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

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==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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<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 1==
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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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~Mintylemon66 ~minor edit by the_eaglercraft_grinder
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==Solution 2==
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Like in the other solution, we find a pattern, except in a more rigorous way.
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Since we start with <math>1</math> and <math>3</math>, the next term is <math>4</math>.
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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…
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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.
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Therefore, there are <math>\boxed{\textbf{(E) }674}</math> evens.
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~e_is_2.71828
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==Solution 3==
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We know that \( L_1 = 1 \) and \( L_2 = 3 \).
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To find \( L_3 \), we put \( n=1 \) into \( L_{n+1} = L_n + L_{n-1} \).
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This gives \( L_3 = 4 \). Continuing for \( L_4 \), \( L_5 \), and \( L_6 \), we get
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\( L_4 = 7 \)
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\( L_5 = 11 \)
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\( L_6 = 18 \).
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We can now make an educated guess that this pattern will always remain the same, and evens will always appear on the \( L_{3n} \)th number. (Proof is below).
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So, we take the \( \lfloor 2023/3 \rfloor \), which gives <math>\boxed{E. 674}</math>
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Note that \( \lfloor x \rfloor \) = to the nearest rounded down whole number.
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~Pinotation
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==Proof that all \( L_{3n} \) is even==
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We have that \( L_1 \) is an odd number and \( L_2 \) is also an odd number. The sum of two odd numbers is even. Now \( L_3 = L_2 + L_1 \), so \( L_3 \) must be even.
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We continue with proof by contradiction. Say \( L_{3n} \), \( n \not\le 1 \) was not even. We take \( L_6 \), for instance, and this should be odd. This means that because \( L_6 = L_5 + L_4 \), either \( L_5 \) or \( L_4 \) must be even, or else the number will be odd.
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So given that \( L_3 \) is even, then \( L_4 = L_3 + L_2 \), which is an even + odd, which is odd. So now \( L_5 \) must be even for this to not work.
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We know that \( L_4 \) is odd, and \( L_3 \) is even.
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We have \( L_5 = L_4 + L_3 \), and that is again an odd + even case, showing that \( L_5 \) is also odd. Therefore, if \( L_{3n} \), \( n \not\le 1 \) is not even, there lies a contradiction.
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Therefore, each \( L_{3n} \) number is even, and that is why we do \( \lfloor 2023/3 \rfloor \) to get <math>\boxed{E. 674}</math>
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~Proof By Pinotation
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==Video Solutions==
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==Video Solution==
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https://youtu.be/EuLkw8HFdk4?si=iNQdS6bI38MUha1I&t=1174
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~Math-X
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https://www.youtube.com/watch?v=cT-0V4a3FYY ~SpreadTheMathLove
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https://www.youtube.com/watch?v=wdNGZpTrjxY ~e_is_2.71828
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==Video Solution==
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https://youtu.be/DmE9mmTx3Fw
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution by Interstigation==
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https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
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==See also==
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{{AMC10 box|year=2023|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 19:04, 22 August 2025

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

Solution 3

We know that \( L_1 = 1 \) and \( L_2 = 3 \).

To find \( L_3 \), we put \( n=1 \) into \( L_{n+1} = L_n + L_{n-1} \).

This gives \( L_3 = 4 \). Continuing for \( L_4 \), \( L_5 \), and \( L_6 \), we get

\( L_4 = 7 \)

\( L_5 = 11 \)

\( L_6 = 18 \).

We can now make an educated guess that this pattern will always remain the same, and evens will always appear on the \( L_{3n} \)th number. (Proof is below).

So, we take the \( \lfloor 2023/3 \rfloor \), which gives $\boxed{E. 674}$

Note that \( \lfloor x \rfloor \) = to the nearest rounded down whole number.

~Pinotation

Proof that all \( L_{3n} \) is even

We have that \( L_1 \) is an odd number and \( L_2 \) is also an odd number. The sum of two odd numbers is even. Now \( L_3 = L_2 + L_1 \), so \( L_3 \) must be even. We continue with proof by contradiction. Say \( L_{3n} \), \( n \not\le 1 \) was not even. We take \( L_6 \), for instance, and this should be odd. This means that because \( L_6 = L_5 + L_4 \), either \( L_5 \) or \( L_4 \) must be even, or else the number will be odd.

So given that \( L_3 \) is even, then \( L_4 = L_3 + L_2 \), which is an even + odd, which is odd. So now \( L_5 \) must be even for this to not work.

We know that \( L_4 \) is odd, and \( L_3 \) is even.

We have \( L_5 = L_4 + L_3 \), and that is again an odd + even case, showing that \( L_5 \) is also odd. Therefore, if \( L_{3n} \), \( n \not\le 1 \) is not even, there lies a contradiction.

Therefore, each \( L_{3n} \) number is even, and that is why we do \( \lfloor 2023/3 \rfloor \) to get $\boxed{E. 674}$

~Proof By Pinotation

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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