Difference between revisions of "2019 AMC 12A Problems/Problem 13"

Revision as of 19:24, 12 February 2019

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 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 $9$ .

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

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 $3$ ), thus we do casework on $6$ .

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

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. WLOG, 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 color each, therefore $6$ ways to color these.

$5$ and $7$ are whatever, therefore $3\times 3$ ways to color these.

$6$ can be in two ways once $2$ is colored, and thus $3$ has also $2$ following $6$ , which leaves another $2$ for $9$ .

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

--DrB_Coach.

@@ Line 36: / Line 36: @@
 All together: <math>6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{E}</math>.
-DrB_Coach.
+--DrB_Coach.
 ==See Also==

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