# 1998 AJHSME Problems/Problem 22

## Problem

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: $23, 18, 9, 81, 76, \ldots .$

Find the $98^\text{th}$ term of the sequence that begins $98, 49, \ldots .$ $\text{(A)}\ 6 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 54$

## Solution

We could start by looking for a pattern. $98, 49, 44, 22, 11, 6, 54, 27, 22, 11, 6, \ldots .$

From here, we see that we have a pattern of $22, 11, 6, 54, 27, \ldots .$ after $98, 49, 44$

Our problem is now really

Find the $95^\text{th}$ term of the sequence that goes $22, 11, 6, 54, 27, 22, 11, 6, 54, 27, 22, \ldots .$

There are 5 terms in each repetition of the pattern, and $95\equiv0\pmod{5}$, so the answer is $\boxed{D}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 