Difference between revisions of "2018 AMC 10A Problems/Problem 20"
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There are only ten squares we get to actually choose, and two independent choices for each, for a total of <math>2^{10} = 1024</math> codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of <math>\fbox{\textbf{(B)} \text{ 1022}}</math> | There are only ten squares we get to actually choose, and two independent choices for each, for a total of <math>2^{10} = 1024</math> codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of <math>\fbox{\textbf{(B)} \text{ 1022}}</math> | ||
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== See Also == | == See Also == | ||
Revision as of 12:56, 11 February 2018
A scanning code consists of a 
grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of 
squares. A scanning code is called 
if its look does not change when the entire square is rotated by a multiple of 
counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
Solution
Draw a square.
There are only ten squares we get to actually choose, and two independent choices for each, for a total of 
codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of 
See Also
| 2018 AMC 10A (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 AMC 10 Problems and Solutions | ||
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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. 
