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

 
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<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==
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==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  
 
<math>3, 2^2, 5^2</math> when multiplied gets <math>300</math>. The multiples of <math>300 - 300, 600, 900, 1200, 1500, 1800, 2100</math>. The number of multiples between 1000 and 2000 is <math>\boxed{\textbf{(C)}\ 3}</math>.
 
<math>3, 2^2, 5^2</math> when multiplied gets <math>300</math>. The multiples of <math>300 - 300, 600, 900, 1200, 1500, 1800, 2100</math>. The number of multiples between 1000 and 2000 is <math>\boxed{\textbf{(C)}\ 3}</math>.
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=18|num-a=20}}
 
{{AMC8 box|year=2003|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 12:42, 8 September 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 $\boxed{\textbf{(C)}\ 3}$.


See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
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

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