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

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

@@ Line 3: / Line 3: @@
 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?
-<math>\textbf{(A) }673\qquad\textbf{(B)} 674\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }1011</math>
+<math>\textbf{(A) }673\qquad\textbf{(B) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674</math>
-On my copy of the AMC 10B, the order of the answers is different, so the correct answer is (E) 674
 ==Solution 1==
-We calculate more terms:
+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>.
-<math>1,3,4,7,11,18,...</math>
-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.
-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 ~minor edit by the_eaglercraft_grinder
 ==Solution 2==
@@ Line 27: / Line 19: @@
 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.
-Therefore, there are <math>\boxed{\textbf{(B) }674}</math> evens.
+Therefore, there are <math>\boxed{\textbf{(E) }674}</math> 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 <math>\boxed{E. 674}</math>
+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 <math>\boxed{E. 674}</math>
+~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==
+{{AMC10 box|year=2023|ab=B|num-b=5|num-a=7}}
+{{MAA Notice}}

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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Difference between revisions of "2023 AMC 10B Problems/Problem 6"

Latest revision as of 19:04, 22 August 2025

Contents

Problem

Solution 1

Solution 2

Solution 3

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

Video Solutions

Video Solution

Video Solution

Video Solution by Interstigation

See also