2002 AMC 10A Problems/Problem 18: Difference between revisions
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== Problem == | == Problem == | ||
A | A <math>3</math>x<math>3</math>x<math>3</math> cube is made of <math>27</math> normal dice. Each die's opposite sides sum to <math>7</math>. What is the smallest possible sum of all of the values visible on the <math>6</math> faces of the large cube? | ||
<math>\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96</math> | <math>\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96</math> | ||
==Solution== | ==Solution== | ||
In a 3x3x3 cube, there are 8 cubes with three faces showing, 12 with two faces showing and 6 with one face showing. The smallest sum with three faces showing is 1+2+3=6, with two faces showing is 1+2=3, and with one face showing is 1. Hence, the smallest possible sum is <math>8(6)+12(3)+6(1)=48+36+6=90</math>. Our answer is thus <math>\boxed{\text{(D)}\ 90} </math>. | In a 3x3x3 cube, there are <math>8</math> cubes with three faces showing, <math>12</math> with two faces showing and <math>6</math> with one face showing. The smallest sum with three faces showing is <math>1+2+3=6</math>, with two faces showing is <math>1+2=3</math>, and with one face showing is <math>1</math>. Hence, the smallest possible sum is <math>8(6)+12(3)+6(1)=48+36+6=90</math>. Our answer is thus <math>\boxed{\text{(D)}\ 90} </math>. | ||
==Video Solution== | |||
https://www.youtube.com/watch?v=1sdXHKW6sqA ~David | |||
==Video Solution by Daily Dose of Math== | |||
https://youtu.be/QSLlfaf50Is | |||
~Thesmartgreekmathdude | |||
==See Also== | ==See Also== | ||
Latest revision as of 17:40, 15 October 2024
Problem
A 
xx cube is made of 
normal dice. Each die's opposite sides sum to 
. What is the smallest possible sum of all of the values visible on the 
faces of the large cube?
Solution
In a 3x3x3 cube, there are 
cubes with three faces showing, 
with two faces showing and with one face showing. The smallest sum with three faces showing is 
, with two faces showing is 
, and with one face showing is 
. Hence, the smallest possible sum is 
. Our answer is thus 
.
Video Solution
https://www.youtube.com/watch?v=1sdXHKW6sqA ~David
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See Also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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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