Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
If 1/3 = 1/m + 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
Another beautiful problem of yours, Max. Congrats!
$$\left\{ \matrix{
m,n\,\,{\rm{distinct}}\,\,\, \ge \,\,{\rm{1}}\,\,\,{\rm{ints}}\,\,\,\,\left( * \right) \hfill \cr
{1 \over 3} = {1 \over m} + {1 \over n}\,\,\,\left( {**} \right) \hfill \cr} \right.$$
$$? = m + n$$
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)
$$\left. {\left\{ \matrix{
\,x = 1\,\,\, \Rightarrow \,\,\,n = 5\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,{1 \over 3} = {1 \over m} + {1 \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,{1 \over m} = {1 \over 3} - {1 \over 5} = {5 \over {15}} - {3 \over {15}} = {2 \over {15}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,m\,\,\,{\rm{not}}\,\,{\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}\,\,\, \hfill \cr
\,x = 2\,\,\, \Rightarrow \,\,\,n = 4\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,{1 \over 3} = {1 \over m} + {1 \over 4}\,\,\,\,\, \Rightarrow \,\,\,\,{1 \over m} = {1 \over 3} - {1 \over 4}\,\, = {4 \over {12}} - {3 \over {12}} = {1 \over {12}}\,\,\,\,\, \Rightarrow \,\,\,\,m = 12\,\,\, \hfill \cr} \right.\,\,} \right\}\,\,\,\, \Rightarrow \,\,\,\,? = 12 + 4 = 16$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
