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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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New post 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  [#permalink]

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New post 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 \(m+n\)?

\(A. 9\)
\(B. 12\)
\(C. 16\)
\(D. 18\)
\(E. 20\)
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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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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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
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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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New post 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  [#permalink]

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New post 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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New post 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, \(m + n = 16.\)

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 &nbs [#permalink] 16 Nov 2018, 19:11
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