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