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

Revision as of 18:45, 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 1

We calculate more terms:

$1,3,4,5,9,14,...$

We find a pattern: if $n$ is a multiple of $3$ , then the term is even, or else it is odd. There are $\lfloor \frac{2023}{3} \rfloor =\boxed{\textbf{(B) }674}$ multiples of $3$ from $1$ to $2023$ .

~Mintylemon66 this is so wrong

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

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

Revision as of 18:45, 15 November 2023

Problem

Solution 1

Solution 2

@@ Line 18: / Line 18: @@
 ~Mintylemon66
+this is so wrong
 ==Solution 2==