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$\large a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d+\dfrac{1}{e}}}}=\dfrac{2011}{1990}$
Find $$\large a+b+c+d+e.$$

Note by Abhay Kumar
9 months, 2 weeks ago

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I assume that $$a,b,c,d,e$$ must be positive integers. Given that, we first note that the equation can be written as

$$a + \dfrac{1}{x} = \dfrac{2011}{1990} \lt 2,$$ where $$x \gt b \ge 1.$$

Thus the only possible value for $$a$$ is $$1,$$ in which case

$$\dfrac{1}{x} = \dfrac{21}{1990} \Longrightarrow x = \dfrac{1990}{21} \Longrightarrow b + \dfrac{1}{y} = 94 + \dfrac{16}{21},$$

where $$y \gt c \ge 1.$$ Thus the only possible value for $$b$$ is $$94,$$ in which case

$$\dfrac{1}{y} = \dfrac{16}{21} \Longrightarrow y = \dfrac{21}{16} \Longrightarrow c + \dfrac{1}{z} = 1 + \dfrac{5}{16},$$

where $$z \gt d \ge 1.$$ Thus the only possible value for $$c$$ is $$1,$$ in which case

$$\dfrac{1}{z} = \dfrac{5}{16} \Longrightarrow z = \dfrac{16}{5} \Longrightarrow d + \dfrac{1}{e} = 3 + \dfrac{1}{5},$$

for which $$d = 3, e = 5$$ are the unique solutions. Thus $$a + b + c + d + e = 1 + 94 + 1 + 3 + 5 = \boxed{104}.$$

Note that $$[1;94,1,3,5]$$ is the "continued fraction" representation of $$\dfrac{2011}{1990}.$$ · 9 months, 2 weeks ago

Very nice! Thank you very much Sir. · 9 months, 2 weeks ago