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>: | ||
+ | \[ | ||
+ | \setlength\extrarowheight{2pt} | ||
+ | \begin{array}{@{}l|l@{}} | ||
+ | 2 & 48,72,108\\ \cline{2-2} | ||
+ | 2 & 24,36,54\\ \cline{2-2} | ||
+ | 3 & 12,18,27\\ \cline{2-2} | ||
+ | \arrayrulecolor{red} | ||
+ | \color{red}3 & 4,6,9\\ \cline{2-2} | ||
+ | \color{red}2 & 4,2,3\\ \cline{2-2} | ||
+ | \multicolumn{1}{c}{} & 2,1,3 | ||
+ | \end{array} | ||
+ | \] | ||
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> |
Revision as of 19:25, 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
:
\[
\setlength\extrarowheight{2pt}
\begin{array}{@{}l|l@{}}
2 & 48,72,108\\ \cline{2-2}
2 & 24,36,54\\ \cline{2-2}
3 & 12,18,27\\ \cline{2-2}
\arrayrulecolor{red}
\color{red}3 & 4,6,9\\ \cline{2-2}
\color{red}2 & 4,2,3\\ \cline{2-2}
\multicolumn{1}{c}{} & 2,1,3
\end{array}
\]
Using this, we can remove all the common factors of
and
that are shared with
:
We must also cancel the same factors in
:
The remaining numbers left of , and
(
and
) yield:
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.