Skip to content / Đến nội dung
ThichHocToan.com

2000 AIME II Question 7

AIME · 2000 AIME II

Question 7

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 $\frac{N}{100}.$