1979 AMC 12 第 22 题

先试着解答 1979 AMC 12 第 22 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 1979 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。

22.

求满足下列方程的整数对 (m,n)(m,n) 的个数:m3+6m2+5m=27n3+9n2+9n+1。 \begin{aligned} m^3+6m^2+5m &=27n^3+9n^2\\ &\quad+9n+1 \end{aligned}\text{。}
Find the number of pairs (m,n)(m,n) of integers which satisfy the equation m3+6m2+5m=27n3+9n2+9n+1. \begin{aligned} m^3+6m^2+5m &=27n^3+9n^2\\ &\quad+9n+1. \end{aligned}
00
11
33
99
无穷多个
infinitely many
答案:A
知识点:整除性模运算因式分解
小提示:
把左边因式分解为 m(m+1)(m+5)m(m+1)(m+5)
Factor the left side as m(m+1)(m+5)m(m+1)(m+5)
大提示:
比较等式两边模 33 的余数
Compare the two sides modulo 33
解答:
左边等于 m(m+1)(m+5)=m(m+1)(m+2)+3m(m+1), \begin{gathered} m(m+1)(m+5)\\ =m(m+1)(m+2)\\ \quad+3m(m+1) \end{gathered}\text{,}对每个整数 mm 都能被 33 整除。右边为 3(9n3+3n2+3n)+13(9n^3+3n^2+3n)+1,与 1(mod3)1\pmod3 同余。因此不存在满足方程的整数对。
因此,正确答案是 A。
The left side is m(m+1)(m+5)=m(m+1)(m+2)+3m(m+1), \begin{gathered} m(m+1)(m+5)\\ =m(m+1)(m+2)\\ \quad+3m(m+1), \end{gathered} which is divisible by 33 for every integer m.m. The right side is 3(9n3+3n2+3n)+1,3(9n^3+3n^2+3n)+1, which is congruent to 1(mod3).1\pmod3. Therefore no integer pair satisfies the equation.
Therefore, the correct answer is A.
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