Difference between revisions of "2002 AMC 10A Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | In a | + | 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>. |
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| + | ==Video Solution== | ||
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| + | https://www.youtube.com/watch?v=1sdXHKW6sqA ~David | ||
==See Also== | ==See Also== | ||
Latest revision as of 20:43, 28 May 2023
Contents
Problem
A 3x3x3 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
