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

Difference between revisions of "2023 AMC 10A Problems/Problem 6"
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(23 intermediate revisions by 14 users not shown)
Line 3: Line 3:
 
<math>\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252</math>
 
<math>\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252</math>
  
−
==Solution==
+
==Solution 1==
  
 
Each of the vertices is counted <math>3</math> times because each vertex is shared by three different edges.  
 
Each of the vertices is counted <math>3</math> times because each vertex is shared by three different edges.  
Line 12: Line 12:
  
 
==Solution 2==
 
==Solution 2==
−
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>.
+
Note that each vertex is counted <math>2\times 3=6</math> times. Thus, the answer is <math>21\times6=\boxed{\textbf{(D) } 126}</math>.
 +
 
 +
~Mathkiddie
 +
 
 +
==Solution 3==
 +
Just set one vertex 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=\boxed{\textbf{(D) } 126}</math>.
  
 
~SirAppel
 
~SirAppel
 +
~minor edits by TiguhBabeHwo
  
−
==Solution 3==
+
==Solution 4==
  
−
Since there are 8 verticies in a cube, there are <math>21/4</math> verticies for two edges. There are 4 edges in a face, and 6 faces in a cube, so the value of the cube is <math>21/4 * 24 = \boxed{\textbf{(D) } 126}</math>
+
Since there are 8 vertices in a cube, there are <math>\dfrac{21}4</math> vertices for two edges. There are <math>4</math> edges per face, and <math>6</math> faces in a cube, so the value of the cube is <math>\dfrac{21}4 \cdot 24 = \boxed{\textbf{(D) } 126}</math>.
  
 +
~DRBStudent ~Failure.net
  
−
~DRBStudent
+
(Minor formatting by Technodoggo)
  
−
==Solution 4 (use an example)==
+
==Solution 5 (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 <math>21\times6=\boxed{\textbf{(D) } 126}</math>
 
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 <math>21\times6=\boxed{\textbf{(D) } 126}</math>
  
−
See Also==
+
~milquetoast
 +
 
 +
==Solution 6==
 +
 
 +
The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers <math>21, 0, 0, 0, 0, 0, 0, 0</math>, which are indeed <math>8</math> integers that add to <math>21</math>. Doing this, we find three edges that have a value of <math>21</math>, and from there, we get three faces with a value of <math>42</math> (while the other three faces have a value of <math>0</math>). Adding the three faces together, we get <math>42+42+42 = \boxed{\textbf{(D) } 126}</math>.
 +
 
 +
~MathHafiz
 +
 
 +
==Solution 7==
 +
Let the vertices be <math>a, b, c, d, e, f, g, h</math>. We know that <math>a+b+c+d+e+f+g+h=21</math>. All the edges in a cube are equal, so we can assign the value <math>x</math> to the edge length. 2 vertices make an edge, so <math>4x=21</math>. This is also the same value of a face, so we can multiply <math>21</math> by <math>6</math> and get <math>\boxed{\textbf{(D) } 126}</math>.
 +
 
 +
~RocketScientist
 +
 
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=BFfQCHMRWXEnUV6k&t=956
 +
~little-fermat
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=3nDWUtwfk6n3ka9V&t=1240
 +
~Math-X
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=9KVpfLeW9ro
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/n9siXKiS9OA
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution (easy to digest) by Power Solve==
 +
https://youtu.be/Od1Spf3TDBs
 +
 
 +
==See Also==
 
{{AMC10 box|year=2023|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2023|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:22, 5 November 2024

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 vertex equal to $21$, it is trivial to see that there are $3$ faces with value $42$, and $42 \cdot 3=\boxed{\textbf{(D) } 126}$.

~SirAppel ~minor edits by TiguhBabeHwo

Solution 4

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 5 (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

Solution 6

The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers $21, 0, 0, 0, 0, 0, 0, 0$, which are indeed $8$ integers that add to $21$. Doing this, we find three edges that have a value of $21$, and from there, we get three faces with a value of $42$ (while the other three faces have a value of $0$). Adding the three faces together, we get $42+42+42 = \boxed{\textbf{(D) } 126}$.

~MathHafiz

Solution 7

Let the vertices be $a, b, c, d, e, f, g, h$. We know that $a+b+c+d+e+f+g+h=21$. All the edges in a cube are equal, so we can assign the value $x$ to the edge length. 2 vertices make an edge, so $4x=21$. This is also the same value of a face, so we can multiply $21$ by $6$ and get $\boxed{\textbf{(D) } 126}$.

~RocketScientist

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=BFfQCHMRWXEnUV6k&t=956 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=3nDWUtwfk6n3ka9V&t=1240 ~Math-X

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=9KVpfLeW9ro

Video Solution

https://youtu.be/n9siXKiS9OA

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution (easy to digest) by Power Solve

https://youtu.be/Od1Spf3TDBs

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

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=2023_AMC_10A_Problems/Problem_6&oldid=231682"