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

(Solution 1)
(Solution 1)
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On my copy of the AMC 10B, the order of the answers is different, so the correct answer is (E) 674  
 
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:
 
 
<math>1,3,4,5,9,14,...</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
 
this is so wrong
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:49, 15 November 2023

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?

$\textbf{(A) }673\qquad\textbf{(B)} 674\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }1011$


On my copy of the AMC 10B, the order of the answers is different, so the correct answer is (E) 674

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 $\boxed{\textbf{(B) }674}$ evens.

~e_is_2.71828