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

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A scanning code consists of a <math>7 \times 7</math> 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 <math>49</math> 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 <math>90 ^{\circ}</math> 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?
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{{duplicate|[[2018 AMC 10A Problems/Problem 20|2018 AMC 10A #20]] and [[2018 AMC 12A Problems/Problem 15|2018 AMC 12A #15]]}}
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==Problem==
 +
 
 +
A scanning code consists of a <math>7 \times 7</math> 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 <math>49</math> squares. A scanning code is called <math>\textit{symmetric}</math> if its look does not change when the entire square is rotated by a multiple of <math>90 ^{\circ}</math> 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?
  
 
<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 1==
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 +
Draw a <math>7 \times 7</math> square.
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<cmath> \begin{tabular}{|c|c|c|c|c|c|c|}
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\hline
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K & J & H & G & H & J & K \
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\hline
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J & F & E & D & E & F & J \
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\hline
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H & E & C & B & C & E & H \
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\hline
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G & D & B & A & B & D & G \
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\hline
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H & E & C & B & C & E & H \
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\hline
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J & F & E & D & E & F & J \
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\hline
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K & J & H & G & H & J & K \
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\hline
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\end{tabular} </cmath>
  
==Solution==
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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!)
  
Draw a <math>7 \times 7</math> square.
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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>\fbox{\textbf{(B)} \text{ 1022}}</math>.
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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) }1022}</math>.
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Note that this problem is very similar to the 1996 AIME Problem 7 https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7.
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==Solution 2==
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<asy>
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size(100pt);
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draw((1,0)--(8,0),linewidth(0.5));
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draw((1,2)--(6,2),linewidth(0.5));
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draw((1,4)--(4,4),linewidth(0.5));
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draw((1,6)--(2,6),linewidth(0.5));
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draw((2,6)--(2,0),linewidth(0.5));
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draw((4,4)--(4,0),linewidth(0.5));
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draw((6,2)--(6,0),linewidth(0.5));
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draw((1,0)--(1,7),dashed+linewidth(0.5));
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draw((1,7)--(8,0),dashed+linewidth(0.5));
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</asy>
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Imagine folding the scanning code along its lines of symmetry. There will be <math>10</math> regions which you have control over coloring. Since we must subtract off <math>2</math> cases for the all-black and all-white cases, the answer is <math>2^{10}-2=\boxed{\textbf{(B) } 1022.}</math>
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==Solution 3==
  
<math> \begin{tabular}{|c|c|c|c|c|c|c|}
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<cmath> \begin{tabular}{|c|c|c|c|c|c|c|}
 
\hline
 
\hline
a & b & c & d & c & b & a \
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T & T & T & X & T & T & T \
 
\hline
 
\hline
b & e & f & g & f & e & b \
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T & T & T & Y& T & T & T \
 
\hline
 
\hline
c & f & h & i & h & f & c \
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T & T & T & Z & T & T & T \
 
\hline
 
\hline
d & g & i & j & i & g & d \
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X & Y & Z & W & Z & Y & X \
 
\hline
 
\hline
c & f & h & i & h & f & c \
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T & T & T & Z & T & T & T \
 
\hline
 
\hline
b & e & f & g & f & e & b \
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T & T & T & Y & T & T & T \
 
\hline
 
\hline
a & b & c & d & c & b & a \
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T & T & T & X & T & T & T \
 
\hline
 
\hline
\end{tabular} </math>
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\end{tabular} </cmath>
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There are <math>3 \times 3</math> squares in the corners of this <math>7 \times 7</math> square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)
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<cmath> \begin{tabular}{|c|c|c|}
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\hline
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A & B & C \
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\hline
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B & D & E \
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\hline
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C & E & F \
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\hline
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\end{tabular}</cmath>
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(Note that coloring one <math>3 \times 3</math> square will also determine the colorings of the other <math>3</math> because the large <math>7 \times 7</math> square must look the same when it is rotated by <math>90^\circ</math>)
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There are <math>6</math> different letters and <math>2</math> choices of color (black and white) for each letter, so there are <math>2^6=64</math> colorings of a proper <math>3 \times 3</math> square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are <math>4</math> different letters, so there are <math>2^4=16</math> different colorings. Multiplying them together gives <math>16 \times 64 = 1024</math>. Going back to the question, we see that "there must be at least one square of each color in this grid of <math>49</math> squares." We must then eliminate <math>2</math> options: an all-black grid and an all-white grid. <math>1024-2= \boxed{\textbf{(B)} 1022} \ \phantom{}</math>  
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-hansenhe
 +
 
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==Video Solution by Pi Academy==
 +
 
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https://youtu.be/lDk1--vGCYc?si=lPYpCtBdfuec72Xm
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 +
~ Pi Academy
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== Video Solution 2==
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https://youtu.be/M22S82Am2zM
 +
 
 +
~IceMatrix
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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>
 
      ~Nosysnow
 
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}
 +
{{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 20:01, 11 October 2024

The following problem is from both the 2018 AMC 10A #20 and 2018 AMC 12A #15, so both problems redirect to this page.

Problem

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 1

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. (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 $4$ 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 $\fbox{\textbf{(B)} \text{ 1022}}$.

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 $\fbox{\textbf{(B) }1022}$.


Note that this problem is very similar to the 1996 AIME Problem 7 https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7.

Solution 2

[asy] size(100pt); draw((1,0)--(8,0),linewidth(0.5)); draw((1,2)--(6,2),linewidth(0.5)); draw((1,4)--(4,4),linewidth(0.5)); draw((1,6)--(2,6),linewidth(0.5)); draw((2,6)--(2,0),linewidth(0.5)); draw((4,4)--(4,0),linewidth(0.5)); draw((6,2)--(6,0),linewidth(0.5)); draw((1,0)--(1,7),dashed+linewidth(0.5)); draw((1,7)--(8,0),dashed+linewidth(0.5)); [/asy]

Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{\textbf{(B) } 1022.}$

Solution 3

\[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]

There are $3 \times 3$ squares in the corners of this $7 \times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)

\[\begin{tabular}{|c|c|c|} \hline A & B & C \\ \hline B & D & E \\ \hline C & E & F \\ \hline \end{tabular}\] (Note that coloring one $3 \times 3$ square will also determine the colorings of the other $3$ because the large $7 \times 7$ square must look the same when it is rotated by $90^\circ$)

There are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \times 64 = 1024$. Going back to the question, we see that "there must be at least one square of each color in this grid of $49$ squares." We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \boxed{\textbf{(B)} 1022} \\ \phantom{}$ -hansenhe

Video Solution by Pi Academy

https://youtu.be/lDk1--vGCYc?si=lPYpCtBdfuec72Xm

~ Pi Academy

Video Solution 2

https://youtu.be/M22S82Am2zM

~IceMatrix


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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