# Difference between revisions of "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 1

The given information means that the number of students in the match has $12$ factors. The least integer with $12$ factors is $2^2\cdot3\cdot5=60\implies\boxed{\textbf{(C) }~60}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 