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

(Added Solution 2)
m (changed headers)
(18 intermediate revisions by 16 users not shown)
Line 1: Line 1:
 +
==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?
 
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>
 
 
==Solution==
 
==Solution==
 +
===Solution 1===
  
 
Draw a <math>7 \times 7</math> square.
 
Draw a <math>7 \times 7</math> square.
Line 25: Line 27:
 
\end{tabular} </math>
 
\end{tabular} </math>
  
Start from the center and label all protruding cells symmetrically.
+
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>. ∎ --anna0kear
+
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)} \text{ 1022}}</math>.
      ~Nosysnow
 
  
==Solution 2==
+
 
 +
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>
 
<asy>
Line 48: Line 52:
  
 
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>
 
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>
 +
 +
===Solution 3===
 +
This <math>7 \times 7</math> square drawn in Solution 1 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|}
 +
\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} </math>
 +
 +
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.
 +
 +
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>.
 +
 +
 +
===Video Solution===
 +
https://youtu.be/M22S82Am2zM
 +
  
 
== See Also ==
 
== See Also ==
Line 54: Line 88:
 
{{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]

Revision as of 19:10, 1 February 2020

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

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


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

This $7 \times 7$ square drawn in Solution 1 satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.

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

In the grid, 10 letters are used: $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$, and $K$. Each of the letters must have its own color, either white or black. This means, for example, all $K$'s must have the same color for the grid to be symmetrical.

So there are $2^{10}$ 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 $2^{10}-2=1024-2=$ $\boxed{\textbf{(B) } 1022}$.


Video Solution

https://youtu.be/M22S82Am2zM


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png