# Difference between revisions of "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.

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

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

## Solution 3 (really fast)

First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving $4.5^3$ (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer $\boxed{\textbf{(D)}\ 729 }$. Also note that if you cannot evaluate $4.5^3$ quickly, a rough approximation of $5*4.5*4*8$ will yield 720, very close to our answer. -rayprati