Difference between revisions of "2019 AMC 12A Problems/Problem 13"
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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 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>. | ||
| − | <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>. |
==See Also== | ==See Also== | ||
Revision as of 20:17, 10 February 2019
Problem
How many ways are there to paint each of the integers 
either red, green, or blue so that each number has a different color from each of its proper divisors?
Solution
The 
and 
can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of or . There are 3 ways to paint each, giving us 
ways to paint both. The 
is the most restrictive number. There are 
ways to paint , but WLOG, let it be painted red. 
cannot be the same color as or 
, so there are 
ways to paint , which automatically determines the color for . 
cannot be painted red, so there are ways to paint , but WLOG, let it be painted blue. There are choices for the color for 
, which is either red or green in this case. Lastly, there are ways to choose the color for 
.
.
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. 
