Question

Butch 和 Sundance 需要离开 Dodge。为了尽快前进，两人按如下方式轮流步行和骑唯一的一匹马 Sparky。开始时 Butch 步行，Sundance 骑马。Sundance 到达路线中第一个拴马桩时，把 Sparky 拴在那里并开始步行；这些拴马桩恰好每隔一英里设置一个。当 Butch 到达 Sparky 时，他骑马直到超过 Sundance，然后在下一个拴马桩留下 Sparky，继续步行；两人如此反复。Sparky、Butch 和 Sundance 的速度分别为每小时 $6$ 、 $4$ ，和 $2.5$ 英里。Butch 和 Sundance 第一次在某个里程标处相遇时，他们离 Dodge 有 $n$ 英里，并且已经行进了 $t$ 分钟。求 $n + t$ 。

Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t.$

答案

答案：279

知识点：路程、速度与时间丢番图方程

难度评级：2460

解答：

Sparky 走一英里需 $10$ 分钟，Butch 步行一英里需 $15$ 分钟，Sundance 步行一英里需 $24$ 分钟。马沿着两人同一路线前进，并且每一英里恰好由其中一人骑过。因此，若 Butch 在 $n$ 英里中步行了 $x$ 英里，骑马走了其余 $n - x$ 英里，则 Sundance 骑马走了这 $x$ 英里，并步行其余 $n - x$ 英里。

当他们在里程标相遇时，两人行进的时间相同，所以 $\begin{aligned} &15x + 10(n - x) \\ &= 10x + 24(n - x), \end{aligned}$ 化简得 $19x = 14n$ 。由于交接发生在里程标处， $x$ 和 $n$ 都是整数；最小正整数解为 $x = 14$ 、 $n = 19$ 。

因此 $t = 15 \cdot 14 + 10 \cdot 5 = 260$ 分钟，所以 $n + t = 19 + 260 = 279$ 。

Walking a mile takes Sparky $10$ minutes, Butch $15,$ and Sundance $24.$ The horse advances along the same route as the men and is ridden over each mile by exactly one of them, so if Butch walks $x$ of the $n$ miles and rides the other $n - x,$ then Sundance rides those $x$ miles and walks the remaining $n - x.$

When they meet at a milepost they have been traveling for the same amount of time, so $\begin{aligned} &15x + 10(n - x) \\ &= 10x + 24(n - x), \end{aligned}$ which simplifies to $19x = 14n.$ Since the handoffs happen at mileposts, $x$ and $n$ are integers, and the smallest positive solution is $x = 14,$ $n = 19.$

Then $t = 15 \cdot 14 + 10 \cdot 5 = 260$ minutes, so $n + t = 19 + 260 = 279.$

2012 AIME I 第 4 题

答案：279

难度评级：2460

解答：

其他年份的第 4 题