1998 AMC 8 Problem 22

Attempt Problem 22 of the 1998 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 8 solutions, or check the answer key.

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22.

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
Answer: D
Concepts:recursionmodular arithmetic
Small hint:
Generate terms until a value repeats.
Big hint:
After the repeat begins, reduce the index modulo the cycle length.
Solution:
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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