2016 AMC 12B Problems/Problem 15

Problem

All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution 1

First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\boxed{\textbf{(D)}\ 729 }$ .

Solution 2

We proceed from the factorization in the above solution. By the AM-GM inequality,

$\frac{a_1+a_2++a_3}{3}\geq\sqrt[3]{a_1a_2a_3}$

Cubing both sides,

$\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}$

Let $a_1=(a+f)$ , $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.

$\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}$

$a+b+c+d+e+f$ is fixed at 27.

$\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}$

$\boxed{\textbf{(D)}\ 729 }\geq{(a+f)(b+c)(d+e)}$

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2016 AMC 12B Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See Also