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

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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, 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 $$m+n$$?

$$A. 9$$
$$B. 12$$
$$C. 16$$
$$D. 18$$
$$E. 20$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Community Reply Senior PS Moderator Status: It always seems impossible until it's done. Joined: 16 Sep 2016 Posts: 737 GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha [#permalink] ### Show Tags 02 Nov 2018, 01:43 3 2 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 $$m+n$$? $$A. 9$$ $$B. 12$$ $$C. 16$$ $$D. 18$$ $$E. 20$$ _________________ Regards, Gladi “Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back) ##### General Discussion GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 If 1/3 = 1/m + 1/n, where m and n are different positive integers, wha [#permalink] ### Show Tags 02 Nov 2018, 12:33 1 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 :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8104 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, 18:02 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 Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 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, 20:19 => $$\frac{1}{3} = \frac{1}{m} + \frac{1}{n}$$ $$=> mn = 3n + 3m$$ $$=> mn – 3m – 3n = 0$$ $$=> mn – 3m – 3n + 9 = 9$$ $$=> (m – 3)(n – 3) = 9$$ 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$79 for 1 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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05 Nov 2018, 00:43
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  [#permalink]

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06 Nov 2018, 09:18
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, 20:11
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?
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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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22 Nov 2018, 21:43
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 $$(m-3, n-3)$$ 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?

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  [#permalink]

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24 Nov 2018, 06:56
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,

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  [#permalink]

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24 Nov 2018, 07:03
1
dave13 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.

$$\frac{1}{3} = \frac{1}{m} + \frac{1}{n}$$
$$\frac{1}{3} = \frac{(m+n)}{mn}$$
$$\frac{mn}{3} = (m+n)$$
$$mn = 3(m+n)$$
$$mn = 3m+3n$$
$$mn -3m= 3n$$
$$m(n-3) = 3n$$
$$m = \frac{3n}{(n-3)}$$

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,

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  [#permalink]

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24 Nov 2018, 11:03
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,
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   [#permalink] 24 Nov 2018, 11:03
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