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

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 $\boxed{\textbf{(C)}\ 3}$ .

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

Page

Toolbox

Search

2003 AMC 8 Problems/Problem 19

Problem

Solution

See Also