AMC 8 1998 Problem 22

AMC8 1998 年第 22 题

余数与同余★★★☆☆
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: If the integer is less than 10, multiply it by 9. Rule 2: If the integer is even and greater than 9, divide it by 2. Rule 3: If the integer is odd and greater than 9, subtract 5 from it. A sample sequence: 23,18,9,81,76,….23, 18, 9, 81, 76, \ldots . Find the 98th98^\text{th} term of the sequence that begins 98,49,….98, 49, \ldots .
Terri 按照以下三条规则生成一个正整数序列。她从一个正整数开始,然后对结果应用适当的规则,并继续以此方式进行。
规则 1:如果整数小于 10,乘以 9。 规则 2:如果整数为偶数且大于 9,除以 2。 规则 3:如果整数为奇数且大于 9,减去 5。
示例序列:23,18,9,81,76,…23, 18, 9, 81, 76, \ldots。
求以 98,49,…98, 49, \ldots 开头的序列的第 98 项。
  1. A.6
  2. B.11
  3. C.22
  4. D.27
  5. E.54

正确答案 · Correct Answer

D

解析 · Solution

Solution 1

We could start by looking for a pattern.
98,49,44,22,11,6,54,27,22,11,6,….98, 49, 44, 22, 11, 6, 54, 27, 22, 11, 6, \ldots.
From here, we see that we have a pattern of 22,11,6,54,27,….22, 11, 6, 54, 27, \ldots. after 98, 49, 44
Our problem is now really
Find the 95-th term of the sequence that goes 22,11,6,54,27,22,11,6,54,27,22,….22, 11, 6, 54, 27, 22, 11, 6, 54, 27, 22, \ldots.
There are 5 terms in each repetition of the pattern, and 95≡0(mod5)95\equiv0\pmod{5}, so the answer is DD

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