2018 AMC 10A Problems/Problem 20
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 1
Draw a square.
Start from the center and label all protruding cells symmetrically.
More specifically, since there are given lines of symmetry ( diagonals, vertical, horizontal) and they split the plot into equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose but then subtract the cases where all are white or all are black. That leaves us with . ∎
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 .
~Nosysnow
Solution 2
Imagine folding the scanning code along its lines of symmetry. There will be regions which you have control over coloring. Since we must subtract off cases for the all-black and all-white cases, the answer is
-EatingStuff
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.