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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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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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Question Stats: 42% (02:28) correct 58% (02:24) wrong based on 174 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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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]

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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,

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,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
General Discussion
GMATH Teacher P
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Joined: 12 Oct 2010
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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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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  [#permalink]

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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.

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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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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=>

$$\frac{1}{3} = \frac{1}{m} + \frac{1}{n}$$
$$=> mn = 3n + 3m$$
$$=> mn – 3m – 3n = 0$$
$$=> mn – 3m – 3n + 9 = 9$$
$$=> (m – 3)(n – 3) = 9$$

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.
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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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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 V
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]

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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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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, $$m + n = 16.$$

Therefore, the answer is 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  [#permalink]

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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, $$m + n = 16.$$

Therefore, the answer is 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  [#permalink]

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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,

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 Manager  P
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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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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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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]

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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 _________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back) 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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