A number theory problem by Budi Utomo

Number Theory Level 3

$\dfrac{21197}{2097} = a + \dfrac1{b + \dfrac2{c + \dfrac3{d + \dfrac4{e+5}}}}$

Given that $a,b,c,d$ and $e$ are positive integers satisfying the equation above, and if the value of the fraction below

$1 + \dfrac a{2 + \dfrac b{3 + \dfrac c{4 + \dfrac d{5+e}}}}$

is equal to $\dfrac xy$ , where $x$ and $y$ are coprime positive integers, find $x+y$ .

The answer is 4292.

1 solution

Chew-Seong Cheong
Jan 16, 2016

$\begin{array} {} \dfrac{21197}{2097} & = 10 \frac{227}{2097} & \Rightarrow a = 10 \\ & = 10 + \dfrac{1}{\frac{2097}{227}} = 10 + \dfrac{1}{9\frac{54}{227}} & \Rightarrow b = 9 \\ & = 10 + \dfrac{1}{9 + \dfrac{2}{\frac{227}{27}}} = 10 + \dfrac{1}{9 + \dfrac{2}{8\frac{11}{27}}} & \Rightarrow c = 8 \\ & = 10 + \dfrac{1}{9 + \dfrac{2}{8 + \dfrac{3}{7\frac{4}{11}}}} & \Rightarrow d = 7 \\ & = 10 + \dfrac{1}{9 + \dfrac{2}{8 + \dfrac{3}{7 + \dfrac{4}{6+5}}}} & \Rightarrow e = 6 \end{array} $

Now, we have:

$\begin{aligned} 1 + \frac{a}{2 + \frac{b}{3 + \frac{c}{4 + \frac{d}{5+e}}}} & = 1 + \frac{10}{2 + \frac{9}{3 + \frac{8}{4 + \frac{7}{5+6}}}} = 1 + \frac{10}{2 + \frac{9}{3 + \frac{8}{\frac{51}{11}}}}\\ = 1 + \frac{10}{2 + \frac{9}{\frac{241}{51}}}= 1 + \frac{10}{\frac{941}{241}} = \frac{3351}{941} \end{aligned}$

Since 941 is a prime, the 3351 and 941 are coprimes therefore $x+y = 3351+941 = \boxed{4292}$

can you plz tell how you moved from 1st to 2nd line??

Chaitnya Shrivastava - 5 years, 5 months ago

I have added more info in the solution.

Chew-Seong Cheong - 5 years, 5 months ago

Very well and thank you

Chaitnya Shrivastava - 5 years, 5 months ago

A number theory problem by Budi Utomo

The answer is 4292.

1 solution

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