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

#Algebra

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Comments

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 $\dfrac{2011}{1990}.$

Brian Charlesworth - 5 years, 6 months ago

Very nice! Thank you very much Sir.

A Former Brilliant Member - 5 years, 6 months ago

Continued fractions method gives a = 1, b = 94, c = 1, d = 3, and e = 5.Hence answer is 104.

Rajen Kapur - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$