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Difference between revisions of "2018 AMC 10A Problems/Problem 20"

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<math>\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}</math>
 
<math>\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}</math>
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==Solution==
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Draw a <math>7 \times 7</math> square.
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<math> \begin{tabular}{|c|c|c|c|c|c|c|}
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\hline
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a & b & c & d & c & b & a \\
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\hline
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b & e & f & g & f & e & b \\
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\hline
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c & f & h & i & h & f & c \\
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\hline
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d & g & i & j & i & g & d \\
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\hline
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c & f & h & i & h & f & c \\
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\hline
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b & e & f & g & f & e & b \\
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\hline
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a & b & c & d & c & b & a \\
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\hline
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\end{tabular} </math>
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There are two choices for each letter, for a total of <math>2^{10} = 1024</math> codes. Two codes must be subtracted for an answer of <math>\fbox{\textbf{(B)} \text{ 1022}}</math>
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      ~Nosysnow
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== See Also ==
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{{AMC10 box|year=2018|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 15:44, 8 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 [i]symmetric[/i] 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.

$\begin{tabular}{|c|c|c|c|c|c|c|} \hline a & b & c & d & c & b & a \\ \hline b & e & f & g & f & e & b \\ \hline c & f & h & i & h & f & c \\ \hline d & g & i & j & i & g & d \\ \hline c & f & h & i & h & f & c \\ \hline b & e & f & g & f & e & b \\ \hline a & b & c & d & c & b & a \\ \hline \end{tabular}$

There are two choices for each letter, for a total of $2^{10} = 1024$ codes. Two codes must be subtracted for an answer of $\fbox{\textbf{(B)} \text{ 1022}}$ 

     ~Nosysnow

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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