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2003 AIME I Problems/Problem 1

Revision as of 19:45, 15 July 2006 by Xantos C. Guin (talk | contribs) (Added problem and solution)

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

$\frac{((3!)!)!}{3!} = \frac{(6!)!}{3!} = \frac{720!}{3!} = \frac{720!}{6} = \frac{720 \cdot 719!}{6} = 120 \cdot 719! = k \cdot n!$

Therefore: $k + n = 120 + 719 = 839$

See also

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