2003 AMC 8 Problems/Problem 19
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 |
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