Digital Fibonacci

If we start writing down the last digits of the Fibonacci num- bers, we notice that they repeat in cycles of length sixty. Since $123456789 = 60(2057613) + 9$ , the last digit of the $123456789$ th Fibonacci number is the last digit of the 9th Fibonacci number, or $4$ .

Writing down the first $10$ fibonacci numbers we get $1,1,2,3,0,3,3,1,4,0...\mod 5$ . The next two numbers would be $4$ and then $4$ again which we can write as or $-1,-1\mod5$ which would lead to the repetition of the above sequence but negative, and the next ten digits would be the negative of those digits, or just the original $10$ digits, so the fibonacci sequence, $\mod5$ , repeats every $20$ numbers,

$123456789\equiv9 \mod 20$

So $F_{123456789}\equiv F_{9}\equiv4 \mod5$

The fibonacci sequence is $1,1,0,1,1,0,...\mod2$ repeating every $3$ ,

since $123456789\equiv0\equiv3 \mod 3$

we have $F_{123456789}\equiv F_{3}\equiv0 \mod 2$

Knowing $F_{123456789}\equiv4 \mod 5$ and $F_{123456789}\equiv0 \mod 2$ ,

the CRT gives $F_{123456789}\equiv\boxed{4}\mod 10$

We can just observe the patterns through the list of fibonacci numbers. As the unit digit 0 appears in cycles of 15. Also if we divide 123456789 by 15 we get a remainder of 9. Then we can observe the patterns of the digits of Fibonacci numbers that are in a cycle of like 9,15+9,15+9+9,........(in fibonacci numbers).Then we can get a pattern of like 8,1,9,4,8,6,2,4,8,6,2,4.........(as the pattern of 8,6,2,4 continues).Thus we divide 12345679 by 60 and then we can observe the remainder which is 9 and thus we put the digit of the 9th fibonacci number.