Difference between revisions of "2003 AMC 8 Problems/Problem 19"
(→Solution 2) |
(→Solution 2) |
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<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>: | ||
− | \ | + | \begin{tabular}{lrrr} |
− | \ | + | \textcolor{red}{3} & \multicolumn{1}{!{\color{red}\vline}r}{12} & 15 & 18\\ |
− | \ | + | \arrayrulecolor{red}\cline{2-4} |
− | + | \textcolor{red}{2} & \multicolumn{1}{!{\color{red}\vline}r}{4} & 5 & 6\\ | |
− | + | \arrayrulecolor{red}\cline{2-4} | |
− | + | & 2 & 5 & 3 | |
− | \ | + | \end{tabular} |
− | \ | ||
− | \ | ||
− | |||
− | \end{ | ||
− | |||
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 19:26, 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
:
\begin{tabular}{lrrr}
\textcolor{red}{3} & \multicolumn{1}{!{\color{red}\vline}r}{12} & 15 & 18\\
\arrayrulecolor{red}\cline{2-4}
\textcolor{red}{2} & \multicolumn{1}{!{\color{red}\vline}r}{4} & 5 & 6\\
\arrayrulecolor{red}\cline{2-4}
& 2 & 5 & 3
\end{tabular}
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.