Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
How many ways are there to paint each of the integers <math>2, 3, \dots, 9</math> either red, green, or blue so that each number has a different color from each of its proper divisors?
+
How many ways are there to paint each of the integers <math>2, 3, \cdots , 9</math> either red, green, or blue so that each number has a different color from each of its proper divisors?
  
−
<math>\textbf{(A)}\ 144\qquad\textbf{(B)}\ 216\qquad\textbf{(C)}\ 256\qquad\textbf{(D)}\ 384\qquad\textbf{(E)}\ 432</math>
+
<math>\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432</math>
 +
 
 +
==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==
 
==Solution 1==
  
−
The <math>5</math> and <math>7</math> can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of <math>5</math> or <math>7</math>. There are 3 ways to paint each, giving us <math>\underline{9}</math> ways to paint both. The <math>2</math> is the most restrictive number. There are <math>\underline{3}</math> ways to paint <math>2</math>, but WLOG, let it be painted red. <math>4</math> cannot be the same color as <math>2</math> or <math>8</math>, so there are <math>\underline{2}</math> ways to paint <math>4</math>, which automatically determines the color for <math>8</math>. <math>6</math> cannot be painted red, so there are <math>\underline{2}</math> ways to paint <math>6</math>, but WLOG, let it be painted blue. There are <math>\underline{2}</math> choices for the color for <math>3</math>, which is either red or green in this case. Lastly, there are <math>\underline{2}</math> ways to choose the color for <math>9</math>.
+
The <math>5</math> and <math>7</math> can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of <math>5</math> or <math>7</math>. There are 3 ways to paint each, giving us <math>\underline{9}</math> ways to paint both. The <math>2</math> is the most restrictive number. There are <math>\underline{3}</math> ways to paint <math>2</math>, but without loss of generality, let it be painted red. <math>4</math> cannot be the same color as <math>2</math> or <math>8</math>, so there are <math>\underline{2}</math> ways to paint <math>4</math>, which automatically determines the color for <math>8</math>. <math>6</math> cannot be painted red, so there are <math>\underline{2}</math> ways to paint <math>6</math>, but WLOG, let it be painted blue. There are <math>\underline{2}</math> choices for the color for <math>3</math>, which is either red or green in this case. Lastly, there are <math>\underline{2}</math> ways to choose the color for <math>9</math>.
  
 
<math>9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{\textbf{(E) }432}</math>.
 
<math>9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{\textbf{(E) }432}</math>.
  
 
==Solution 2==
 
==Solution 2==
−
We note that the primes can be colored any of the <math>3</math> colors since they don't have any proper divisors other than <math>1</math>, which is not in the list. Furthermore, <math>6</math> is the only number in the list that has <math>2</math> distinct prime factors (namely, <math>2</math> and <math>3</math>), thus we do casework on <math>6</math>.  
+
We note that the primes can be colored any of the <math>3</math> colors since they don't have any proper divisors other than <math>1</math>, which is not in the list. Furthermore, <math>6</math> is the only number in the list that has <math>2</math> distinct prime factors (namely, <math>2</math> and <math>3</math>), so we do casework on <math>6</math>.  
  
−
Case 1: <math>2</math> and <math>3</math> are the same colors
+
'''Case 1''': <math>2</math> and <math>3</math> are the same color
  
 
In this case, we have <math>3</math> primes to choose the color for (<math>2</math>, <math>5</math>, and <math>7</math>). Afterwards, <math>4</math>, <math>6</math>, and <math>9</math> have two possible colors, which will determine the color of <math>8</math>. Thus, there are <math>3^3\cdot 2^3=216</math> possibilities here.  
 
In this case, we have <math>3</math> primes to choose the color for (<math>2</math>, <math>5</math>, and <math>7</math>). Afterwards, <math>4</math>, <math>6</math>, and <math>9</math> have two possible colors, which will determine the color of <math>8</math>. Thus, there are <math>3^3\cdot 2^3=216</math> possibilities here.  
  
−
Case 2: <math>2</math> and <math>3</math> are different colors
+
'''Case 2''': <math>2</math> and <math>3</math> are different colors
  
−
In this case, we have <math>4</math> primes to color. WLOG, we'll color the <math>2</math> first, then the <math>3</math>. Then there are <math>3</math> color choices for <math>2,5,7</math>, and <math>2</math> color choices for <math>3</math>. This will determine the color of <math>6</math> as well. After that, we only need to choose the color for <math>4</math> and <math>9</math>, which each have <math>2</math> choices. Thus, there are <math>3^3\cdot 2^3=216</math> possibilities here as well.  
+
In this case, we have <math>4</math> primes to color. Without loss of generality, we'll color the <math>2</math> first, then the <math>3</math>. Then there are <math>3</math> color choices for <math>2,5,7</math>, and <math>2</math> color choices for <math>3</math>. This will determine the color of <math>6</math> as well. After that, we only need to choose the color for <math>4</math> and <math>9</math>, which each have <math>2</math> choices. Thus, there are <math>3^3\cdot 2^3=216</math> possibilities here as well.  
  
 
Adding up gives <math>216+216=\boxed{\textbf{(E) }432}</math>.
 
Adding up gives <math>216+216=\boxed{\textbf{(E) }432}</math>.
−
 
−
 
  
 
==Solution 3==
 
==Solution 3==
  
−
<math>2,4,8</math> require different color each, therefore <math>6</math> ways to color these.
+
<math>2,4,8</math> require different colors each, so there are <math>6</math> ways to color these.
  
−
<math>5</math> and <math>7</math> are whatever, therefore <math>3\times 3</math> ways to color these.
+
<math>5</math> and <math>7</math> can be any color, so there are <math>3\times 3</math> ways to color these.
  
−
<math>6</math> can be in two ways once <math>2</math> is colored, and thus <math>3</math> has also <math>2</math> following <math>6</math>, which leaves another <math>2</math> for <math>9</math>.
+
<math>6</math> can have <math>2</math> colors once <math>2</math> is colored, and thus <math>3</math> also has <math>2</math> colors following <math>6</math>, which leaves another <math>2</math> for <math>9</math>.
  
 
All together: <math>6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{E}</math>.
 
All together: <math>6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{E}</math>.
−
 
−
--DrB_Coach.
 
  
 
==See Also==
 
==See Also==

Latest revision as of 00:31, 3 November 2024

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_13&oldid=231311"