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 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 
.
Solution 2
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)
Now take the lowest common multiple of 
and :
\begin{tabular}{lrrr}
\textcolor{red}{3} & \multicolumn{1}{!{\color{red}\vline}r}{12} & 15 & 18\\
\arrayrulecolor{red}\cline{2-4}
\textcolor{red}{2} & \multicolumn{1}{!{\color{red}\vline}r}{4} & 5 & 6\\
\arrayrulecolor{red}\cline{2-4}
& 2 & 5 & 3
\end{tabular}
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)
We must also cancel the same factors in :
![\[1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}\]](http://latex.artofproblemsolving.com/9/1/a/91a0f5b9d96655f3e33031ac93560d7f1babdc06.png)
The remaining numbers left of 
, and (
and 
) yield:
![\[3, 5, 15\]](http://latex.artofproblemsolving.com/9/8/c/98ca6a6e1ac5fa54917c1a281f2d976ddd46fd7e.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
