Question

已知求小于 $\frac{N}{100}$ 的最大整数。 $\begin{aligned} &\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} \\ &\quad {}+ \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} \\ &\quad {}+ \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{1!18!}, \end{aligned}$

Given that $\begin{aligned} &\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} \\ &\quad {}+ \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} \\ &\quad {}+ \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{1!18!}, \end{aligned}$ find the greatest integer that is less than $\frac{N}{100}.$

答案

答案：137

知识点：二项式定理阶乘对称性（代数）

难度评级：2360

解答：

两边同乘 $19!$ 。左边每一项变为 $\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}$ ，其中 $k = 2, \ldots, 9$ ；右边变为 $19N$ 。

因为 $\binom{19}{k} = \binom{19}{19 - k}$ ，这一行二项式系数的前半部分之和为 $\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18}$ 。所以 $\begin{aligned} \sum_{k=2}^{9}\binom{19}{k} &= 2^{18} - \binom{19}{0} \\ &\quad {}- \binom{19}{1} \\ &= 262144 - 20 \\ &= 262124. \end{aligned}$

因此 $N = \frac{262124}{19} = 13796$ ，所以 $\frac{N}{100} = 137.96$ ，小于它的最大整数是 $137$ 。

Multiply both sides by $19!.$ Each term on the left becomes $\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}$ for $k = 2, \ldots, 9,$ while the right side becomes $19N.$

Since $\binom{19}{k} = \binom{19}{19 - k},$ the first half of the binomial row sums to $\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18},$ so $\begin{aligned} \sum_{k=2}^{9}\binom{19}{k} &= 2^{18} - \binom{19}{0} \\ &\quad {}- \binom{19}{1} \\ &= 262144 - 20 \\ &= 262124. \end{aligned}$

Hence $N = \frac{262124}{19} = 13796,$ so $\frac{N}{100} = 137.96,$ and the greatest integer less than this is $137.$

2000 AIME II 第 7 题

答案：137

难度评级：2360

解答：

其他年份的第 7 题