Difference between revisions of "2003 AMC 8 Problems/Problem 19"
(→Solution 2) |
(→Solution 2) |
||
Line 18: | Line 18: | ||
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>2000</math> to ensure that we don't exceed our range: | + | 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> | <cmath>2000 = 2 * 2 * 5 * \cancel{2} * \cancel{2} * \cancel{5} * \cancel{5}</cmath> | ||
− | The product of the remaining factors of <math> | + | 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>. | ||
Line 27: | Line 30: | ||
~Hawk2019 | ~Hawk2019 | ||
− | (Note that <math>3, 5,</math> and <math>15</math> are all less than <math>20</math>. Had they been larger than <math>20</math> they wouldn't have been between <math>1000</math> and <math>2000</math>) | + | (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}} |
Revision as of 14:43, 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 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.