MathRevolution wrote:

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Math Revolution GMAT math practice question]

If \(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\), where \(m\) and \(n\) are different positive integers, what is the value of \(m+n\)?

\(A. 9\)

\(B. 12\)

\(C. 16\)

\(D. 18\)

\(E. 20\)

\(\left\{ \matrix{

m,n\,\,{\rm{distinct}}\,\,\, \ge \,\,{\rm{1}}\,\,\,{\rm{ints}}\,\,\,\,\left( * \right) \hfill \cr

{1 \over 3} = {1 \over m} + {1 \over n}\,\,\,\left( {**} \right) \hfill \cr} \right.\)

\(? = m + n\)

1st way ("the smart guy/girl approach"):

\({1 \over 3}\,\,\, = \,\,\,{4 \over {12}}\,\,\,\, = \,\,\,{1 \over {12}} + {3 \over {12}}\,\,\, = \,\,\,{1 \over {12}} + {1 \over 4}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 12 + 4 = 16\)

2nd way ("the analytical approach"):

\(\left( * \right)\,\, \cap \,\,\left( {**} \right)\,\,\,\,\mathop \Rightarrow \limits^{{\rm{WLOG}}\,\,} \,\,\,\,\left\{ \matrix{

\,n = 6 - x\,\,\,,\,\,\,x \in \left\{ {1,2} \right\} \hfill \cr

\,m = 6 + y\,\,,\,\,\,\,y \ge 1\,\,{\mathop{\rm int}} \hfill \cr} \right.\)

\(\left( {\,\,{1 \over 6} + {1 \over 6} = {1 \over 3}\,\,\,\,\,\,;\,\,\,\,\,\,\,{1 \over 6} + {1 \over { \ge 7}} \ne {1 \over 3}\,\,} \right)\)

(WLOG = without loss of generality)

\(\left\{ \matrix{

\,x = 1\,\,\, \Rightarrow \,\,\,n = 5\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,{1 \over 3} = {1 \over m} + {1 \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,{1 \over m} = {1 \over 3} - {1 \over 5} = {5 \over {15}} - {3 \over {15}} = {2 \over {15}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,m\,\,\,{\rm{not}}\,\,{\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}\, \hfill \cr

\,x = 2\,\,\, \Rightarrow \,\,\,n = 4\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,{1 \over 3} = {1 \over m} + {1 \over 4}\,\,\,\,\, \Rightarrow \,\,\,\,{1 \over m} = {1 \over 3} - {1 \over 4}\,\, = {4 \over {12}} - {3 \over {12}} = {1 \over {12}}\,\,\,\,\, \Rightarrow \,\,\,\,m = 12\,\,\, \hfill \cr} \right.\)

\(? = m + n = 12 + 4 = 16\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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