\[\large a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d+\dfrac{1}{e}}}}=\dfrac{2011}{1990}\]

Find \(\large a+b+c+d+e.\)

Find \(\large a+b+c+d+e.\)

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TopNewestI 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 · 1 year, 3 months ago

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– Abhay Kumar · 1 year, 3 months ago

Very nice! Thank you very much Sir.Log in to reply

Continued fractions method gives a = 1, b = 94, c = 1, d = 3, and e = 5.Hence answer is 104. – Rajen Kapur · 1 year, 3 months ago

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