2003 AMC 8 Problems/Problem 19
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
to ensure that we don't exceed our range:
The product of the remaining factors of
is equal to
, while the remaining numbers left of
, and
(
and
) yield the numbers:
Thus, counting these numbers we get our answer of:
.
~Hawk2019
(Note that and
are all less than
. Had they been larger 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 |
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