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In how many different ways can 2/15 be represented as 1/A + 1/B, if A and B are positive integers with A less than or equal to B

 Jan 21, 2019
edited by Guest  Jan 21, 2019
 #1
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+1

Is this what you mean?

 

1/8  + 1 / 120    = 2/15
1/9  +  1 / 45     = 2/15
1/10  +  1 / 30   = 2/15
1/12  +  1 / 20   = 2/15

1/15  +  1/15     = 2/15

 Jan 22, 2019
edited by Guest  Jan 22, 2019
 #2
avatar+24093 
+8

In how many different ways can 2/15 be represented as 1/A + 1/B,

if A and B are positive integers with A less than or equal to B

 

\(\text{For odd $n>2$ there is always at least one decomposition into exactly two unit fractions: $\dfrac{2}{n} = \dfrac{1}{A} + \dfrac{1}{B}$ } \\ \text{Finding all possibilities.}\\ \text{The prime factorization of $n^2$ results in all possible decompositions into two unit fractions.} \)

 

\(n=15\ \text{is odd} \\ n^2 =225 \\ \text{All divisors of $n^2=225$ are: $1,\ 3,\ 5,\ 9,\ 15,\ 25,\ 45,\ 75,\ 225$}\\ \text{Let $n^2=p\times q$ }\)

 

\(\begin{array}{|r|r|r|c|c|c|c|c|c|c|c|c| } \hline & p & q & n^2 & s & t & r & k & A & B & A\le B & \\ & & & =p\cdot q & = \frac{p+q}{2} & = \frac{p-q}{2} & =\frac{t}{2} & = \frac{15+\sqrt{15^2+t^2} }{2} & =k-r & =k+r & \\ \hline 1. & 225 & 1 & 225=225 \cdot 1 & 113 & 112 & 56 & 64 & 8 & 120 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{8} + \dfrac{1}{120}} \\ \hline 2. & 75 & 3 & 225= 73 \cdot 3 & 39 & 36 & 18 & 27 & 9 & 45 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{9} + \dfrac{1}{45}} \\ \hline 3. & 45 & 5 & 225= 45 \cdot 5 & 25 & 20 & 10 & 20 & 10 & 30 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{10} + \dfrac{1}{30}} \\ \hline 4. & 25 & 9 & 225= 25 \cdot 9 & 17 & 8 & 4 & 16 & 12 & 20 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{12} + \dfrac{1}{20}} \\ \hline 5. & 15 & 15 & 225= 15 \cdot 15 & 15 & 0 & 0 & 15 & 15 & 15 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{15} + \dfrac{1}{15}} \\ \hline \end{array}\)

 

laugh

 Jan 22, 2019
edited by heureka  Jan 22, 2019
edited by heureka  Jan 22, 2019

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