Difference between revisions of "2015 AMC 8 Problems/Problem 22"
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===Solution 1=== | ===Solution 1=== | ||
| − | + | As we read through this text, we find that the given information means that the number of students in the match has <math>12</math> factors, since each arrangement is a factor. The smallest integer with <math>12</math> factors is <math>2^2\cdot3\cdot5=60\implies\boxed{\textbf{(C) }~60}</math>. | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=21|num-a=23}} | {{AMC8 box|year=2015|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:30, 25 November 2015
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?
Solution 1
As we read through this text, we find that the given information means that the number of students in the match has 
factors, since each arrangement is a factor. The smallest integer with factors is 
.
See Also
| 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. 
