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
a & b & c & d & c & b & a \
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K & J & H & G & H & J & K \
 
\hline
 
\hline
b & e & f & g & f & e & b \
+
J & F & E & D & E & F & J \
 
\hline
 
\hline
c & f & h & i & h & f & c \
+
H & E & C & B & C & E & H \
 
\hline
 
\hline
d & g & i & j & i & g & d \
+
G & D & B & A & B & D & G \
 
\hline
 
\hline
c & f & h & i & h & f & c \
+
H & E & C & B & C & E & H \
 
\hline
 
\hline
b & e & f & g & f & e & b \
+
J & F & E & D & E & F & J \
 
\hline
 
\hline
a & b & c & d & c & b & a \
+
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>
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Start from the center and label all protruding cells symmetrically.
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 +
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
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 +
 
 
== See Also ==
 
== See Also ==
  

Revision as of 12:57, 11 February 2018

A scanning code consists of a $7 \times 7$ 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 $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ 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?

$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

Solution

Draw a $7 \times 7$ square.

$\begin{tabular}{|c|c|c|c|c|c|c|} \hline K & J & H & G & H & J & K \\ \hline J & F & E & D & E & F & J \\ \hline H & E & C & B & C & E & H \\ \hline G & D & B & A & B & D & G \\ \hline H & E & C & B & C & E & H \\ \hline J & F & E & D & E & F & J \\ \hline K & J & H & G & H & J & K \\ \hline \end{tabular}$

Start from the center and label all protruding cells symmetrically.

More specifically, since there are $4$ given lines of symmetry ($2$ diagonals, $1$ vertical, $1$ horizontal) and they split the plot into $8$ equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has $10$ distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose $2^10=1024$ but then subtract the $2$ 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 $2^{10} = 1024$ 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 $\fbox{\textbf{(B)} \text{ 1022}}$.

      ~Nosysnow


See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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