Question

Alice 和 Bob 玩一个圆上的游戏，圆周被 $12$ 个等距点分成若干段。这些点按顺时针方向编号为 $1$ 到 $12$ 。两人都从点 $12$ 开始。Alice 顺时针移动，Bob 逆时针移动。每轮游戏中，Alice 顺时针移动 $5$ 个点，Bob 逆时针移动 $9$ 个点。当他们停在同一点时游戏结束。需要多少轮？

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?

$6$

$8$

$12$

$14$

$24$

答案：A

知识点：模运算最小公倍数

难度评级：1450

解答：

一轮后，Alice 顺时针移动 $5$ 个点，Bob 逆时针移动 $9$ 个点。他们的相对移动为每轮 $5+9=14$ 个点。

当 $14k$ 是 $12$ 的倍数时他们相遇，其中 $k$ 是轮数。

因为 $14k\equiv2k\pmod{12}$ ，满足 $12\mid2k$ 的最小正整数 $k$ 是 $6$ 。

所以正确答案是 A。

After one turn, Alice has moved $5$ points clockwise and Bob has moved $9$ points counterclockwise. Their relative movement is $5+9=14$ points per turn.

They meet when $14k$ is a multiple of $12$ , where $k$ is the number of turns.

Since $14k\equiv2k\pmod{12}$ , the smallest positive $k$ with $12\mid2k$ is $6$ .

Thus, A is the correct answer.

2005 AMC 8 第 20 题

答案：A

难度评级：1450

解答：

其他年份的第 20 题