When Fibonacci meets Euler

$a_{ 1 }= 0.1$

$a_{ 2 }= 0.01$

$a_{ 3 }= a_{ 2 }/10 + a_{ 1 }/100$

$a_{ 4 }= a_{ 3 }/10 + a_{ 2 }/100 \\ .\\ .\\ .\\ a_{ n }= a_{ n-1 }/10 + a_{ n-2 }/100$

$\\ .\\ .\\ .\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$

$\displaystyle \sum _{ i=1 }^{ \infty }{ a_{ i } } = S= \frac{S}{10} + \frac{S}{100} + \frac{ 9a_{ 1 }}{10} +a_{ 2 } \implies S = 10/89$

We want to tranformate the fraction $\frac { 10 }{ 89 }$ in one of the type $\frac { 10y }{ 99...999 }$ .

The smaller number that satisfy $89 \times y = 999...99$ will be $1/10$ of the number that represents the period.

$y = ({10 }^{ x }-1)/89 )$ such that $10y$ will have $x$ algarisms.

By Euler's Theorem:

${ 10 }^{ 88 } \equiv 1 \pmod{89} \implies { 10 }^{ 88/n }\equiv 1 \pmod{89}$

$88/n = x \implies x \in D(88) = \{ 1, 2, 4, 8, 11, 22, 44, 88\}$

Testing:
${ 10 }^{ 1 } \equiv 10 \pmod{89}$
${ 10 }^{ 2 } \equiv 11 \pmod{89}$
${ 10 }^{ 4 } \equiv 32 \pmod{89}$
${ 10 }^{ 8 }\equiv 45 \pmod{89}$
${ 10 }^{ 11 } \equiv 55 \pmod{89}$
${ 10 }^{ 22 } \equiv (-1) \pmod{89}$
${ 10 }^{ 44 }\equiv 1 \pmod{89}$

$\boxed{x=44}$