2018 AIME II Problem 2

Question

Let

a_0 = 2,

a_1 = 5,

and

a_2 = 8,

and for

n \gt 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} \cdot a_{2020} \cdot a_{2022}.

Answer

Compute terms one at a time; each term depends only on the previous three, so once a triple of consecutive terms repeats, the sequence is periodic

The sequence repeats with period

10,

so reduce the indices

2018,

2020,

2022

modulo

10

Computing successive terms gives

\begin{gathered} 2,\ 5,\ 8,\ 5,\ 6,\ 10,\ 7,\ 4,\ 7,\ 6,\ \\ 2,\ 5,\ 8,\ \ldots \end{gathered}

Since

(a_{10}, a_{11}, a_{12})

= (2, 5, 8)

= (a_0, a_1, a_2)

and each term depends only on the previous three, the sequence is periodic with period

10.

Therefore

a_{2018} = a_8 = 7,

a_{2020} = a_0 = 2,

and

a_{2022} = a_2 = 8,

so the product is

7 \cdot 2 \cdot 8 = 112.

Problem 2 in other years