2016 AMC 12B 第 15 题

先试着解答 2016 AMC 12B 第 15 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2016 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

15.

数字 22,33,44,55,66,77 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 2,2, 3,3, 4,4, 5,5, 6,6, 77 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?
312312
343343
625625
729729
16801680
答案:D
知识点:因式分解算术-几何平均不等式
小提示:
将三对相对面分组为 (a,b),(c,d),(e,f)(a,b),(c,d),(e,f)。
Group the three pairs of opposite faces as (a,b),(c,d),(e,f)(a,b),(c,d),(e,f)
大提示:
八个顶点乘积之和可分解为 (a+b)(c+d)(e+f)(a+b)(c+d)(e+f);总和固定时,乘积在因子尽量相等时最大。
The sum of the eight vertex products factors as (a+b)(c+d)(e+f);(a+b)(c+d)(e+f); with a fixed total, a product is largest when the factors are equal
解答:
将相对面配成 (a,b),(c,d),(e,f)(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)(a+b)(c+d)(e+f)。三个因子的总和固定为 2+3+4+5+6+7=272+3+4+5+6+7=27,乘积在三者相等、各为 99 时最大。这可由 (2,7),(3,6),(4,5)(2,7),(3,6),(4,5) 达到,得到 9⋅9⋅9=7299\cdot9\cdot9=729。
所以正确答案是 D。
Pair the opposite faces as (a,b),(c,d),(e,f).(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).(a+b)(c+d)(e+f). The three factors have fixed total 2+3+4+5+6+7=27,2+3+4+5+6+7=27, and a product with fixed sum is largest when the factors are equal, at 99 each. This balance is achievable with (2,7),(3,6),(4,5),(2,7),(3,6),(4,5), giving 9⋅9⋅9=729.9\cdot9\cdot9=729.
Thus, the correct answer is D.
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.