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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 24"

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==Solution==
 
==Solution==
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Since we want the sum of the edges of each face to be <math>2</math>, we need there to be <math>2</math> <math>1</math>s and <math>2</math> <math>0</math>s on each face. Through experimentation
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Since we want the sum of the edges of each face to be <math>2</math>, we need there to be <math>2</math> <math>1</math>s and <math>2</math> <math>0</math>s on each face. Through experimentation, we find that either <math>2, 4,</math> or all of them have <math>1</math>s adjacent to <math>1</math>s and <math>0</math>s adjacent to <math>0</math> on each face. WLOG, let the first face (counterclockwise) be <math>0,0,1,1</math>. In this case we are trying to have all of them be adjacent to each other. First face: <math>0,0,1,1</math>. Second face: <math>2</math> choices: <math>1,0,0,1</math> or <math>0,0,1,1</math>. After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply <math>2</math> by <math>4</math> to get a total of <math>8</math> different arrangements.
  
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I am still working on the solution - in the meantime PLEASE DO NOT EDIT.
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Secondly: <math>4</math> of the faces have all of them adjacent and <math>2</math> of the faces do not: WLOG counting counterclockwise, we have <math>0,0,1,1</math>. Then, we choose the other face next to it. There are two cases, which are <math>0,1,0,1</math> and <math>1,0,1,0</math>. Therefore, this subcase has <math>4</math> different arrangements. Then, we can choose the face at front to be <math>1,0,1,0</math>. This has <math>4</math> cases. The sides can either be <math>0,1,1,0</math> or <math>1,1,0,0</math>. Therefore, we have another <math>8</math> cases.
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Summing these up, we have <math>8+4+8</math> = <math>20</math>. Therefore, our answer is <math>\boxed {\textbf{(E)}20}</math>
  
 
~Arcticturn
 
~Arcticturn
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Remark: It is very easy to get disorganized when counting, so when doing this problem, make sure to draw a diagram of the cube. Labeling is a bit harder, since we often confuse one side with another. Try doing the problem by labeling sides on the lines (literally letting the lines pass through your <math>0</math>s and <math>1</math>s.) I found that to be very helpful when solving this problem.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:44, 23 November 2021

Problem

Each of the $12$ edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$?

$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$

Solution

Since we want the sum of the edges of each face to be $2$, we need there to be $2$ $1$s and $2$ $0$s on each face. Through experimentation, we find that either $2, 4,$ or all of them have $1$s adjacent to $1$s and $0$s adjacent to $0$ on each face. WLOG, let the first face (counterclockwise) be $0,0,1,1$. In this case we are trying to have all of them be adjacent to each other. First face: $0,0,1,1$. Second face: $2$ choices: $1,0,0,1$ or $0,0,1,1$. After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply $2$ by $4$ to get a total of $8$ different arrangements.

Secondly: $4$ of the faces have all of them adjacent and $2$ of the faces do not: WLOG counting counterclockwise, we have $0,0,1,1$. Then, we choose the other face next to it. There are two cases, which are $0,1,0,1$ and $1,0,1,0$. Therefore, this subcase has $4$ different arrangements. Then, we can choose the face at front to be $1,0,1,0$. This has $4$ cases. The sides can either be $0,1,1,0$ or $1,1,0,0$. Therefore, we have another $8$ cases.

Summing these up, we have $8+4+8$ = $20$. Therefore, our answer is $\boxed {\textbf{(E)}20}$

~Arcticturn

Remark: It is very easy to get disorganized when counting, so when doing this problem, make sure to draw a diagram of the cube. Labeling is a bit harder, since we often confuse one side with another. Try doing the problem by labeling sides on the lines (literally letting the lines pass through your $0$s and $1$s.) I found that to be very helpful when solving this problem.

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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

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