2018 AIME II Problems/Problem 2

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Problem

Let $a_{0} = 2$, $a_{1} = 5$, and $a_{2} = 8$, and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$($a_{n-1}$ + $a_{n-2}$ + $a_{n-3}$) is divided by $11$. Find $a_{2018}$$a_{2020}$$a_{2022}$.

2018 AIME II (ProblemsAnswer KeyResources)
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