Difference between revisions of "2005 AMC 10A Problems/Problem 22"
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Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=\boxed{\textbf{(D)} 668}</math> | Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=\boxed{\textbf{(D)} 668}</math> | ||
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+ | ~ LaTeX edits my dolphin7 | ||
==See Also== | ==See Also== |
Revision as of 14:53, 25 March 2020
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
Since the least common multiple , the elements that are common to
and
must be multiples of
.
Since and
, several multiples of
that are in
won't be in
, but all multiples of
that are in
will be in
. So we just need to find the number of multiples of
that are in
.
Since every
rd element of
will be a multiple of
Therefore the answer is
~ LaTeX edits my dolphin7
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.