Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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Find the least common multiple of <math>15, 20, 25</math> by turning the numbers into their prime factorization. <cmath>15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2</cmath> Gather all necessary multiples | Find the least common multiple of <math>15, 20, 25</math> by turning the numbers into their prime factorization. <cmath>15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2</cmath> Gather all necessary multiples | ||
<math>3, 2^2, 5^2</math> when multiplied gets <math>300</math>. The multiples of <math>300 - 300, 600, 900, 1200, 1500, 1800, 2100</math>. The number of multiples between 1000 and 2000 is <math>\boxed{\textbf{(C)}\ 3}</math>. | <math>3, 2^2, 5^2</math> when multiplied gets <math>300</math>. The multiples of <math>300 - 300, 600, 900, 1200, 1500, 1800, 2100</math>. The number of multiples between 1000 and 2000 is <math>\boxed{\textbf{(C)}\ 3}</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Using the previous solution, turn <math>15, 20,</math> and <math>25</math> into their prime factorizations. | ||
+ | <cmath>15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2</cmath> | ||
+ | Notice that <math>1000</math> can be prime factorized into: | ||
+ | <cmath>1000 = 2 * 2 * 2 * 5 * 5 * 5</cmath> | ||
+ | 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> | ||
+ | Of the remaining numbers (<math>3</math> and <math>5</math>), the following numbers can be made: | ||
+ | <cmath>3, 5, 15</cmath> | ||
+ | Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>. | ||
==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 14:24, 28 July 2024
Contents
Problem
How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
Solution
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
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
:
Of the remaining numbers (
and
), the following numbers can be made:
Thus, counting these numbers we get our answer of:
.
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.