Difference between revisions of "2015 AMC 8 Problems/Problem 22"
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Since we know the number must be a multiple of <math>15</math>, we can eliminate <math>A</math>. We also know that after <math>12</math> days, the students can't find any more arrangements, meaning the number has <math>12</math> factors. Now, we just list the factors of every number, starting with <math>30</math>: | Since we know the number must be a multiple of <math>15</math>, we can eliminate <math>A</math>. We also know that after <math>12</math> days, the students can't find any more arrangements, meaning the number has <math>12</math> factors. Now, we just list the factors of every number, starting with <math>30</math>: | ||
<cmath>30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6</cmath> | <cmath>30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6</cmath> | ||
| − | <cmath>60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\ | + | <cmath>60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot10</cmath> |
<math>60</math> has <math>12</math> factors, so the answer is <math>\boxed{\textbf{(C) } 60}</math>. | <math>60</math> has <math>12</math> factors, so the answer is <math>\boxed{\textbf{(C) } 60}</math>. | ||
Revision as of 11:29, 2 January 2024
Contents
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?
Solution 1
The text suggests that the number of students in the group has 
factors, since each arrangement is a factor. The smallest integer with factors is 
.
Solution 2
Since we know the number must be a multiple of 
, we can eliminate 
. We also know that after days, the students can't find any more arrangements, meaning the number has factors. Now, we just list the factors of every number, starting with 
:
![\[30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6\]](http://latex.artofproblemsolving.com/e/8/a/e8af5982e203be2f7b064ac712a376099c031e4e.png)
![\[60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot10\]](http://latex.artofproblemsolving.com/c/0/5/c0545467e968bdfdf5647d51345a0621f2c62dc3.png)
has factors, so the answer is 
.
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=5241
~ pi_is_3.14
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. 
