2005 AMC 8 Problems/Problem 20

Problem

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?

Solution

Alice moves steps and Bob moves steps, where is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, , is a multiple of . Since this number must be a multiple of , as stated in the previous sentence, has a factor , must have a factor of . The smallest number of turns that is a multiple of is .

Another solution (Modular Arithmetic) Step 1: Write their positions after 𝑘 k turns.

Alice: moves 5 steps clockwise each turn. Position after 𝑘 k turns:

𝐴 ( 𝑘 ) = 12 + 5 𝑘 ( m o d 12 ) A(k)=12+5k(mod12)

Bob: moves 9 steps counterclockwise each turn. Moving 9 steps counterclockwise is the same as moving +3 clockwise (since 12 − 9 = 3 12−9=3). Position after 𝑘 k turns:

𝐵 ( 𝑘 ) = 12 + 3 𝑘 ( m o d 12 ) B(k)=12+3k(mod12) Step 2: Find when they meet.

They meet when:

𝐴 ( 𝑘 ) ≡ 𝐵 ( 𝑘 ) ( m o d 12 ) A(k)≡B(k)(mod12)

So:

12 + 5 𝑘 ≡ 12 + 3 𝑘 ( m o d 12 ) 12+5k≡12+3k(mod12)

Cancel the 12’s:

5 𝑘 ≡ 3 𝑘 ( m o d 12 ) 5k≡3k(mod12) 2 𝑘 ≡ 0 ( m o d 12 ) 2k≡0(mod12) Step 3: Solve for 𝑘 k.

That means 2 𝑘 2k is divisible by 12:

2 𝑘 ≡ 0 ( m o d 12 ) ⇒ 𝑘 ≡ 0 ( m o d 6 ) 2k≡0(mod12)⇒k≡0(mod6)

So the smallest positive 𝑘 k is:

𝑘 = 6 k=6

✅ They meet after 6 turns → Answer: (A) 6 - timi821

See Also

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