Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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<cmath>1000 = 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath> | <cmath>1000 = 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath> | ||
We now get the following factors from both of them: | We now get the following factors from both of them: | ||
− | <cmath>3, 2, \text{and} 5</cmath> | + | <cmath>3, 2, \text{and } 5</cmath> |
Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>. | Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>. | ||
Revision as of 18:32, 29 July 2024
Contents
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. Gather all necessary multiples when multiplied gets . The multiples of . The number of multiples between 1000 and 2000 is .
Solution 2
Using the previous solution, turn and into their prime factorizations. Notice that can be prime factorized into: Now take the lowest common multiple of and which is : Using 's prime factorization, we can cancel the following factors that are common in both and : We now get the following factors from both of them: Thus, counting these numbers we get our answer of: .
~Hawk2019
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.