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If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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02 Nov 2018, 00:32
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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 \(m+n\)? \(A. 9\) \(B. 12\) \(C. 16\) \(D. 18\) \(E. 20\)
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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02 Nov 2018, 00:43
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. \(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\) \(\frac{1}{3} = \frac{(m+n)}{mn}\) \(\frac{mn}{3} = (m+n)\) \(mn = 3(m+n)\) \(mn = 3m+3n\) \(mn 3m= 3n\) \(m(n3) = 3n\) \(m = \frac{3n}{(n3)}\) 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 wrote: [ 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 \(m+n\)? \(A. 9\) \(B. 12\) \(C. 16\) \(D. 18\) \(E. 20\)
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If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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02 Nov 2018, 11:33
MathRevolution wrote: [ 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.
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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03 Nov 2018, 17:02
MathRevolution wrote: [ 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 \(m+n\)? \(A. 9\) \(B. 12\) \(C. 16\) \(D. 18\) \(E. 20\) Since 1/3 = 4/12, we see that 4/12 = 3/12 + 1/12 = 1/4 + 1/12; thus, m = 4 and n = 12 (or m = 12 and n = 4). Therefore, m + n = 16. Alternate Solution: We have 1/3 = 1/6 + 1/6, but since m and n are different, m = n = 6 is not possible. On the other hand, both m and n cannot be greater than 6 at the same time, because otherwise, 1/m + 1/n would be a number that is less than 1/3. Furthermore, none of the numbers can be less than or equal to 3, because otherwise, 1/m + 1/n would be a number that is greater than 1/3. Therefore, either of m or n must equal 4 or 5. Let’s try m = 4. 1/3 = 1/4 + 1/n 1/n = 1/3  1/4 = 1/12 n = 12 If m = 4 and n = 12, we have 1/4 + 1/12 = 4/12 = 1/3, which is the desired result. Thus, m + n = 4 + 12 = 16. Answer: C
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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04 Nov 2018, 19:19
=> \(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\) \(=> mn = 3n + 3m\) \(=> mn – 3m – 3n = 0\) \(=> mn – 3m – 3n + 9 = 9\) \(=> (m – 3)(n – 3) = 9\) The possible pairs of values \((m3, n3)\) giving a product of \(9\) are \((1,9), (9,1)\) and \((3,3)\). Case 1: \(m – 3 = 1, n – 3 = 9.\) We have \(m = 4, n = 12\) and \(m + n = 16\) Case 2: \(m – 3 = 9, n – 3 = 1\). We have \(m = 12, n = 4\) and \(m + n = 16.\) Case 3: \(m – 3 = 3, n – 3 = 3\) We have \(m = 6, n = 6.\) Since \(m = n\), this case does not satisfy the original condition. Thus, \(m + n = 16.\) Therefore, the answer is C. Answer: C
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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04 Nov 2018, 23:43
Hi. What is wrong with the following solution:
1/3 > 5/15 > 2/15 + 3/15 > 2/15 + 1/5
which gives us 15+5 > 20
why is this not the correct answer?



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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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06 Nov 2018, 08:18
Mansoor50 wrote: Hi. What is wrong with the following solution:
1/3 > 5/15 > 2/15 + 3/15 > 2/15 + 1/5
which gives us 15+5 > 20
why is this not the correct answer? 1/3 is expressed as a sum of two fractions with numerator 1, which are reciprocals of integers. 2/15 is not that kind of fraction.
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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16 Nov 2018, 19:11
MathRevolution wrote: =>
\(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\) \(=> mn = 3n + 3m\) \(=> mn – 3m – 3n = 0\) \(=> mn – 3m – 3n + 9 = 9\) \(=> (m – 3)(n – 3) = 9\)
The possible pairs of values \((m3, n3)\) giving a product of \(9\) are \((1,9), (9,1)\) and \((3,3)\).
Case 1: \(m – 3 = 1, n – 3 = 9.\) We have \(m = 4, n = 12\) and \(m + n = 16\)
Case 2: \(m – 3 = 9, n – 3 = 1\). We have \(m = 12, n = 4\) and \(m + n = 16.\)
Case 3: \(m – 3 = 3, n – 3 = 3\) We have \(m = 6, n = 6.\) Since \(m = n\), this case does not satisfy the original condition.
Thus, \(m + n = 16.\)
Therefore, the answer is C. Answer: C Hi Math Revolution, Could you please explain how you went from mn – 3m – 3n = 0 to mn – 3m – 3n + 9 = 9?



