BasicSequence AIME Intermediate
2018


Problem - 4117

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 $a_{2018}$ • $a_{2020}$ • $a_{2022}$.


Answer     112

When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.

After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$.

$a_{0} = 2$$a_{1} = 5$$a_{2} = 8$$a_{3} = 5$$a_{4} = 6$$a_{5} = 10$$a_{6} = 7$$a_{7} = 4$$a_{8} = 7$$a_{9} = 6$$a_{10} = 2$$a_{11} = 5$$a_{12} = 8$$a_{13} = 5$

We can simplify the expression we need to solve to $a_{8}$ • $a_{10}$ • $a_{2}$.

Our answer is $7$ • $2$ • $8$ $= \boxed{112}$.

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