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

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==See also==
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{{AMC10 box|year=2023|ab=B|num-b=5|num-a=7}}
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Revision as of 20:29, 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?

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

Solution 1

We calculate more terms:

$1,3,4,7,11,18,...$

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

~Mintylemon66

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

~e_is_2.71828

See also

2023 AMC 10B (ProblemsAnswer KeyResources)
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

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