Difference between revisions of "2015 AMC 8 Problems/Problem 22"

Revision as of 03:02, 29 December 2022

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 $2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}$ .

Video Solution

https://youtu.be/E2om3xIvtYM

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=5241

~ pi_is_3.14

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

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