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: $abc$ , $abe$ , $bcd$ , $bde$ , $acf$ , $cdf$ , $cef$ , and $def$ . Let us find the sum of these products:

We notice $b$ is a factor of the first four terms, and $f$ is factor is the last four terms.

Now, we can factor even more:

We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make $(7+2)$ , $(6+3)$ , and $(5+4)$ all factors.

Thus our answer is $\textbf{(D)}\ 729$ .

Solution 2=

We first find the factorization $(b+f)(a+d)(c+e)$ . 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 $\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}$

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2016 AMC 10B Problems/Problem 17

Contents

Problem

Solution 1

Solution 2=

See Also