Difference between revisions of "2002 AMC 10A Problems/Problem 18"
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| − | 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> | ||
Revision as of 23:08, 21 September 2024
Contents
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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
