
=> 2^{2m+n}=2^4
=> 2m+n = 4
If m = 1 and n =2, then m + n = 3.
If m = 0 and n = 4, then m + n = 4.
Since it does not yield a unique solution, condition 1) is not sufficient.

Therefore, the answer is B.
Answer: B