Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 8 Problems/Problem 19"

(→‎Solution 2)
(→‎Solution 2)
Line 17: Line 17:
 
<cmath>1000 = 2 * 2 * 2 * 5 * 5 * 5</cmath>
 
<cmath>1000 = 2 * 2 * 2 * 5 * 5 * 5</cmath>
 
Now take the lowest common multiple of <math>15,20</math> and <math>25</math>:
 
Now take the lowest common multiple of <math>15,20</math> and <math>25</math>:
−
\[
+
\begin{tabular}{lrrr}
−
\setlength\extrarowheight{2pt}
+
\textcolor{red}{3} & \multicolumn{1}{!{\color{red}\vline}r}{12} & 15 & 18\\
−
\begin{array}{@{}l|l@{}}
+
\arrayrulecolor{red}\cline{2-4}
−
2 & 48,72,108\\ \cline{2-2}  
+
\textcolor{red}{2} &   \multicolumn{1}{!{\color{red}\vline}r}{4} & 5 & 6\\
−
2 & 24,36,54\\  \cline{2-2}
+
\arrayrulecolor{red}\cline{2-4}
−
3 & 12,18,27\\  \cline{2-2}
+
&   2 & 5 & 3
−
\arrayrulecolor{red}
+
\end{tabular}
−
\color{red}3 & 4,6,9\\ \cline{2-2}
 
−
\color{red}2 & 4,2,3\\ \cline{2-2}
 
−
\multicolumn{1}{c}{} & 2,1,3
 
−
\end{array}
 
−
\]
 
 
Using this, we can remove all the common factors of <math>15, 20,</math> and <math>25</math> that are shared with <math>1000</math>:
 
Using this, we can remove all the common factors of <math>15, 20,</math> and <math>25</math> that are shared with <math>1000</math>:
 
<cmath> 3 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}</cmath>
 
<cmath> 3 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}</cmath>

Revision as of 18:26, 29 July 2024

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 $25$: \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 $15, 20,$ 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_19&oldid=223924"