Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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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> | ||
− | We must also cancel the same factors in <math>1000</math> | + | We must also cancel the same factors in <math>1000</math>: |
<cmath>1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}</cmath> | <cmath>1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}</cmath> | ||
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The remaining numbers left of <math>15, 20</math>, and <math>25</math> (<math>3</math> and <math>5</math>) yield: | The remaining numbers left of <math>15, 20</math>, and <math>25</math> (<math>3</math> and <math>5</math>) yield: | ||
<cmath>3, 5, 15</cmath> | <cmath>3, 5, 15</cmath> | ||
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~Hawk2019 | ~Hawk2019 | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=18|num-a=20}} | {{AMC8 box|year=2003|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:19, 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:
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.