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Difference between revisions of "2023 AMC 10B Problems/Problem 6"

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Let <math>L_{1}=1, L_{2}=3</math>, and <math>L_{n+2}=L_{n+1}+L_{n}</math> for <math>n\geq 1</math>. How many terms in the sequence <math>L_{1}, L_{2}, L_{3},...,L_{2023}</math> are even?
 
Let <math>L_{1}=1, L_{2}=3</math>, and <math>L_{n+2}=L_{n+1}+L_{n}</math> for <math>n\geq 1</math>. How many terms in the sequence <math>L_{1}, L_{2}, L_{3},...,L_{2023}</math> are even?
  
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<math>\textbf{(A) }673\qquad\textbf{(B)} 674\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }1011</math>
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<math>\textbf{(A) }673\qquad\textbf{(B) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674</math>
  
 
==Solution 1==
 
==Solution 1==
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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{(E) }674}</math> multiples of <math>3</math> from <math>1</math> to <math>2023</math>.  
  
 
~Mintylemon66
 
~Mintylemon66
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When we take <math> \frac{2023}{3}</math> we get <math>674</math> with a remainder of one. So we have <math>674</math> full cycles, and an extra odd at the end.  
 
When we take <math> \frac{2023}{3}</math> we get <math>674</math> with a remainder of one. So we have <math>674</math> full cycles, and an extra odd at the end.  
  
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Therefore, there are <math>\boxed{\textbf{(B) }674}</math> evens.
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Therefore, there are <math>\boxed{\textbf{(E) }674}</math> evens.
  
 
~e_is_2.71828
 
~e_is_2.71828

Revision as of 21:29, 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?

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

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

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