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2016 AMC 10B Problem 17

2016 AMC 10B Problem 17AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 3 of 5 · about 4 min · position P16-20

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?

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

Answer

Key insight

Each vertex takes one face from each pair of opposite faces, so the eight products sum to (a+b)(c+d)(e+f); the pair sums total 27, so the max is 9^3.

Solution

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 $\boxed{\textbf{(D)}\ 729}$ .

Why this works

The cube's vertices are exactly the "one from each pair" selections, so a sum of eight triple products factors into a product of three binomials. Once factored, the geometry disappears and the problem becomes: split $27$ into three parts with the largest product, which balance (AM-GM) decides. Look for this factoring whenever every combination of independent choices is summed.

Alternative approach

Sanity-check the choices: $729 = 9^3$ and $625$ , $343$ , $312$ are smaller than what the balanced pairing gives, while $1680 = 5 \cdot 6 \cdot 7 \cdot 8$ exceeds the AM-GM bound $9^3$ , so it is impossible.

The trap

Trying to place the numbers so the three largest meet at one vertex, which does not maximize the total; only the pairing into opposite faces matters.

Common mistakes

Trying to place the numbers so the three largest meet at one vertex, which does not maximize the total; only the pairing into opposite faces matters.
Pairing adjacent-looking numbers such as $(2,3), (4,5), (6,7)$ , which gives $5 \cdot 9 \cdot 13 = 585$ , well below the balanced $729$ .

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Exploit symmetry to reduce work or pair up objects