Question

数字 $2,3,4,5,6,7$ 被分配到一个立方体的六个面上，每面一个数。对立方体的每个顶点，计算包含该顶点的三个面上的数字之积。这八个乘积之和的最大可能值是多少？

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?

$312$

$343$

$625$

$729$

$1680$

答案：D

知识点：因式分解算术-几何平均不等式

难度评级：1800

解答：

将相对面配成 $(a,b),(c,d),(e,f)$ 每个顶点乘积从每对相对面中取一个数，所以八个乘积之和可分解为 $(a+b)(c+d)(e+f)$ 三个因子的总和固定为 $2+3+4+5+6+7=27$ 乘积在三者相等、各为 $9$ 时最大。这可由 $(2,7),(3,6),(4,5)$ 达到，得到 $9\cdot9\cdot9=729$ 。

所以正确答案是 D。

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.

2016 AMC 12B 第 15 题

答案：D

难度评级：1800

解答：

其他年份的第 15 题