Find remainder of $F_n$ when divided by $5$ 
Let $\{ F_n\}$ be the sequence of numbers defined by $F_1=1=F_2;\, F_{n+1}=F_n+F_{n-1}$ for $n \geq 2$. Let $f_n$ be the remainder left when $F_n$ is divided by $5$. Then $f_{2000}$ equals
(A) $0$ $~~~~~~~~~~~~~~~~$ (B) $1$ $~~~~~~~~~~~~~~~~$ (C) $2$$~~~~~~~~~~~~~~~~$ (D) $3$

I found that $F(1)=1$, $F(2)=1$, $F(3)=2$, $F(4)=3$, $F(5)=5$, $F(6)=8$, $F(7)=13$, $F(8)=21$, $F(9)=34$, and $F(10)=55$. But I need a systematic pattern to find $F_n$.
 A: $\rm{\bf Hint}\ \ \  mod\ 5\!:\,\ F_{k}\!\equiv\color{#0A0}0\:\Rightarrow\:F_{k+5}\!\equiv\color{#C00}0\ \ \ by\ \ \  \begin{array}{|c|c|c|c|c|c|} \hline
\rm k&\rm k\!+\!1 &\rm k\!+\!2 &\rm k\!+\!3 &\rm k\!+\!4 &\rm k\!+\!5\\ \hline
 \color{#0A0}0&\rm n       &\rm    n    &\rm    2n   &\rm       3n  &\rm \color{#C00}0 \\\hline\end{array}$
A: Now the pattern for the remainder is,
$f_{1}=1$, $f_{2}={1}$, $f_{3}={2}$, $f_{4}={3}$ now we have $f_{5}=0$. So this is our first zero. If you proceed in this manner you will see that $f_{10}=0$ and $f_{15}=0$ and so on. So, $f_{2000}=0$.
A: $$F_n=F_{n-1}+F_{n-2}=F_{n-2}+F_{n-3}+F_{n-2}=2F_{n-2}+F_{n-1}$$
$$=2(F_{n-3}+F_{n-4})+F_{n-3}=3F_{n-3}+2F_{n-4}$$
$$=3(F_{n-4}+F_{n-5})+2F_{n-4}=5F_{n-4}+3F_{n-5}$$
$$\text{ So,} F_n\equiv 3F_{n-5}\pmod 5 $$
As $F_1=F_2=1\implies F_0=F_2-F_1=0\equiv0\pmod 5$
$\implies 5\mid F_n$ if $5\mid n$

Alternatively,  from Lemma#$5$ of this , $F_n\mid F_m \iff n\mid m\text{ or } n=2$ for $m\ge n\ge 1$
We find $F_2=F_1=1,F_3=2,F_4=3, F_5=5$ 
So, $F_5\mid F_m \iff 5\mid m$ for $m\ge1$
A: Hint:
Show by induction that $F_n \equiv 2n3^n\mod 5$.
