Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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| − | == | + | ==Problem== |
| − | + | How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors? | |
| − | |||
| − | <math>\ | + | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> |
| + | ==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 | ||
| + | <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== | ||
{{AMC8 box|year=2003|num-b=18|num-a=20}} | {{AMC8 box|year=2003|num-b=18|num-a=20}} | ||
| + | {{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?
Solution 1
Find the least common multiple of 
by turning the numbers into their prime factorization. ![\[15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2\]](http://latex.artofproblemsolving.com/6/1/9/6192e8caacff181a6fc8f6dc9cf58937b6cf8e4d.png)
Gather all necessary multiples
when multiplied gets 
. The multiples of 
. The number of multiples between 1000 and 2000 is 
.
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. 
