Define a sequence recursively by $F_0=0$, $F_1=1$, and $F_n=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by 3, for all $n \ge 2$. Thus the sequence starts $0, 1, 1, 2, 0, 2, ...$. What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$?

For anyone stuck, don't try to list terms forever. The remainders mod 3 repeat in a cycle. Find the period, then add one full cycle's sum and multiply.

The sequence starts 0,1,1,2,0,2,2,1,... I noticed the sum of the first 8 is 9. Since 2024-2017+1=8, does that mean the answer is just 9? Feels too easy for 655-705.

Why is the answer not 8? I got 8 because I summed 0+1+1+2+0+2+2+1=9? Wait, I think I misadded. Let me recheck.

Took me a minute to realize 2017 is not a multiple of the period. I initially thought the sum would be from the start of the cycle. 2017 mod 8 = 1, so the block starts at the second term. Still same sum though because it's a full cycle.