Difference between revisions of "2023 AMC 10B Problems/Problem 6"
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==Problem== | ==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}... | + | 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)} 674\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }1011</math> | ||
Revision as of 16:58, 15 November 2023
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
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.
