1998 AMC 8 第 22 题

先试着解答 1998 AMC 8 第 22 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 1998 AMC 8 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。

22.

Terri 按三条规则生成一个正整数序列。她从一个正整数开始,然后对结果应用合适的规则,并持续这样做。
规则 11:如果整数小于 1010,则乘以 99。
规则 22:如果整数是偶数且大于 99,则除以 22。
规则 33:如果整数是奇数且大于 99,则减去 55。
例如,样例序列为 2323,1818,99,8181,7676,…\ldots。
求以 9898、4949、…\ldots 开头的这个序列的第 9898 项。
Terri produces a sequence of positive integers by following three rules. She starts with a positive integer, then applies the appropriate rule to the result, and continues in this fashion.
Rule 1:1: If the integer is less than 10,10, multiply it by 9.9.
Rule 2:2: If the integer is even and greater than 9,9, divide it by 2.2.
Rule 3:3: If the integer is odd and greater than 9,9, subtract 55 from it.
For example, consider the sample sequence 23,23, 18,18, 9,9, 81,81, 76,76, ….\ldots.
Find the 9898th term of the sequence that begins 98,98, 49,49, ….\ldots.
66
1111
2222
2727
5454
答案:D
知识点:递推模运算
小提示:
生成各项,直到某个值重复
Generate terms until a value repeats.
大提示:
循环开始后,把项数按循环长度取余
After the repeat begins, reduce the index modulo the cycle length.
解答:
序列开始为
98,49,44,22,11,6,54,27,22,…。 \begin{gathered} 98,49,44,22,11, \\ 6,54,27,22,\ldots \end{gathered}\text{。}
前三项之后,循环 (22,11,6,54,27)(22,11,6,54,27) 重复。因为 98−3=9598-3=95 是 55 的倍数,第 9898 项是循环中的第五项,即 2727。
所以正确答案是 D。
The sequence begins
98,49,44,22,11,6,54,27,22,…. \begin{gathered} 98,49,44,22,11, \\ 6,54,27,22,\ldots. \end{gathered}
After the first three terms, the cycle (22,11,6,54,27)(22,11,6,54,27) repeats. Since 98−3=9598-3=95 is a multiple of 5,5, the 9898th term is the fifth term of the cycle, 27.27.
Thus, the correct answer is D .
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