Question

Fibonacci 数列 $1, 1, 2, 3, 5, 8, 13, 21, \ldots$ 以两个 $1$ 开始，之后每一项都是前两项之和。十个数字中，哪一个最晚出现在 Fibonacci 数列某一项的个位上？

The Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, 21, \ldots$ starts with two $1$ s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

$0$

$4$

$6$

$7$

$9$

答案：C

知识点：斐波那契数列个位数字找规律

难度评级：1240

解答：

只记录个位数，得到 $\begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered}$

依次查看每个数字第一次出现的位置，数字 $6$ 是十个数字中最后出现的。

所以正确答案是 C。

Recording only the units digits gives the sequence $\begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered}$

Scanning for the first appearance of each digit, the digit $6$ is the last of the ten digits to show up.

Thus, the correct answer is C.

2000 AMC 10 第 6 题

答案：C

难度评级：1240

解答：

其他年份的第 6 题