2003 AIME I Problems/Problem 1

Problem

Given that

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

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

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

2003 AIME I (Problems • Answer Key • Resources)
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2003 AIME I Problems/Problem 1

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