2019 AMC 12A Problems/Problem 13

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

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 WLOG, 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 $9$ .

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

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2019 AMC 12A Problems/Problem 13

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