Difference between revisions of "2003 AMC 8 Problems/Problem 19"
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<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | ||
− | ==Solution== | + | ==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 | 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== | + | ==Solution 2== |
Using the previous solution, turn <math>15, 20,</math> and <math>25</math> into their prime factorizations. | Using the previous solution, turn <math>15, 20,</math> and <math>25</math> into their prime factorizations. | ||
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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> and <math>2000</math> to ensure that we don't exceed our range: | |
+ | <cmath>1000 = 2 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5} * \cancel{5}</cmath> | ||
+ | |||
+ | <cmath>2000 = 2 * 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath> | ||
+ | The product of the remaining factors of <math>1000</math> is <math>2</math>, while the product of the remaining factors of <math>2000</math> is <math>20</math>. | ||
+ | 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> | ||
Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>. | Thus, counting these numbers we get our answer of: <math>\boxed{\textbf{(C)}\ 3}</math>. | ||
+ | |||
+ | |||
+ | ~Hawk2019 | ||
+ | |||
+ | (Note that <math>3, 5,</math> and <math>15</math> are all less than <math>20</math>, but greater than <math>2</math>. Had they been larger than <math>20</math> or less than <math>2</math>, they wouldn't have been between <math>1000</math> and <math>2000</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}} |
Latest revision as of 14:44, 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 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
and
to ensure that we don't exceed our range:
The product of the remaining factors of
is
, while the product of the remaining factors of
is
.
The remaining numbers left of
, and
(
and
) yield:
Thus, counting these numbers we get our answer of:
.
~Hawk2019
(Note that and
are all less than
, but greater than
. Had they been larger than
or less than
, they wouldn't have been between
and
)
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.