Difference between revisions of "2021 Fall AMC 12B Problems/Problem 20"
Ehuang0531 (talk | contribs) (→Solution 1 (Direct Counting): I know that I did something like this on the test and got the question right. But somehow as I'm writing this out, my solution makes less and less sense to me. Whelp.) |
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<math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ | <math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ | ||
10 \qquad\textbf{(E)}\ 11</math> | 10 \qquad\textbf{(E)}\ 11</math> | ||
| + | |||
| + | ==Solution 1 (Direct Counting)== | ||
| + | |||
| + | Divide the <math>2 \times 2 \times 2</math> 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: Each layer contains 3 cubes of one color and 1 cube of the other color. | ||
| + | |||
| + | Split this cube into two layers; the sole white cube on one layer must be on the opposite corner of the sole black cube on the other layer, otherwise there will be some way to spit the cube into two layers such that there are 2 cubes of each color on each layer. | ||
| + | |||
| + | Case 3: One layer contains 0 white cubes and the other layer contains 4 white cubes. | ||
| + | |||
| + | 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>6 + 1 + 0 = \boxed{\textbf{(A)}\ 7}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 18:36, 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: Each layer contains 3 cubes of one color and 1 cube of the other color.
Split this cube into two layers; the sole white cube on one layer must be on the opposite corner of the sole black cube on the other layer, otherwise there will be some way to spit the cube into two layers such that there are 2 cubes of each color on each layer.
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. 
