2002 AMC 10A Problems/Problem 18

Problem

A $3$ x $3$ x $3$ cube is made of $27$ normal dice. Each die's opposite sides sum to $7$ . What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?

$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$

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 $8(6)+12(3)+6(1)=48+36+6=90$ . Our answer is thus $\boxed{\text{(D)}\ 90}$ .

Video Solution

https://www.youtube.com/watch?v=1sdXHKW6sqA ~David

Video Solution by Daily Dose of Math

https://youtu.be/QSLlfaf50Is

~Thesmartgreekmathdude

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.

Page

Toolbox

Search

2002 AMC 10A Problems/Problem 18

Contents

Problem

Solution

Video Solution

Video Solution by Daily Dose of Math

See Also