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

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2018 AMC 10A Problems/Problem 20|2018 AMC 10A #20]] and [[2018 AMC 12A Problems/Problem 15|2018 AMC 12A #15]]}}
+==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>
 ==Solution 1==
 Draw a <math>7 \times 7</math> square.
-<math> \begin{tabular}{|c|c|c|c|c|c|c|}
+<cmath> \begin{tabular}{|c|c|c|c|c|c|c|}
 \hline
 K & J & H & G & H & J & K \\
@@ Line 23: / Line 26: @@
 K & J & H & G & H & J & K \\
 \hline
-\end{tabular} </math>
+\end{tabular} </cmath>
 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 <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>\boxed{(B)}, 1022</math>. ∎
+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>.
-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) }1022}</math>.
-Note that this problem is very similar to the 1996 AIME Problem 7.
+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==
@@ Line 52: / Line 55: @@
 ==Solution 3==
-This <math>7 \times 7</math> square drawn in <math>Solution 1</math> satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.
-<math> \begin{tabular}{|c|c|c|c|c|c|c|}
+<cmath> \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
-K & J & H & G & H & J & K \\
+X & Y & Z & W & Z & Y & X \\
+\hline
+T & T & T & Z & T & T & T \\
 \hline
-J & F & E & D & E & F & J \\
+T & T & T & Y & T & T & T \\
 \hline
-H & E & C & B & C & E & H \\
+T & T & T & X & T & T & T \\
 \hline
-G & D & B & A & B & D & G \\
+\end{tabular} </cmath>
+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)
+<cmath> \begin{tabular}{|c|c|c|}
 \hline
-H & E & C & B & C & E & H \\
+A & B & C \\
 \hline
-J & F & E & D & E & F & J \\
+B & D & E \\
 \hline
-K & J & H & G & H & J & K \\
+C & E & F \\
 \hline
-\end{tabular} </math>
+\end{tabular}</cmath>
+(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>)
+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>
+-hansenhe
+==Video Solution by Pi Academy==
+https://youtu.be/lDk1--vGCYc?si=lPYpCtBdfuec72Xm
+~ Pi Academy
-In the grid, 10 letters are used: <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, <math>H</math>, <math>J</math>, and <math>K</math>. Each of the letters must have its own color, either white or black. This means, for example, all <math>K</math>'s must have the same color for the grid to be symmetrical.
+== Video Solution 2==
+https://youtu.be/M22S82Am2zM
-So there are <math>2^{10}</math> 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 <math>2^{10}-2=1024-2=</math> <math>\boxed{\textbf{(B) } 1022}</math>.
+~IceMatrix
-~OlutosinNGA
 == See Also ==
 {{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}
-{{AMC12 box|year=2018|ab=|num-b=14|num-a=16}}
+{{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}
 {{MAA Notice}}
 [[Category:Intermediate Combinatorics Problems]]

2018 AMC 10A (Problems • Answer Key • Resources)
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2018 AMC 12A (Problems • Answer Key • Resources)
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