2000 AMC 12 第 4 题

先试着解答 2000 AMC 12 第 4 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2000 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

4.

Fibonacci 数列 11,11,22,33,55,88,1313,2121,…\ldots 以两个 11 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, 1,1, 2,2, 3,3, 5,5, 8,8, 13,13, 21,21, …\ldots starts with two 11s, 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?
00
44
66
77
99
答案:C
知识点:斐波那契数列个位数字系统列举
小提示:
只追踪每一项的个位数字。
Track only the units digit of each term
大提示:
列出个位数字,直到十个数字都出现,并注意哪个最后出现。
List the units digits until all ten digits have appeared, and note which shows up last
解答:
个位数字序列开始为 1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,… \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}
查看这个列表,数字 66 是十个数字中最后出现的。
因此,正确答案是 C。
The sequence of units digits begins 1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,… \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 this list, the digit 66 is the last of the ten digits to appear.
Thus, the correct answer is C.
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