Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2021 Fall AMC 10A Problems/Problem 24

Revision as of 20:44, 23 November 2021 by Arcticturn (talk | contribs) (→‎Solution)

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_24&oldid=165715"