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

## Solution 2

Using the previous solution, turn $15, 20,$ and $25$ into their prime factorizations. $$15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2$$ Notice that $1000$ can be prime factorized into: $$1000 = 2 * 2 * 2 * 5 * 5 * 5$$ Now take the lowest common multiple of $15,20$ and $25$ which is $300$: $$\text{LCM}(15,20,25) = 300$$ Using $300$'s prime factorization, we can cancel the following factors that are common in both $300$ and $1000$: $$300 = 3 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}$$ $$1000 = 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}$$ We now get the following factors from both of them: $$3, 2, \text{and } 5$$ Thus, counting these numbers we get our answer of: $\boxed{\textbf{(C)}\ 3}$.

~Hawk2019