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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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22 Nov 2018, 20:43
csaluja wrote: MathRevolution wrote: =>
\(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\) \(=> mn = 3n + 3m\) \(=> mn – 3m – 3n = 0\) \(=> mn – 3m – 3n + 9 = 9\) \(=> (m – 3)(n – 3) = 9\)
The possible pairs of values \((m3, n3)\) giving a product of \(9\) are \((1,9), (9,1)\) and \((3,3)\).
Case 1: \(m – 3 = 1, n – 3 = 9.\) We have \(m = 4, n = 12\) and \(m + n = 16\)
Case 2: \(m – 3 = 9, n – 3 = 1\). We have \(m = 12, n = 4\) and \(m + n = 16.\)
Case 3: \(m – 3 = 3, n – 3 = 3\) We have \(m = 6, n = 6.\) Since \(m = n\), this case does not satisfy the original condition.
Thus, \(m + n = 16.\)
Therefore, the answer is C. Answer: C Hi Math Revolution, Could you please explain how you went from mn – 3m – 3n = 0 to mn – 3m – 3n + 9 = 9? Hi csaluja, It has been done by adding 9 to both sides of the equation : mn – 3m – 3n = 0 New Equation becomes  mn – 3m – 3n + 9 = 9. Then solve for m & n.
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If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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24 Nov 2018, 05:56
Gladiator59 wrote: 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. \(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\) \(\frac{1}{3} = \frac{(m+n)}{mn}\) \(\frac{mn}{3} = (m+n)\)\(mn = 3(m+n)\) \(mn = 3m+3n\) \(mn 3m= 3n\) \(m(n3) = 3n\) \(m = \frac{3n}{(n3)}\) 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 wrote: [ 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 \(m+n\)? \(A. 9\) \(B. 12\) \(C. 16\) \(D. 18\) \(E. 20\) hi Gladiator59 what prompted you to multiply both sides by mn and not to cross multiply like this \(\frac{1}{3} = \frac{(m+n)}{mn}\) > \(3(m+n)=mn\) ? thanks



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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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24 Nov 2018, 06:03
dave13 wrote: Gladiator59 wrote: 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. \(\frac{1}{3} = \frac{1}{m} + \frac{1}{n}\) \(\frac{1}{3} = \frac{(m+n)}{mn}\) \(\frac{mn}{3} = (m+n)\)\(mn = 3(m+n)\) \(mn = 3m+3n\) \(mn 3m= 3n\) \(m(n3) = 3n\) \(m = \frac{3n}{(n3)}\) 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 wrote: [ 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 \(m+n\)? \(A. 9\) \(B. 12\) \(C. 16\) \(D. 18\) \(E. 20\) hi Gladiator59 what prompted you to multiply both sides by mn and not to cross multiply like this \(\frac{1}{3} = \frac{(m+n)}{mn}\) > \(3(m+n)=mn\) ? thanks Hi dave13, Both are same, he did cross multiplication but in two different steps to express m in terms of n, by doing so we can try to predict a few possible values of n. Hope this helps.
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha
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24 Nov 2018, 10:03
Well, that is exactly what I have done. Just to show the intermediate steps, I have written it in two steps instead of one. Best, Gladi dave13 wrote: hi Gladiator59 what prompted you to multiply both sides by mn and not to cross multiply like this \(\frac{1}{3} = \frac{(m+n)}{mn}\) > \(3(m+n)=mn\) ? thanks
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha &nbs
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