Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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− | == | + | ==Problem== |
− | + | How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors? | |
− | |||
− | <math>\ | + | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> |
+ | ==Solution 1== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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> | ||
+ | Now take the lowest common multiple of <math>15,20</math> and <math>25</math> which is <math>300</math>: | ||
+ | <cmath>\text{LCM}(15,20,25) = 300</cmath> | ||
+ | Using <math>300</math>'s prime factorization, we can cancel the following factors that are common in both <math>300</math> and <math>1000</math>: | ||
+ | <cmath> 300 = 3 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath> | ||
+ | <cmath>1000 = 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath> | ||
+ | We now get the following factors from both of them: | ||
+ | <cmath>3, 2, \text{and } 5</cmath> | ||
+ | Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>. | ||
+ | |||
+ | |||
+ | ~Hawk2019 | ||
+ | |||
+ | ==See Also== | ||
{{AMC8 box|year=2003|num-b=18|num-a=20}} | {{AMC8 box|year=2003|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:32, 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
which is
:
Using
's prime factorization, we can cancel the following factors that are common in both
and
:
We now get the following factors from both of them:
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.