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2019 AMC 12A Problems/Problem 13

Revision as of 20:55, 17 February 2019 by Sevenoptimus (talk | contribs) (Fixed formatting and removed irrelevant attribution)

Problem

How many ways are there to paint each of the integers $2, 3, \dots, 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?

$\textbf{(A)}\ 144\qquad\textbf{(B)}\ 216\qquad\textbf{(C)}\ 256\qquad\textbf{(D)}\ 384\qquad\textbf{(E)}\ 432$

Solution 1

The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$. There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$, but without loss of generality, let it be painted red. $4$ cannot be the same color as $2$ or $8$, so there are $\underline{2}$ ways to paint $4$, which automatically determines the color for $8$. $6$ cannot be painted red, so there are $\underline{2}$ ways to paint $6$, but WLOG, let it be painted blue. There are $\underline{2}$ choices for the color for $3$, which is either red or green in this case. Lastly, there are $\underline{2}$ ways to choose the color for $9$.

$9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{\textbf{(E) }432}$.

Solution 2

We note that the primes can be colored any of the $3$ colors since they don't have any proper divisors other than $1$, which is not in the list. Furthermore, $6$ is the only number in the list that has $2$ distinct prime factors (namely, $2$ and $3$), so we do casework on $6$.

Case 1: $2$ and $3$ are the same color

In this case, we have $3$ primes to choose the color for ($2$, $5$, and $7$). Afterwards, $4$, $6$, and $9$ have two possible colors, which will determine the color of $8$. Thus, there are $3^3\cdot 2^3=216$ possibilities here.

Case 2: $2$ and $3$ are different colors

In this case, we have $4$ primes to color. Without loss of generality, we'll color the $2$ first, then the $3$. Then there are $3$ color choices for $2,5,7$, and $2$ color choices for $3$. This will determine the color of $6$ as well. After that, we only need to choose the color for $4$ and $9$, which each have $2$ choices. Thus, there are $3^3\cdot 2^3=216$ possibilities here as well.

Adding up gives $216+216=\boxed{\textbf{(E) }432}$.

Solution 3

$2,4,8$ require different colors each, so there are $6$ ways to color these.

$5$ and $7$ can be any color, so there are $3\times 3$ ways to color these.

$6$ can have $2$ colors once $2$ is colored, and thus $3$ also has $2$ colors following $6$, which leaves another $2$ for $9$.

All together: $6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{E}$.

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

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