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)
(One intermediate revision by the same user not shown)
Line 5: Line 5:
 
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math>
  
−
==Solution==
+
==Solution 1==
  
 
Find the least common multiple of <math>15, 20, 25</math> by turning the numbers into their prime factorization. <cmath>15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2</cmath> Gather all necessary multiples  
 
Find the least common multiple of <math>15, 20, 25</math> by turning the numbers into their prime factorization. <cmath>15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2</cmath> Gather all necessary multiples  
Line 18: Line 18:
 
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>
−
We must also cancel the same factors in <math>2000</math> to ensure that we don't exceed our range:
+
We must also cancel the same factors in <math>1000</math> and <math>2000</math> to ensure that we don't exceed our range:
 +
<cmath>1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}</cmath>
 +
 
 
<cmath>2000 = 2 * 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath>
 
<cmath>2000 = 2 * 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath>
−
The product of the remaining factors of <math>2000</math> is equal to <math>20</math>, while the remaining numbers left of <math>15, 20</math>, and <math>25</math> (<math>3</math> and <math>5</math>) yield the numbers:
+
The product of the remaining factors of <math>1000</math> is <math>2</math>, while the product of the remaining factors of <math>2000</math> is <math>20</math>.
 +
The remaining numbers left of <math>15, 20</math>, and <math>25</math> (<math>3</math> and <math>5</math>) yield:
 
<cmath>3, 5, 15</cmath>
 
<cmath>3, 5, 15</cmath>
 
Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>.
 
Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>.
Line 27: Line 30:
 
~Hawk2019
 
~Hawk2019
  
−
(Note that <math>3, 5,</math> and <math>15</math> are all less than <math>20</math>. Had they been larger than <math>20</math> they wouldn't have been between <math>1000</math> and <math>2000</math>)
+
(Note that <math>3, 5,</math> and <math>15</math> are all less than <math>20</math>, but greater than <math>2</math>. Had they been larger than <math>20</math> or less than <math>2</math>, they wouldn't have been between <math>1000</math> and <math>2000</math>)
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=18|num-a=20}}
 
{{AMC8 box|year=2003|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:44, 28 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: \[1000 = 2 * 2 * 2 * 5 * 5 * 5\] 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 $1000$ and $2000$ to ensure that we don't exceed our range: \[1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}\]

\[2000 = 2 * 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}\] The product of the remaining factors of $1000$ is $2$, while the product of the remaining factors of $2000$ is $20$. 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

(Note that $3, 5,$ and $15$ are all less than $20$, but greater than $2$. Had they been larger than $20$ or less than $2$, they wouldn't have been between $1000$ and $2000$)

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=223821"