Difference between revisions of "2021 Fall AMC 10A Problems/Problem 24"
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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. | 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. | ||
| − | Summing these up, we have <math>8+4+8</math> = <math>20</math>. Therefore, our answer is <math>\boxed {\textbf{(E)}20}</math> | + | 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 | ||
Revision as of 12:09, 25 November 2021
Problem
Each of the 
edges of a cube is labeled 
or 
. 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 
faces of the cube equal to 
?
Solution
Since we want the sum of the edges of each face to be , we need there to be s and s on each face. Through experimentation, we find that either 
or all of them have s adjacent to s and s adjacent to on each face. WLOG, let the first face (counterclockwise) be 
. In this case we are trying to have all of them be adjacent to each other. First face: . Second face: choices: 
or . After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply by 
to get a total of 
different arrangements.
Secondly: of the faces have all of them adjacent and of the faces do not: WLOG counting counterclockwise, we have . Then, we choose the other face next to it. There are two cases, which are 
and 
. Therefore, this subcase has different arrangements. Then, we can choose the face at front to be . This has cases. The sides can either be 
or 
. Therefore, we have another cases.
Summing these up, we have 
= 
. Therefore, our answer is 
~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 s and 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
