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.