2016 AMC 12B Problem 15

Question

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?

312

343

625

729

1680

Answer: D

Concepts:factoring AM-GM Inequality

Difficulty rating: 1800

Small hint:

Group the three pairs of opposite faces as

(a,b),(c,d),(e,f)

Big hint:

The sum of the eight vertex products factors as

(a+b)(c+d)(e+f);

with a fixed total, a product is largest when the factors are equal

Solution:

Pair the opposite faces as

(a,b),(c,d),(e,f).

Each vertex product uses one face from each pair, so the sum of all eight products factors as

(a+b)(c+d)(e+f).

The three factors have fixed total

2+3+4+5+6+7=27,

and a product with fixed sum is largest when the factors are equal, at

9

each. This balance is achievable with

(2,7),(3,6),(4,5),

giving

9\cdot9\cdot9=729.

Thus, the correct answer is D.