Problem 20 · 2005 AMC 8
Hard
Number Theory
modular-meeting
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
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Answer: A — 6 turns.
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Hint 1 of 2
They walk toward each other around the loop. Each turn they close the gap by 5 + 9 = 14 points. They land together once that closing distance is a whole number of full laps — a multiple of 12.
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Hint 2 of 2
So you need the smallest k with 14k a multiple of 12 — that's the least common multiple of 14 and 12, divided by 14.
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Approach: relative motion — close the gap by a whole lap
- Going opposite directions, Alice and Bob eat up 5 + 9 = 14 points of separation each turn. They start together, so they reunite exactly when their combined motion is a whole number of laps of 12.
- Need 14k = multiple of 12. The smallest such total is lcm(14, 12) = 84, so k = 84 ÷ 14 = 6 turns.
- Why this transfers: two bodies moving in opposite directions around a loop have their speeds add (relative speed), and they meet when the combined travel equals a full circuit — far cleaner than tracking each position separately.
Another way — track positions with modular arithmetic:
- After k turns Alice sits at +5k and Bob at −9k (mod 12). They coincide when 5k ≡ −9k, i.e. 14k ≡ 0 (mod 12).
- Divide through by 2: 7k ≡ 0 (mod 6). Since 7 and 6 share no factor, k must be a multiple of 6 ⇒ smallest is 6.
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