Problem 22 · 1998 AJHSME
Stretch
Algebra & Patterns
find-the-cycle
Terri builds a sequence of positive integers by these rules: if the integer is less than 10, multiply it by 9; if it is even and greater than 9, divide it by 2; if it is odd and greater than 9, subtract 5. Find the 98th term of the sequence that begins 98, 49, … .
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Answer: D — 27.
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Hint 1 of 2
Nobody computes 98 terms by hand. These rule-driven sequences always trap into a repeating loop — generate terms only until you SEE a number come back, and you've found the cycle.
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Hint 2 of 2
Once the loop's length is L, terms repeat every L steps. Find where the cycle starts, then use the leftover after dividing to jump straight to the 98th term.
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Approach: generate until it loops, then use the cycle length to skip ahead
- Run the rules: 98 →(÷2) 49 →(−5) 44 →(÷2) 22 →(÷2) 11 →(−5) 6 →(×9) 54 →(÷2) 27 →(−5) 22 … The 22 has returned, so from term 4 on it cycles 22, 11, 6, 54, 27 — a loop of length 5.
- Terms 4, 5, 6, 7, 8 are positions 0, 1, 2, 3, 4 of the cycle. For term 98, step in from term 4: that's 98 − 4 = 94 steps, and 94 leaves a leftover of 4 when shared into groups of 5. Position 4 of the cycle is 27.
- Why this transfers: any 'find the very-far term' of a rule-based sequence is really 'find the cycle, then take the step-count's leftover (its remainder) as your position.' The big index 98 never needs all 98 terms.
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