2003 AMC 8 Problems/Problem 19
Contents
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 
.
Solution
Using the previous solution, turn 
and 
into their prime factorizations.
Notice that 
can be prime factorized into:
![\[1000 = 2 * 2 * 2 * 5 * 5 * 5\]](http://latex.artofproblemsolving.com/5/e/2/5e2218492e7de4e2d627cae0c5c1cb5d19d84c0e.png)
Using this, we can remove all the common factors of and that are shared with :
![\[3 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}\]](http://latex.artofproblemsolving.com/7/5/4/754f5c60ee4f20cbb4f9141a25201b61340b9785.png)
Of the remaining numbers (
and 
), the following numbers can be made:
![\[3, 5, 15\]](http://latex.artofproblemsolving.com/9/8/c/98ca6a6e1ac5fa54917c1a281f2d976ddd46fd7e.png)
Thus, counting these numbers we get our answer of: .
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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