2005 AMC 12B 第 10 题

先试着解答 2005 AMC 12B 第 10 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 2005 AMC 12B 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。

10.

一个数列的第一项是 20052005。每一项之后的下一项等于前一项各位数字的立方和。该数列的第 20052005 项是多少?
The first term of a sequence is 2005.2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 20052005th term of the sequence?
2929
5555
8585
133133
250250
答案:E
知识点:数字递推找规律
小提示:
计算前几项并寻找循环。
Compute the first several terms and look for a repeating cycle
大提示:
一旦某个值重复出现,后面的各项就以同样的周期循环;用项的序号对该周期取模即可。
Once a value repeats, the later terms repeat with the same period; use the term index modulo that period
解答:
数列开始为 2005,133,55,250,133,…2005, 133, 55, 250, 133, \ldots,因为 23+03+03+53=1332^3 + 0^3 + 0^3 + 5^3 = 133,13+33+33=551^3 + 3^3 + 3^3 = 55,53+53=2505^3 + 5^3 = 250,且 23+53+03=1332^3 + 5^3 + 0^3 = 133。
在首项 20052005 之后,数列以 133,55,250133, 55, 250 为周期 33 循环。
当 n≥2n \ge 2 时,第 nn 项是序列 133,55,250133, 55, 250 中索引为 ((n−2) mod 3)((n-2)\bmod 3) 的一项。因为 2005−2=2003≡2(mod3)2005 - 2 = 2003 \equiv 2 \pmod 3,所以第 20052005 项是 250250。
所以正确答案是 E。
The sequence begins 2005,133,55,250,133,…2005, 133, 55, 250, 133, \ldots since 23+03+03+53=133,2^3 + 0^3 + 0^3 + 5^3 = 133, 13+33+33=55,1^3 + 3^3 + 3^3 = 55, 53+53=250,5^3 + 5^3 = 250, and 23+53+03=133.2^3 + 5^3 + 0^3 = 133.
After the initial 2005,2005, the terms cycle through 133,55,250133, 55, 250 with period 3.3.
Term nn for n≥2n \ge 2 is the ((n−2) mod 3)((n-2)\bmod 3)th entry of 133,55,250.133, 55, 250. Since 2005−2=2003≡2(mod3),2005 - 2 = 2003 \equiv 2 \pmod 3, the 20052005th term is 250.250.
Thus, the correct answer is E.
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