AIME 2003 I · 第 1 题

AIME 2003 I — Problem 1

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Given that

\frac{((3!)!)!}{3!} = k \cdot n!,

where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$

解析

Solution

Note that

${{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.$ Because $120\cdot719!<720!$ , we can conclude that $n < 720$ . Thus, the maximum value of $n$ is $719$ . The requested value of $k+n$ is therefore $120+719=\boxed{839}$ .

~yofro