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In how many different ways can \(\frac{2}{15}\) be represented as \(\frac{1}{a} + \frac{1}{b}\), if \(a\) and \(b\) are positive integers with \(a \ge b\)?

 

I chose to use Simon's Favorite Factoring trick. 

I let 2/15 = 1/a + 1/b. Multiplied both sides by 15/2 and ab: ab = 15/2*(a) + 15/2*(b).

Subtracted 15/2*(a) + 15/2*(b) from both sides: ab - 15/2*(a) - 15/2*(b) = 0. Then added 225/4: ab - 15/2*(a) - 15/2(b) + 225/4 = 225/4. I factored the left side: (a - 15/2)*(b - 15/2) = 225/4. And finally, I'm stuck on what to do next. :(

 Jun 21, 2018
 #1
avatar+98196 
+2

2/15  =

 

1/8  + 1 / 120

 

1/ 9  +   1 / 45

 

1/10  +  1  /30

 

1/ 12  + 1 / 20

 

 

I can generate the first through something called "Egyptian Fractions"....I don't know how the other three are obtained

 

You can read here : http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html

 

 

cool cool cool

 Jun 21, 2018
 #2
avatar+35 
+2

Ummm, ok thanks. Though I think it would've been better to multiply both sides by four. And then, since \(225=3^2\cdot5^2\), there are \(3\cdot3=9\) positive integer divisors of \(225 \). Of these, \(\boxed{5}\) yield ordered pairs of divisors \((2a-15,2b-15)\) for which \(a \geq b\)

Toytrain  Jun 21, 2018
 #3
avatar+21869 
+2

In how many different ways can
\(\frac{2}{15} \)
be represented as
\(\frac{1}{a} + \frac{1}{b}\),
if and are positive integers with
\(a \ge b\) ?

 

Because \(n = 15\) is odd:

we calculate \(n^2 = 225\)

And all divisors of \(n^2 = 225 \) are: \(1,3,5,9,15,25,45,75,225\)

\(\text{Let $n^2 = p\times q=225 $ } \)

 

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

 

\(\begin{array}{|r|r|c|} \hline & \frac{1}{a} + \frac{1}{b} & a \ge b\ \\ \hline 1. & \frac{1}{120} + \frac{1}{8} & \checkmark \\ 2. & \frac{1}{45} + \frac{1}{9} & \checkmark \\ 3. & \frac{1}{30} + \frac{1}{10} & \checkmark \\ 4. & \frac{1}{20} + \frac{1}{12} & \checkmark \\ 5. & \frac{1}{15} + \frac{1}{15} & \checkmark \\ \hline \end{array} \)

 

 

laugh

 Jun 21, 2018

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