Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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Notice that <math>1000</math> can be prime factorized into: | Notice that <math>1000</math> can be prime factorized into: | ||
<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> 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: | |
− | + | <math>3, 2, 5</math> | |
− | |||
− | Using | ||
− | <cmath> 3 | ||
− | |||
− | <cmath>1000 = 2 * \cancel{2} * \cancel{2 | ||
− | |||
− | |||
− | |||
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>. | ||
Revision as of 19:32, 29 July 2024
Contents
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.
Gather all necessary multiples
when multiplied gets
. The multiples of
. The number of multiples between 1000 and 2000 is
.
Solution 2
Using the previous solution, turn and
into their prime factorizations.
Notice that
can be prime factorized into:
Now take the lowest common multiple of
and
which is
:
Using
's prime factorization, we can cancel the following factors that are common in both
and
:
We now get the following factors from both of them:
Thus, counting these numbers we get our answer of:
.
~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.