Question

有 $10$ 匹马，名字分别为 Horse $1$ 、Horse $2$ 、 $\ldots$ 、Horse $10$ 。它们的名字来自它们绕圆形跑道跑一圈所需的分钟数：Horse $k$ 恰好用 $k$ 分钟跑一圈。在时刻 $0$ ，所有马都在跑道起点。它们沿同一方向开始奔跑，并以各自恒定速度一直绕跑道跑。所有 $10$ 匹马再次同时到达起点的最小正时间（分钟）为 $S\gt0$ ，且 $S=2520$ 。设 $T\gt0$ 是至少 $5$ 匹马再次同时在起点的最小正时间（分钟）。 $T$ 的各位数字之和是多少？

There are $10$ horses, named Horse $1,$ Horse $2,$ $\ldots,$ Horse $10.$ They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S\gt0,$ in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520.$ Let $T\gt0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$

$2$

$3$

$4$

$5$

$6$

答案：B

知识点：最小公倍数整除性

难度评级：1630

小提示：

Horse $k$ 在时刻 $t$ 回到起点，当且仅当 $k$ 整除 $t$ 。

Horse $k$ is back at the start at time $t$ exactly when $k$ divides $t$

大提示：

找到最小的 $t$ ，使它在 $1$ 到 $10$ 中至少有 $5$ 个因数

Find the smallest $t$ having at least $5$ divisors among $1$ through $10$

解答：

Horse $k$ 在时刻 $t$ 位于起点，恰好当 $t$ 是 $k$ 的倍数。因此我们需要最小的 $t$ ，使它在 $1,2,\ldots,10$ 中至少有 $5$ 个因数。

小于 $12$ 的正整数至多有 $4$ 个这样的因数，而 $12$ 可被 $1,2,3,4$ 和 $6$ 整除。因此 $T=12$ ，它的各位数字之和是 $1+2=3$ 。

所以正确答案是 B。

Horse $k$ is at the starting point at time $t$ precisely when $t$ is a multiple of $k.$ So we want the smallest $t$ with at least $5$ divisors among $1,2,\ldots,10.$

The positive integers below $12$ have at most $4$ divisors, while $12$ is divisible by $1,2,3,4,$ and $6.$ Thus $T=12,$ and the sum of its digits is $1+2=3.$

Thus, the correct answer is B.

2017 AMC 12A 第 12 题

答案：B

难度评级：1630

小提示：

大提示：

解答：

其他年份的第 12 题