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02 Nov 2018, 01:32
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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 \(m+n\)?
\(A. 9\)
\(B. 12\)
\(C. 16\)
\(D. 18\)
\(E. 20\)
If \(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\), where \(m\) and \(n\) are different positive integers, what is the value of \(m+n\)?
\(A. 9\)
\(B. 12\)
\(C. 16\)
\(D. 18\)
\(E. 20\)
02 Nov 2018, 01:43
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So, we have been given one equation with two unknowns. Hence we cannot solve it without considering the additional info given about m & n that they are two different integers.
\(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\)
\(\frac{1}{3} = \frac{(m+n)}{mn}\)
\(\frac{mn}{3} = (m+n)\)
\(mn = 3(m+n)\)
\(mn = 3m+3n\)
\(mn -3m= 3n\)
\(m(n-3) = 3n\)
\(m = \frac{3n}{(n-3)}\)
Minimum value of n is 4 as m is a positive integer.
n=4 -> m = 12 ( and we are done!)
Hence m + n = 4 + 12 = 16
Hence Option (C) is our choice!
Best,
Gladi
\(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\)
\(\frac{1}{3} = \frac{(m+n)}{mn}\)
\(\frac{mn}{3} = (m+n)\)
\(mn = 3(m+n)\)
\(mn = 3m+3n\)
\(mn -3m= 3n\)
\(m(n-3) = 3n\)
\(m = \frac{3n}{(n-3)}\)
Minimum value of n is 4 as m is a positive integer.
n=4 -> m = 12 ( and we are done!)
Hence m + n = 4 + 12 = 16
Hence Option (C) is our choice!
Best,
Gladi
MathRevolution
General Discussion
02 Nov 2018, 12:33
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MathRevolution
\(\left\{ \matrix{\\
m,n\,\,{\rm{distinct}}\,\,\, \ge \,\,{\rm{1}}\,\,\,{\rm{ints}}\,\,\,\,\left( * \right) \hfill \cr \\
{1 \over 3} = {1 \over m} + {1 \over n}\,\,\,\left( {**} \right) \hfill \cr} \right.\)
\(? = m + n\)
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.
