Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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Revision as of 23:47, 4 July 2013
Problem
How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
Solution
Find the least common multiple of 
by turning the numbers into their prime factorization. ![\[15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2\]](http://latex.artofproblemsolving.com/6/1/9/6192e8caacff181a6fc8f6dc9cf58937b6cf8e4d.png)
Gather all necessary multiples
when multiplied gets 
. The multiples of 
. The number of multiples between 1000 and 2000 is 
.
See Also
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
