Difference between revisions of "2021 Fall AMC 12B Problems/Problem 20"
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==Solution 1 (Direct Counting)== | ==Solution 1 (Direct Counting)== | ||
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Divide the <math>2 \times 2 \times 2</math> cube into two layers. | Divide the <math>2 \times 2 \times 2</math> cube into two layers. | ||
| − | Case 1: Each layer contains 2 | + | Case 1: Each layer contains 2 cubes of each color. |
| − | There are 2 ways the | + | There are 2 ways that the two cubes of each color can be arranged in each layer: adjacent or diagonal to each other. There are five situations: both adjacent and on top of each other, both adjacent but the two layers diagonal to each other, one adjacent and one diagonal (it doesn't matter diagonal which way), both diagonal and the two layers across from each other, and both diagonal but the two layers on top of each other. In the last case, this is the same as the second case. So we have four arrangements here. |
Case 2: One layer contains 1 white cube and the other layer contains 3 white cubes. | Case 2: One layer contains 1 white cube and the other layer contains 3 white cubes. | ||
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Only 1 possible <math>2 \times 2 \times 2</math> cube can result from this case. However, if we divide up this <math>2 \times 2 \times 2</math> cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other. | Only 1 possible <math>2 \times 2 \times 2</math> cube can result from this case. However, if we divide up this <math>2 \times 2 \times 2</math> cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other. | ||
| + | |||
| + | Therefore, our answer is <math>4 + 3 + 0 = \boxed{\textbf{(A)}\ 7}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 18:29, 25 November 2021
- The following problem is from both the 2021 Fall AMC 12B #20 and 2021 Fall AMC 12B #24, so both problems redirect to this page.
Problem
A cube is constructed from 
white unit cubes and black unit cubes. How many different ways are there to construct the 
cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
Solution 1 (Direct Counting)
Divide the cube into two layers.
Case 1: Each layer contains 2 cubes of each color.
There are 2 ways that the two cubes of each color can be arranged in each layer: adjacent or diagonal to each other. There are five situations: both adjacent and on top of each other, both adjacent but the two layers diagonal to each other, one adjacent and one diagonal (it doesn't matter diagonal which way), both diagonal and the two layers across from each other, and both diagonal but the two layers on top of each other. In the last case, this is the same as the second case. So we have four arrangements here.
Case 2: One layer contains 1 white cube and the other layer contains 3 white cubes.
Assume that we view the cube by rotating it such that there is a white cube in the upper-left of the front layer. If the sole black cube on the back layer is also on the upper-left, then if we split this into left and right layers, the right layer has only two white cubes and is covered in case 1. The other three places for the black cube make it such that no matter how we separate the cube into two layers, each layer will always contain one cube of one color and three cubes of another color. So, we add on another 3 ways the cube could be.
Case 3: One layer contains 0 white cubes and the other layer contains 4 white cubes.
Only 1 possible cube can result from this case. However, if we divide up this cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other.
Therefore, our answer is 
.
See Also
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| 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 12 Problems and Solutions | |
| 2021 Fall AMC 10B (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. 
