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: \[1,3,4,7,11,18,\ldots.\] We find a pattern: if $n+2$ is a multiple of $3$, then the term is even, or else it is odd. There are $\left\lfloor \frac{2023}{3} \right\rfloor =\boxed{\textbf{(E) }674}$ multiples of $3$ from $1$ to $2023$.

~Mintylemon66 ~minor edit by the_eaglercraft_grinder

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

Video Solution

https://youtu.be/EuLkw8HFdk4?si=iNQdS6bI38MUha1I&t=1174

~Math-X

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

https://www.youtube.com/watch?v=wdNGZpTrjxY ~e_is_2.71828

Video Solution

https://youtu.be/DmE9mmTx3Fw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

See also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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