2018 AIME II 第 2 题

Question

设

a_0 = 2

、

a_1 = 5

、

a_2 = 8

，对

n \gt 2

，递归定义

a_n

为

4(a_{n-1} + a_{n-2} + a_{n-3})

除以

11

的余数。求

a_{2018} \cdot a_{2020} \cdot a_{2022}

。

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}.

答案

答案：112

知识点：递推模运算找规律

难度评级：1970

小提示：

逐项计算；每一项只依赖前三项，所以一旦某个连续三项的三元组重复，数列就开始周期重复

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

大提示：

这个数列的周期为

10

，所以把指标

2018

、

2020

、

2022

对

10

取余

The sequence repeats with period

10,

so reduce the indices

2018,

2020,

2022

modulo

10

解答：

逐项计算得

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

因为

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

= (2, 5, 8)

= (a_0, a_1, a_2)

，且每一项只由前三项决定，所以数列以

10

为周期。

因此

a_{2018} = a_8 = 7

、

a_{2020} = a_0 = 2

、

a_{2022} = a_2 = 8

，乘积为

7 \cdot 2 \cdot 8 = 112

。

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.

2018 AIME II 第 2 题

答案：112

难度评级：1970

小提示：

大提示：

解答：

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