2016 AMC 10B Problem 17

Step 1: see how the corners are built.

1. A cube's six faces come in three pairs of opposite faces: top and bottom, front and back, left and right. Call their numbers $(a, b)$, $(c, d)$ and $(e, f)$.
2. Every corner touches exactly one face from each pair. For example, the top-front-left corner touches top, front and left. There are $2 \cdot 2 \cdot 2 = 8$ ways to pick one face from each pair, and each way is one of the $8$ corners.

Step 2: the sum is a product.

3. Multiply out $(a + b)(c + d)(e + f)$. You get $8$ terms, one for every way of picking one letter from each bracket: $ace, acf, ade, \ldots, bdf$. Those are exactly the $8$ corner products. So
$$\text{sum of the corner products} = (a + b)(c + d)(e + f).$$

Step 3: make the product as big as possible.

1. The three pair sums always add to $2 + 3 + 4 + 5 + 6 + 7 = 27$, no matter how the numbers are placed.
2. Three positive numbers with a fixed total have the largest product when they are equal. For example, $8 \cdot 9 \cdot 10 = 720$ is less than $9 \cdot 9 \cdot 9 = 729$. So the best we can hope for is three sums of $9$ each.
3. We can get that: put $2$ opposite $7$, $3$ opposite $6$, and $4$ opposite $5$.

The greatest sum is $9 \cdot 9 \cdot 9 = 729$.

The answer is $\boxed{\textbf{(D)}\ 729}$.

When you add up the results of every possible "one from each group" choice, the total factors into a product of group sums. That turns a picture problem into a number problem: split $27$ into three parts with the largest product. Equal parts always win.

Use the answer choices. The best product of three numbers adding to $27$ is $9^3 = 729$, so $1680$ is impossible. The pairing $2$–$7$, $3$–$6$, $4$–$5$ reaches $729$, and $729$ is bigger than $625$, $343$ and $312$, so it is the answer.

Group the six faces into three pairs of opposite faces, carrying numbers $(a, b)$, $(c, d)$, $(e, f)$. A vertex touches exactly one face from each pair, and the eight vertices realize all eight choices of one face per pair. Therefore the eight vertex products are precisely the eight terms in the expansion of
$$(a + b)(c + d)(e + f),$$
and their sum equals this product.

The three pair sums add to $2 + 3 + \cdots + 7 = 27$, whatever the arrangement. A product of three positive numbers with fixed sum $27$ is largest when the numbers are equal, each $9$ (AM-GM). That is achievable: pair $2$ with $7$, $3$ with $6$, and $4$ with $5$ on opposite faces.

The maximum is $9 \cdot 9 \cdot 9 = 729$.

The answer is $\boxed{\textbf{(D)}\ 729}$.