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

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==Problem==
 
==Problem==
  
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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?
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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?
  
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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>
  
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==Solution==
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==Solution 1==
  
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We calculate more terms:
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We calculate more terms: <cmath>1,3,4,7,11,18,\ldots.</cmath> We find a pattern: if <math>n+2</math> is a multiple of <math>3</math>, then the term is even, or else it is odd. There are <math>\left\lfloor \frac{2023}{3} \right\rfloor =\boxed{\textbf{(E) }674}</math> multiples of <math>3</math> from <math>1</math> to <math>2023</math>.
  
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<math>1,3,4,5,9,14,...</math>
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~Mintylemon66 ~minor edit by the_eaglercraft_grinder
  
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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.
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==Solution 2==
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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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Like in the other solution, we find a pattern, except in a more rigorous way.
 +
Since we start with <math>1</math> and <math>3</math>, the next term is <math>4</math>.
  
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~Mintylemon66
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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 <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{(E) }674}</math> 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==
 +
{{AMC10 box|year=2023|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 20:31, 9 July 2024

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 (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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