Difference between revisions of "2023 AMC 10B Problems/Problem 6"
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− | + | ==Problem== | |
− | + | 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) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674</math> | |
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | 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 ~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 <math>1</math> and <math>3</math>, the next term is <math>4</math>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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
Contents
Problem
Let , and for . How many terms in the sequence are even?
Solution 1
We calculate more terms: We find a pattern: if is a multiple of , then the term is even, or else it is odd. There are multiples of from to .
~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 and , the next term is .
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 we get with a remainder of one. So we have full cycles, and an extra odd at the end.
Therefore, there are 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
~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.