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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Revision as of 23:26, 27 January 2019
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?
Contents
Solution 1
Draw a square.
Start from the center and label all protruding cells symmetrically. (Note that "I" is left out of this labelling, so there are only 10 labels, not 11, as ending in K would suggest!)
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 .
Note that this problem is very similar to the 1996 AIME Problem 7.
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
Solution 3
This square drawn in satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.
In the grid, 10 letters are used: , , , , , , , , , and . Each of the letters must have its own color, either white or black. This means, for example, all 's must have the same color for the grid to be symmetrical.
So there are ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is .
~OlutosinNGA
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 12 (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 |
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