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Difference between revisions of "2023 AMC 10A Problems/Problem 6"

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~Mathkiddie
 
~Mathkiddie
  
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==Solution 2==
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==Solution 3==
 
Just set one vertice equal to <math>21</math>, it is trivial to see that there are <math>3</math> faces with value <math>42</math>, and <math>42 \cdot 3=126</math>.
 
Just set one vertice equal to <math>21</math>, it is trivial to see that there are <math>3</math> faces with value <math>42</math>, and <math>42 \cdot 3=126</math>.
  

Revision as of 18:46, 10 November 2023

Problem

An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$. What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$

Solution 1

Each of the vertices is counted $3$ times because each vertex is shared by three different edges. Each of the edges is counted $2$ times because each edge is shared by two different faces. Since the sum of the integers assigned to all vertices is $21$, the final answer is $21\times3\times2=\boxed{\textbf{(D) } 126}$

~Mintylemon66

Solution 2

Note that each vertex is counted $2\times 3=6$ times. Thus, the answer is $21\times6=\boxed{\textbf{(D) } 126}$.

~Mathkiddie

Solution 3

Just set one vertice equal to $21$, it is trivial to see that there are $3$ faces with value $42$, and $42 \cdot 3=126$.

~SirAppel

Solution 3

Since there are 8 vertices in a cube, there are $\dfrac{21}4$ vertices for two edges. There are $4$ edges per face, and $6$ faces in a cube, so the value of the cube is $\dfrac{21}4 \cdot 24 = \boxed{\textbf{(D) } 126}$.

~DRBStudent ~Failure.net

(Minor formatting by Technodoggo)

Solution 4 (use an example)

Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\times6=\boxed{\textbf{(D) } 126}$

~milquetoast


Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=FgRMzUuISdwtlKdn&t=1160 ~Math-X


See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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