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

We use the definition of a factorial to get

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

We certainly can't make $n$ any larger if $k$ is going to stay an integer, so the answer is $k + n = 120 + 719 = \boxed{839}$ .

Solution 2 (Alcumus)

Note that ${{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.$ Because $120\cdot719!<720!$ , conclude that $n$ must be less than 720, so 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

Contents

Problem

Solution

Solution 2 (Alcumus)

See also