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# If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha

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Math Revolution GMAT Instructor
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If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha  [#permalink]

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02 Nov 2018, 00:32
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85% (hard)

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44% (02:35) correct 56% (02:02) wrong based on 103 sessions

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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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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" PS Forum Moderator Status: It always seems impossible until it's done. Joined: 16 Sep 2016 Posts: 383 GMAT 1: 740 Q50 V40 Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha [#permalink] ### Show Tags 02 Nov 2018, 00:43 1 1 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 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 $$m+n$$? $$A. 9$$ $$B. 12$$ $$C. 16$$ $$D. 18$$ $$E. 20$$ GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 484 If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha [#permalink] ### Show Tags 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. _________________ Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Nov with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount! Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4170 Location: United States (CA) Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha [#permalink] ### Show Tags 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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6529 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha [#permalink] ### Show Tags 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 $$(m-3, n-3)$$ 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha  [#permalink]

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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?
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6529
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha  [#permalink]

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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  [#permalink]

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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 $$(m-3, n-3)$$ 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, $$m + n = 16.$$

Hi Math Revolution,

Could you please explain how you went from mn – 3m – 3n = 0 to mn – 3m – 3n + 9 = 9?
Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha &nbs [#permalink] 16 Nov 2018, 19:11
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