Fibonacci Sequence in amazing!

Let, $p(x)$ be the period for the fibonacci sequence in $\text{mod x}$

It is easy to find that $p(2)=3$ $\implies [1,1,0]$ , and $p(5)=20 \implies [1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0]$

To find $p(10)$ , we need to find when the fibonacci sequence in $\text{ mod 2}$ and $\text{ mod 5}$ meets (since $a \text{ (mod 10)}= a \text{ (mod 2)}$ and $a \text{ (mod 10)}= a \text{ (mod 5)}$ ), so it is equal with $LCM(p(2),p (5))$ , then $p(x)=LCM (3,20)=60$

Generally, $p(x_1 \times x_2\times ... \times x_i)=LCM(p (x_1),p (x_2),...,p (x_n) )$

solution in python:

>> a='11'
>> while a[-2:]not in a[:-1]:a+=str((int(a[-2])+int(a[-1]))%10)

>> a
'11235831459437077415617853819099875279651673033695493257291011'
>> print(len(a)-2)
60