Question

六月一日，一组学生排成若干行，每行 $15$ 人。六月二日，同一组学生排成一长行。六月三日，同一组学生每行只站一人。六月四日，同一组学生每行 $6$ 人。这个过程一直持续到六月十二日，每天每行人数都不同。然而到六月十三日，他们找不到新的排列方式。这组学生最少可能有多少人？

On June 1, a group of students is standing in rows, with $15$ students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with $6$ students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

$21$

$30$

$60$

$90$

$1080$

答案：C

知识点：因数个数最小公倍数分类讨论

难度评级：1460

视频讲解：

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文字解答：

每行可站的人数正好是学生总数的正因数。因为六月一日至六月十二日有不同排列，而六月十三日没有新的排列方式，所以总人数必须恰好有 $12$ 个正因数。

总人数必须同时能被 $15$ 和 $6$ 整除，因此能被 $\operatorname{lcm}(15,6)=30=2\cdot3\cdot5$ 整除。但三十只有 $8$ 个因数。

最小的 $30$ 的倍数且有 $12$ 个因数的是 $60=2^2\cdot3\cdot5$ ，其因数个数为 $(2+1)(1+1)(1+1)=12$ 。

所以正确答案是 C。

The possible numbers of students per row are exactly the positive divisors of the total number of students. Since June 1 through June 12 give different arrangements and June 13 gives no new one, the total number of students must have exactly $12$ positive divisors.

The number must be divisible by both $15$ and $6$ , hence by $\operatorname{lcm}(15,6)=30=2\cdot3\cdot5$ . This number has only $8$ divisors.

The smallest multiple of $30$ with $12$ divisors is $60=2^2\cdot3\cdot5$ , which has $(2+1)(1+1)(1+1)=12$ divisors.

Thus, C is the correct answer.

2015 AMC 8 第 22 题

答案：C

难度评级：1460

视频讲解：

文字解答：

其他年份的第 22 题