# 2002 AMC 10A Problems/Problem 18

## Problem

A 3x3x3 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 $\boxed{\text{(D)}\ 90}$.

## See Also

 2002 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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. Invalid username
Login to AoPS