Please help me to find the needed sum:
If $a,b,c$ are positive integers such that $\;a + \dfrac{1}{b+\large \frac 1c} = \dfrac{37}{16},\;$ find the value of $\;(a+b+c)$.
Thanks!
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amWhy
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2Can you at least find $a$? – Empy2 Jul 11 '13 at 16:30
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Did you expand your expression out to get a single fraction with polynomials in $a,b,c$ top and bottom? – Lubin Jul 11 '13 at 16:37
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Yea expanded everything and got terribly stuck! – Raymond Chua Wei Ren Jul 11 '13 at 16:45
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Have you studied continued fractions? – Thomas Andrews Jul 11 '13 at 16:48
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Nope! Thanks guys! – Raymond Chua Wei Ren Jul 11 '13 at 16:57
3 Answers
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$$\;\color{blue}{\bf a} + \dfrac{1}{\color{red}{\bf b}+1/\color{green}{\bf c}} = \dfrac{37}{16} = 2 + \dfrac 5{16} = 2 + \dfrac 1{\frac{16}{5}} = \color{blue}{\bf 2} + \dfrac 1{\color{red}{\bf 3} + 1/\color{green}{\bf 5}}$$
Now, match up: $\;\color{blue}{\bf a = \;?}\;\quad \color{red}{\bf b = \;?},\quad \color{green}{\bf c = \;?}$
And then find the sum: $$\quad a + b + c = \;\;?$$
amWhy
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Wow, totally didn't think of this solution at all! Thanks! – Raymond Chua Wei Ren Jul 11 '13 at 16:42
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1You still need to argue that this solution is unique, which is not very hard but missing ;) – N. S. Jul 11 '13 at 16:47
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2Leaving out the sum at the end after you've written explicitly the values of $a,b,c$ seems a bit humorous. – Thomas Andrews Jul 11 '13 at 16:50
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This is asking for the Continued Fraction expansion of $\frac{37}{16}$. We can use the Euclid-Wallis Algorithm to compute the expansion: $$ \begin{array}{r} &&2&3&5\\\hline 1&0&1&-3&16\\ 0&1&-2&7&-37\\ 37&16&5&1&0\\ \end{array} $$ The continued fraction is above the horizontal line: $$ \frac{37}{16}=2+\cfrac1{3+\cfrac1{5}} $$ Then, $a+b+c=2+3+5=10$.
