2015 AMC 8 Problems/Problem 22

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?

$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

Solution

As we read through this text, we find that the given information means that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}$.

Video Solution

https://youtu.be/E2om3xIvtYM

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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