# 2015 AMC 8 Problems/Problem 22

## 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

The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}$.

## Solution 2

Since we know the number must be a multiple of $15$, we can eliminate $A$. We also know that after $12$ days, the students can't find any more arrangements, meaning the number has $12$ factors. Now, we just list the factors of every number, starting with $30$: $$30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6$$ $$60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot12$$ $60$ has $12$ factors, so the answer is $\boxed{\textbf{(C) } 60}$.

~savannahsolver

## Video Solution by OmegaLearn

~ pi_is_3.14

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