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

Revision as of 17:56, 1 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 by Math-X (First understand the problem!!!)

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 7: / Line 7: @@
 ==Solution 1==
-We calculate more terms: <cmath>1,3,4,7,11,18,\ldots.</cmath> 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>\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>.
+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>.
-~Mintylemon66
+~Mintylemon66 ~minor edit by the_eaglercraft_grinder
 ==Solution 2==

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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