2024 AMC 12A 第 6 题

先试着解答 2024 AMC 12A 第 6 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 2024 AMC 12A 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。

6.

三个整数的乘积为 6060。这三个整数的正的和的最小可能值是多少?
The product of three integers is 60.60. What is the least possible positive sum of the three integers?
22
33
55
66
1313
答案:B
知识点:因数最优化分类讨论
小提示:
整数可以为负;两个负因数会让乘积仍为正
Integers may be negative; the product stays positive with two negative factors
大提示:
将三个数写成 −p,−q,r-p,-q,r,其中 pqr=60pqr=60,在保持和为正的条件下最小化 r−p−qr-p-q。
Writing the numbers as −p,−q,r-p,-q,r with pqr=60,pqr=60, minimize r−p−qr-p-q while keeping it positive
解答:
要得到较小的正和,可以取两个负整数 −p,−q-p,-q 和一个正整数 rr,其中 pqr=60pqr=60。不计顺序时,6060 的正因数三元组为 (1,1,60),(1,2,30),(1,3,20),(1,4,15),(1,5,12),(1,6,10),(2,2,15),(2,3,10),(2,5,6),(3,4,5)。 \begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5) \end{gathered}\text{。} 把最大的因数取作 rr 时,r−p−qr-p-q 的正值最小。表中得到的正值为 58,27,16,10,6,3,11,558,27,16,10,6,3,11,5;最后两个三元组给出负值。因此最小的正和为 10−1−6=310-1-6=3。(三个正整数之和至少为 33,而三个负整数的乘积不可能为正。)因此正确答案是 B。
To obtain a small positive sum, use two negative integers −p,−q-p,-q and one positive integer r,r, where pqr=60.pqr=60. Up to order, the positive factor triples of 6060 are (1,1,60),(1,2,30),(1,3,20),(1,4,15),(1,5,12),(1,6,10),(2,2,15),(2,3,10),(2,5,6),(3,4,5). \begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5). \end{gathered} A positive value of r−p−qr-p-q is smallest when the largest factor is chosen as r.r. The positive values from the list are 58,27,16,10,6,3,11,5;58,27,16,10,6,3,11,5; the last two triples give negative values. Thus the least positive sum is 10−1−6=3.10-1-6=3. (Three positive integers have sum at least 3,3, and three negative integers cannot have positive product.) Thus, the correct answer is B.
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