Difference between revisions of "2019 AMC 12A Problems/Problem 13"
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− | + | d\textbf{(C)}\ 256\qquad\textbf{(D)}\ 384\qquad\textbf{(E)}\ 432$ | |
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==Solution 1== | ==Solution 1== |
Revision as of 14:43, 9 November 2023
d\textbf{(C)}\ 256\qquad\textbf{(D)}\ 384\qquad\textbf{(E)}\ 432$
Contents
[hide]Solution 1
The and
can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of
or
. There are 3 ways to paint each, giving us
ways to paint both. The
is the most restrictive number. There are
ways to paint
, but without loss of generality, let it be painted red.
cannot be the same color as
or
, so there are
ways to paint
, which automatically determines the color for
.
cannot be painted red, so there are
ways to paint
, but WLOG, let it be painted blue. There are
choices for the color for
, which is either red or green in this case. Lastly, there are
ways to choose the color for
.
.
Solution 2
We note that the primes can be colored any of the colors since they don't have any proper divisors other than
, which is not in the list. Furthermore,
is the only number in the list that has
distinct prime factors (namely,
and
), so we do casework on
.
Case 1: and
are the same color
In this case, we have primes to choose the color for (
,
, and
). Afterwards,
,
, and
have two possible colors, which will determine the color of
. Thus, there are
possibilities here.
Case 2: and
are different colors
In this case, we have primes to color. Without loss of generality, we'll color the
first, then the
. Then there are
color choices for
, and
color choices for
. This will determine the color of
as well. After that, we only need to choose the color for
and
, which each have
choices. Thus, there are
possibilities here as well.
Adding up gives .
Solution 3
require different colors each, so there are
ways to color these.
and
can be any color, so there are
ways to color these.
can have
colors once
is colored, and thus
also has
colors following
, which leaves another
for
.
All together: .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.