# Difference between revisions of "2016 AMC 10B Problems/Problem 17"

## 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

Let us call the six sides of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$, $c$ is opposite $e$, and $b$ is opposite $f$. Thus, for the eight vertices, we have the following products: $abc,abe,bcd,bde,acf,cdf,aef,$ and $def$. Let us find the sum of these products: $$abc+abe+bcd+bde+acf+cdf+aef+def$$ We notice $b$ is a factor of the first four terms, and $f$ is a factor of the last four terms. $$b(ac+ae+cd+de)+f(ac+ae+cd+de)$$ Now, we can factor even more:

\begin{align*} & (b+f)(ac+ae+cd+de) \\ = &(b+f)(a(c+e)+d(c+e)) \\ = &(b+f)(a+d)(c+e) \end{align*} We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is $(7+2)$,$(6+3)$, and $(5+4)$ all factors. \begin{align*} & (7+2)(6+3)(5+4) \\ = & 9 \cdot 9 \cdot 9 \\ = & 729. \end{align*} Thus our answer is $\textbf{\boxed{(D)729}}$.

## Solution 2

We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice. Thus our answer is $\textbf{\boxed{(D)729}}$.

## Solution 2

We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3$. To maximize the factorization, we get the answer is $\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}$

## Solution 3 (Cheap Solution)

### Solution

Create a pairing that seems to intuitively be the optimal value, or, in other words, put a number and it's complement (the number that's the difference of 9 and this number) on opposite sides. $1680$ is way too high using reasonability after you do this so you put $\boxed{\textbf{D}}$.

~ pi_is_3.14