AMC10 2016 10B, Problem 17

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?

Solution: Let us call the six faces 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 $\fbox{729}$.

Answer: 729