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Difference between revisions of "2003 AIME I Problems/Problem 1"

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Given that
 
Given that
  
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<math> \frac{((3!)!)!}{3!} = k \cdot n!, </math>
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<center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center>
  
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where <math> k </math> and <math> n </math> are positive integers and <math> n </math> is as large as possible, find <math> k + n. </math>
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where <math> k </math> and <math> n </math> are [[positive integer]]s and <math> n </math> is as large as possible, find <math> k + n. </math>
  
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== Solution ==
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==Solution ==
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<math> \frac{((3!)!)!}{3!} = \frac{(6!)!}{3!} = \frac{720!}{3!} = \frac{720!}{6} = \frac{720 \cdot 719!}{6} = 120 \cdot 719! = k \cdot n! </math>
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Note that<cmath>{{\left((3!)!\right)!}\over{3!}}=
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{{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.</cmath>Because <math>120\cdot719!<720!</math>, we can conclude that <math>n < 720</math>. Thus, the maximum value of <math>n</math> is <math>719</math>. The requested value of <math>k+n</math> is therefore <math>120+719=\boxed{839}</math>.
  
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Therefore: <math> k + n = 120 + 719 = 839 </math>
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~yofro
  
 
== See also ==
 
== See also ==
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* [[2003 AIME I Problems]]
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{{AIME box|year=2003|n=I|before=First Question|num-a=2}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 07:27, 21 November 2023

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

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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