Question

六月 $1$ 日，一组学生排成若干行，每行 $15$ 人。六月 $2$ 日，同一组学生排成一长行。六月 $3$ 日，同一组学生每行只站一人。六月 $4$ 日，同一组学生每行 $6$ 人。这个过程一直持续到六月 $12$ 日，每天每行人数都不同。然而到六月 $13$ 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

小提示：

不同的每行人数就是总人数的因数

The different row sizes are the divisors of the group size.

大提示：

需要最小的同时为 $6$ 和 $15$ 的倍数，并且恰好有 $12$ 个因数

You need the smallest multiple of both $6$ and $15$ with exactly $12$ divisors.

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 $1$ 日至六月 $12$ 日有不同排列，而六月 $13$ 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。而这个数只有 $8$ 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