AMC 8 2005 Problem 20
AMC8 2005 年第 20 题
余数与同余★★★☆☆
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?
Alice 和 Bob 玩了一个游戏，游戏中有一个圆的周长被 12 个等距点分成 12 等份。这些点按顺时针方向从 1 到 12 编号。两个人都从第 12 个点开始。Alice 顺时针移动，而 Bob 逆时针移动。
在游戏的每个回合，Alice 顺时针移动 5 个点，Bob 逆时针移动 9 个点。当他们停在同一点上时，比赛结束。则到游戏结束一共需要经过多少个回合？
- A.6
- B.8
- C.12
- D.14
- E.24
正确答案 · Correct Answer
A
解析 · Solution
Solution 1
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 12. Since this number must be a multiple of 12, as stated in the previous sentence, 14 has a factor 2, must have a factor of 6. The smallest number of turns that is a multiple of 6 is 6. 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
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