2023 AMC 10B Problems/Problem 6
Contents
Problem
Let 
, and 
for 
. How many terms in the sequence 
are even?
Solution 1
We calculate more terms: ![\[1,3,4,7,11,18,\ldots.\]](http://latex.artofproblemsolving.com/f/5/b/f5b6e4325ea8441396d49251de0e32fb3809a05d.png)
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. 
