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Difference between revisions of "2023 AMC 10B Problems/Problem 6"
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We find a pattern: if <math>n</math> is a multiple of <math>3</math>, then the term is even, or else it is odd.  
 
We find a pattern: if <math>n</math> is a multiple of <math>3</math>, then the term is even, or else it is odd.  
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There are <math>\lfloor \frac{2023}{3} \rfloor =\boxed{\textbf{(B) }674</math> multiples of <math>3</math> from <math>1</math> to <math>2023</math>.  
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There are <math>\lfloor \frac{2023}{3} \rfloor =\boxed{\textbf{(B) }674}</math> multiples of <math>3</math> from <math>1</math> to <math>2023</math>.  
  
 
~Mintylemon66
 
~Mintylemon66

Revision as of 17:10, 15 November 2023

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?

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

Solution

We calculate more terms:

$1,3,4,5,9,14,...$

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

~Mintylemon66

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