2005 AMC 10A Problems/Problem 22
Problem
Let be the set of the
smallest positive multiples of
, and let
be the set of the
smallest positive multiples of
. How many elements are common to
and
?
Solution 1
Since the least common multiple of and
is
, the elements that are common to
and
are all multiples of
. Moreover, as the largest element of
is
, while that of
is
, which is larger, several of the multiples of
that are in
will not be in
, whereas all the multiples of
that are in
will be in
.
Thus we only need to find the number of multiples of that are in
, and so we observe that as
, these multiples of
are precisely every
rd element of
. It follows that there are
such elements.
Solution 2
As in Solution , we find that the elements common to
and
are precisely the multiples of
. As
has exactly
elements, these must range from
to
, and similarly the elements of
range from
to
. This means any element
must satisfy both
and
, which reduces to simply
.
Accordingly, as and
, the multiples of
in the required interval range from
to
, so there are precisely
of them.
Video Solution
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AMC 10 Problems and Solutions |
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