2003 AMC 8 Problems/Problem 19

Problem

How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution 1

Find the least common multiple of $15, 20, 25$ by turning the numbers into their prime factorization. \[15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2\] Gather all necessary multiples $3, 2^2, 5^2$ when multiplied gets $300$. The multiples of $300 - 300, 600, 900, 1200, 1500, 1800, 2100$. The number of multiples between 1000 and 2000 is $\boxed{\textbf{(C)}\ 3}$.


See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png