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: $1000 = 2 * 2 * 2 * 5 * 5 * 5$ Using this, we can remove all the common factors of $15, 20,$ and $25$ that are shared with $1000$ : $3 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}$ We must also cancel the same factors in $1000$ : $1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}$

The remaining numbers left of $15, 20$ , and $25$ ( $3$ and $5$ ) yield: $3, 5, 15$ Thus, counting these numbers we get our answer of: $\boxed{\textbf{(C)}\ 3}$ .

~Hawk2019

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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2003 AMC 8 Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

See Also