2023 AMC 10B Problem 6

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 $\boxed{\textbf{(E)}\ 674}$.

A linear recursion taken modulo $2$ (or modulo any $m$) depends only on the previous residues, so it must eventually cycle; here the cycle appears immediately and has length $3$. Reducing to residues turns a question about huge numbers into counting positions in a short repeating pattern, exactly as with Fibonacci numbers.

• Rounding 2023/3 the wrong way (675) or assuming the last term is even; 2023 is not a multiple of 3, so the last even term is L_2022.
• Counting the odd terms ($2023 - 674 = 1349$) or the positions $n \equiv 1 \pmod 3$ by mistake.