Difference between revisions of "2001 AMC 10 Problems/Problem 8"
(→Problem) |
(→Solution) |
||
| Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
| − | We need to find the least common multiple of the | + | We need to find the least common multiple of the three numbers given. |
| − | |||
| − | <math> 3 \ | + | <math>\textrm{LCM}(15, 20, 25) = \textrm{LCM}(3 \cdot 5, 2^2 \cdot 5, 5^2) = 2^2 \cdot 3 \cdot 5^2 = 300</math> |
| − | + | 300 minutes equals 5 hours. So the bell will ring 5 hours past 12:00, which is 17:00. | |
| − | |||
| − | |||
| − | |||
| − | |||
== See Also == | == See Also == | ||
Revision as of 14:02, 13 July 2023
Problem
A church rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells every 25 minutes. If they all ring their bells at noon on the same day, at what time will they next all ring their bells together? (Answer in the form AB:CD without am or pm, such as 08:00)
Solution
We need to find the least common multiple of the three numbers given.
300 minutes equals 5 hours. So the bell will ring 5 hours past 12:00, which is 17:00.
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
