2023 AMC 10B Problem 6

The terms get huge fast, but we only care whether each one is odd or even. Adding two numbers, whether the answer is odd or even depends only on whether the two numbers were odd or even.

1. Write out the first few terms.
$$1,\ 3,\ 4,\ 7,\ 11,\ 18,\ 29,\ 47,\ 76,\ \ldots$$
2. Mark odd (O) and even (E).

3. See why the pattern keeps going. Each term is the sum of the two before it:
- odd $+$ odd $=$ even,
- odd $+$ even $=$ odd,
- even $+$ odd $=$ odd,

and then we are back to "odd, odd", so the same three steps happen again. The pattern O, O, E repeats forever.
4. So a term is even exactly when its position is a multiple of $3$.
5. Count the multiples of $3$ from $1$ to $2023$. They are $3, 6, 9, \ldots, 2022$. Since $2022 = 3 \times 674$, there are $674$ of them. ($2023$ itself is not a multiple of $3$.)

The answer is $\boxed{\textbf{(E)}\ 674}$.

In a sequence where each term is built from the terms just before it, "odd or even" also follows a rule that depends only on the terms just before it. There are only a few possible odd/even pairs, so the pattern has to repeat. Once you find the repeating block, the question becomes counting positions, which is easy even for a far-away term like number $2023$.

Only parity matters, and the parity of each term is determined by the parities of the two before it. The first few terms are
$$1,\ 3,\ 4,\ 7,\ 11,\ 18,\ 29,\ 47,\ 76,\ \dots$$
with parities odd, odd, even, odd, odd, even, $\dots$ Because odd $+$ odd $=$ even, odd $+$ even $=$ odd, and even $+$ odd $=$ odd, the pattern (odd, odd, even) repeats forever with period $3$.

So $L_n$ is even exactly when $n$ is a multiple of $3$. The multiples of $3$ from $1$ to $2023$ are $3, 6, \dots, 2022$, and there are
$$\left\lfloor \frac{2023}{3} \right\rfloor = 674$$
of them.

The answer is $\boxed{\textbf{(E)}\ 674}$.