2018 Running

Algebra Level 3

Find a + b a + b where the continued fraction has recurring 2-0-1-8 pattern as shown below: a b = 4 + 1 2 + 1 0 + 1 1 + 1 8 + 1 2 + 1 0 + 1 1 + 1 8 + \sqrt{\frac{a}{b}} = 4 + \cfrac{1}{2 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{8+\cfrac{1}{2+\cfrac{1}{0+\cfrac{1}{1+\cfrac{1}{8 + _\ddots}}}}} } } }

a a and b b are coprime positive integers.


The answer is 59.

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1 solution

Zico Quintina
May 9, 2018

We write a b = 4 + x \sqrt{\dfrac{a}{b}} = 4 + x , so that

x = 1 2 + 1 0 + 1 1 + 1 8 + 1 2 + 1 0 + 1 1 + 1 8 + . . . . x = 1 2 + 1 0 + 1 1 + 1 8 + x = 1 2 + 1 0 + x + 8 x + 9 = 1 2 + x + 9 x + 8 = x + 8 3 x + 25 \begin{aligned} x &= \cfrac{1}{2 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{8 + \cfrac{1}{2 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{8 + .... }}}}}}}} \\ \\ x &= \cfrac{1}{2 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{8 + x }}}} \\ \\ &= \cfrac{1}{2 + \cfrac{1}{0 + \dfrac{x + 8}{x + 9}}} \\ \\ &= \cfrac{1}{2 + \dfrac{x + 9}{x + 8}} \\ \\ &= \dfrac{x + 8}{3x + 25} \end{aligned}

This leads to 3 x 2 + 24 x 8 = 0 3x^2 + 24x - 8 = 0 , which yields 3 ( x + 4 ) 2 56 = 0 3(x+4)^2 - 56 = 0 and x = - 4 ± 56 3 x = \text{-}4 \pm \sqrt{ \dfrac{56}{3} } . We discard the negative root as the continued fraction is clearly positive.

Thus a b = x + 4 = 56 3 \sqrt{ \dfrac{a}{b} } = x + 4 = \sqrt{ \dfrac{56}{3} } , so a + b = 59 a + b = \boxed{59} .

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