Question

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?

$21$

$30$

$60$

$90$

$1080$

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Written solution:

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 Problem 22

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Problem 22 in Other Years