2023 AMC 10B Problems/Problem 6

Problem

Let $L_{1}=1, L_{2}=3$ , and $L_{n+2}=L_{n+1}+L_{n}$ for $n\geq 1$ . How many terms in the sequence $L_{1}, L_{2}, L_{3},...,L_{2023}$ are even?

$\textbf{(A) }673\qquad\textbf{(B) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674$

Solution 1

We calculate more terms:

$1,3,4,7,11,18,...$

We find a pattern: if $n$ is a multiple of $3$ , then the term is even, or else it is odd. There are $\lfloor \frac{2023}{3} \rfloor =\boxed{\textbf{(E) }674}$ multiples of $3$ from $1$ to $2023$ .

~Mintylemon66

Solution 2

Like in the other solution, we find a pattern, except in a more rigorous way. Since we start with $1$ and $3$ , the next term is $4$ .

We start with odd, then odd, then (the sum of odd and odd) even, (the sum of odd and even) odd, and so on. Basically the pattern goes: odd, odd, even, odd odd, even, odd, odd even…

When we take $\frac{2023}{3}$ we get $674$ with a remainder of one. So we have $674$ full cycles, and an extra odd at the end.

Therefore, there are $\boxed{\textbf{(E) }674}$ evens.

~e_is_2.71828

Video Solutions

https://www.youtube.com/watch?v=cT-0V4a3FYY ~SpreadTheMathLove

https://youtu.be/wdNGZpTrjxY ~e_is_2.71828

2023 AMC 10B (Problems • Answer Key • Resources)
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2023 AMC 10B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

Video Solutions

See also