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

(Solution 2)
 
(One intermediate revision by the same user not shown)
Line 10: Line 10:
 
<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>.
  
==Solution 2==
 
 
Using the previous solution, turn <math>15, 20,</math> and <math>25</math> into their prime factorizations.
 
<cmath>15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2</cmath>
 
Notice that <math>1000</math> can be prime factorized into:
 
<cmath>1000 = 2 * 2 * 2 * 5 * 5 * 5</cmath>
 
Now take the lowest common multiple of <math>15,20</math> and <math>25</math> which is <math>300</math>:
 
<cmath>\text{LCM}(15,20,25) = 300</cmath>
 
Using <math>300</math>'s prime factorization, we can cancel the following factors that are common in both <math>300</math> and <math>1000</math>:
 
<cmath> 300 = 3 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath>
 
<cmath>1000 = 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath>
 
We now get the following factors from both of them:
 
<cmath>3, 2, \text{and} 5</cmath>
 
Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>.
 
 
 
~Hawk2019
 
  
 
==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}}

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

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