Difference between revisions of "2017 AMC 8 Problems/Problem 20"
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There are <math>5</math> options for the last digit as the integer must be odd. The first digit now has <math>8</math> options left (it can't be <math>0</math> or the same as the last digit). The second digit also has <math>8</math> options left (it can't be the same as the first or last digit). Finally, the third digit has <math>7</math> options (it can't be the same as the three digits that are already chosen). | There are <math>5</math> options for the last digit as the integer must be odd. The first digit now has <math>8</math> options left (it can't be <math>0</math> or the same as the last digit). The second digit also has <math>8</math> options left (it can't be the same as the first or last digit). Finally, the third digit has <math>7</math> options (it can't be the same as the three digits that are already chosen). | ||
| − | Since there are <math>9,000</math> total integers, our answer is <cmath>\frac{ | + | Since there are <math>9,000</math> total integers, our answer is <cmath>\frac{C \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.</cmath> |
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
Revision as of 03:09, 8 December 2024
Solution
There are 
options for the last digit as the integer must be odd. The first digit now has 
options left (it can't be 
or the same as the last digit). The second digit also has options left (it can't be the same as the first or last digit). Finally, the third digit has 
options (it can't be the same as the three digits that are already chosen).
Since there are 
total integers, our answer is ![\[\frac{C \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.\]](http://latex.artofproblemsolving.com/1/b/6/1b6be56292fee733bc1c0407eca0d9c9bb9e0c97.png)
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/tJm9KqYG4fU?t=3114
~savannahsolver
https://www.youtube.com/watch?v=2G9jiu5y5PM ~David
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
