2017 AMC 10A Problem 13

Index $2017$ is far too big to reach by hand, so look for a repeating pattern.

1. Each new term is the sum of the two before it, then take the remainder after dividing by $3$. Keep going from the start:

For example, $F_6 =$ remainder of $0 + 2 = 2$, and $F_8 =$ remainder of $2 + 1 = 3$, which is $0$.
2. At $n = 8, 9$ the pair $0, 1$ appears again, exactly like $F_0, F_1$. Each term depends only on the two before it, so from here the list repeats the same $8$ terms forever: $0, 1, 1, 2, 0, 2, 2, 1$.
3. The question adds $8$ terms in a row ($F_{2017}$ through $F_{2024}$). Any $8$ terms in a row cover each spot of the cycle exactly once, just starting at a different place. So their sum is the sum of one full cycle:
$$0 + 1 + 1 + 2 + 0 + 2 + 2 + 1 = 9.$$

The answer is $\boxed{\textbf{(D)}\ 9}$.

When each term is built from the two before it using remainders, there are only a few possible pairs, so the list must eventually repeat. Watch for the starting pair to come back; that gives the cycle length. A block of terms as long as the cycle always adds up to the same total, wherever it starts.

If you want to check which terms these are: $2017 = 8 \times 252 + 1$, so $F_{2017}$ matches $F_1 = 1$. The eight terms are then $F_1$ through $F_8$ in the cycle: $1, 1, 2, 0, 2, 2, 1, 0$, which add to $9$.

Continue the sequence, reducing each sum mod $3$:
$$0,\ 1,\ 1,\ 2,\ 0,\ 2,\ 2,\ 1,\ 0,\ 1,\ 1,\ 2,\ \ldots$$
After eight terms the pair $(0, 1)$ reappears at $F_8, F_9$. Since each term depends only on the previous two, the sequence repeats from there with period $8$.

The requested sum has exactly eight consecutive terms, so it covers one full period, in some rotation. Its value is the sum of one period:
$$0 + 1 + 1 + 2 + 0 + 2 + 2 + 1 = 9.$$

The answer is $\boxed{\textbf{(D)}\ 9}$.