1997 AMC 12 第 28 题

先试着解答 1997 AMC 12 第 28 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 1997 AMC 12 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。

28.

有多少个整数有序三元组 (a,b,c)(a,b,c) 满足 ∣a+b∣+c=19,ab+∣c∣=97? \begin{aligned} |a+b|+c&=19,\\ ab+|c|&=97 \end{aligned}\text{?}
How many ordered triples of integers (a,b,c)(a,b,c) satisfy ∣a+b∣+c=19,ab+∣c∣=97? \begin{aligned} |a+b|+c&=19,\\ ab+|c|&=97? \end{aligned}
00
44
66
1010
1212
答案:E
知识点:Diophantine equationsabsolute valuesfactorization
小提示:
令 s=∣a+b∣s=|a+b|,则 c=19−sc=19-s,再根据 cc 的正负分类
Set s=∣a+b∣s=|a+b|, so c=19−sc=19-s, and split according to the sign of cc
大提示:
当 c<0c\lt0 时,将所得方程改写为乘积等于 117117 的形式
When c<0c\lt0, rewrite the resulting equations as products equal to 117117
解答:
若 c≥0c\ge0,代入并因式分解后,没有任何数对符合 a+ba+b 所需的符号。因此 c<0c\lt0,所以 s>19s\gt19 且 ab+s=116ab+s=116。若 a+b=sa+b=s,则 (a+1)(b+1)=117(a+1)(b+1)=117,得到无序数对 {0,116}\{0,116\}、{2,38}\{2,38\}、{8,12}\{8,12\}。若 a+b=−sa+b=-s,则 (a−1)(b−1)=117(a-1)(b-1)=117,得到 {−116,0}\{-116,0\}、{−38,−2}\{-38,-2\}、{−12,−8}\{-12,-8\}。每个无序数对都有两种次序,因此共有 6⋅2=126\cdot2=12 个三元组。正确答案是 E。
If c≥0,c\ge0, substitution and factoring lead to no pair consistent with the required sign of a+b.a+b. Thus c<0,c\lt0, so s>19s\gt19 and ab+s=116.ab+s=116. If a+b=s,a+b=s, then (a+1)(b+1)=117,(a+1)(b+1)=117, producing the unordered pairs {0,116},\{0,116\}, {2,38},\{2,38\}, {8,12}.\{8,12\}. If a+b=−s,a+b=-s, then (a−1)(b−1)=117,(a-1)(b-1)=117, producing {−116,0},\{-116,0\}, {−38,−2},\{-38,-2\}, {−12,−8}.\{-12,-8\}. Each unordered pair has two orders, giving 6⋅2=126\cdot2=12 triples. The correct answer is E.
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