Difference between revisions of "2002 AMC 10A Problems/Problem 18"
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==Solution== | ==Solution== | ||
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>. | 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== |
Revision as of 19:43, 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 |
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