AMC 10 · 2016 · #17

Grade 6 geometry-3d
spatial-visualizationoptimizationpolynomial-factoring extremal-construction ↑ Prerequisites: spatial-visualization
📏 Medium solution 💡 3 insights
Problem
The six numbers 2, 3, 4, 5, 6, 7 are placed one per face on a cube. At each of the cube's eight vertices, multiply together the three face-numbers that meet there. Add up all eight of those products. Find the largest total this sum can be.

Pick an answer.

(A)
312
(B)
343
(C)
625
(D)
729
(E)
1680

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The question asks for the greatest possible value, so this is an optimization — tool #14 (Extreme Principle) is the spine. But before we can push the sum to its maximum, we need a clean formula for it. Tool #17 (Visualize Spatial Relationships) shows that each vertex picks one face from each pair of opposite faces, and tool #4 (Introduce a Variable) lets us name the opposite pairs and factor the messy eight-term sum into the tidy product (a+a')(b+b')(c+c'). Once the sum is a product of three pieces whose total is fixed, the Extreme Principle tells us to make the pieces equal, and tool #6 (Guess and Check) confirms an equal split is actually buildable and that lopsided splits score lower.

1STEP 1

See what meets at a vertex

The three faces at any corner are never opposites — a vertex takes one number from each of the three opposite pairs.

vertex product = (from pair 1)×(from pair 2)×(from pair 3)
2STEP 2

Name the opposite pairs

Label the opposite pairs (a,a'), (b,b'), (c,c'); the eight vertex products each pick one letter from every pair.

pairs: (a,a'), (b,b'), (c,c')
3STEP 3

Factor the eight-term sum

Group and factor: the eight products collapse into the single product (a+a')(b+b')(c+c') of the three pair-sums.

Σ = (a+a')(b+b')(c+c')
4STEP 4

Make the three sums equal

The three pair-sums always total 27, and a fixed sum multiplies to the most when split evenly — here 9 × 9 × 9.

(a+a')+(b+b')+(c+c') = 27 → max product at 9· 9· 9 = 9³ = 729
5STEP 5

Build it and check

The pairing (2,7), (3,6), (4,5) makes every pair sum to 9, so the total reaches 9 · 9 · 9 = 729 — choice (D).

(2+7)(3+6)(4+5) = 9· 9· 9 = 729 = (D)
Answer
729
The sum equals (a+a')(b+b')(c+c') with the three factors totaling 27, so the product can be at most 9³ = 729; choice (E) 1680 is impossible because it exceeds this ceiling, and (A) 312, (B) 343, (C) 625 all come from unbalanced pairings that fall short. Sanity check the build: pairing (2,7),(3,6),(4,5) gives 9· 9· 9 = 729, matching (D) exactly. Note 343 = 7³ and 729 = 9³ are both perfect cubes, a hint that the answer is a balanced product.
💡Key takeaway

The cube-sum is just (top+bottom)(front+back)(left+right), and three numbers with a fixed total multiply to the most when you split them evenly — so make every opposite pair add to 9.

  • See what meets at a vertex
  • Name the opposite pairs
  • Factor the eight-term sum
  • Make the three sums equal
  • Build it and check