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 $\boxed{\text{(D)}\ 90}$ .

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2002 AMC 10A Problems/Problem 18

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