2019 AMC 12A Problems/Problem 13

Problem

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

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

Diagram

$[asy] /*Diagram by Technodoggo 2 November 2024*/ import graph; unitsize(2cm); draw((-1,0)--(0,0),Arrow(15)); draw((-0.5,-0.866)--(-1,0),Arrow(15)); draw((-0.5,-0.866)--(0,0),Arrow(15)); draw((0.5,-0.866)--(0,0),Arrow(15)); draw((0.5,-0.866)--(1,0),Arrow(15)); draw((1.5,-0.866)--(1,0),Arrow(15)); filldraw(circle((0,0),0.2),white);label("$2$",(0,0)); filldraw(circle((-1,0),0.2),white);label("$4$",(-1,0)); filldraw(circle((1,0),0.2),white);label("$3$",(1,0)); filldraw(circle((-0.5,-0.866),0.2),white);label("$8$",(-0.5,-0.866)); filldraw(circle((0.5,-0.866),0.2),white);label("$6$",(0.5,-0.866)); filldraw(circle((-0.5,0.866),0.2),white);label("$5$",(-0.5,0.866)); filldraw(circle((0.5,0.866),0.2),white);label("$7$",(0.5,0.866)); filldraw(circle((1.5,-0.866),0.2),white);label("$9$",(1.5,-0.866)); [/asy]$

(Diagram by Technodoggo)

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

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

Page

Toolbox

Search