Difference between revisions of "2018 AMC 10A Problems/Problem 20"
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<math> \begin{tabular}{|c|c|c|c|c|c|c|} | <math> \begin{tabular}{|c|c|c|c|c|c|c|} | ||
\hline | \hline | ||
− | + | K & J & H & G & H & J & K \\ | |
\hline | \hline | ||
− | + | J & F & E & D & E & F & J \\ | |
\hline | \hline | ||
− | + | H & E & C & B & C & E & H \\ | |
\hline | \hline | ||
− | + | G & D & B & A & B & D & G \\ | |
\hline | \hline | ||
− | + | H & E & C & B & C & E & H \\ | |
\hline | \hline | ||
− | + | J & F & E & D & E & F & J \\ | |
\hline | \hline | ||
− | + | K & J & H & G & H & J & K \\ | |
\hline | \hline | ||
\end{tabular} </math> | \end{tabular} </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> | + | Start from the center and label all protruding cells symmetrically. |
+ | |||
+ | More specifically, since there are <math>4</math> given lines of symmetry (<math>2</math> diagonals, <math>1</math> vertical, <math>1</math> horizontal) and they split the plot into <math>8</math> equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has <math>10</math> distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose <math>2^10=1024</math> but then subtract the <math>2</math> cases where all are white or all are black. That leaves us with <math>\box{(B)}, 1022</math>. ∎ --anna0kear | ||
+ | |||
+ | 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>. | ||
+ | ~Nosysnow | ||
+ | |||
+ | |||
== See Also == | == See Also == | ||
Revision as of 13:57, 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.
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 $\box{(B)}, 1022$ (Error compiling LaTeX. Unknown error_msg). ∎ --anna0kear
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
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.