2003 AMC 8 Problems/Problem 19: Difference between revisions
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Problem
How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
Solution 1
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 | ||
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