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

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We certainly can't make <math>n</math> any larger if <math>k</math> is going to stay an integer, so the answer is <math> k + n = 120 + 719 = \boxed{839} </math>.
 
We certainly can't make <math>n</math> any larger if <math>k</math> is going to stay an integer, so the answer is <math> k + n = 120 + 719 = \boxed{839} </math>.
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==Solution 2 (Alcumus)==
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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>, conclude that <math>n</math> must be less than 720, so the maximum value of <math>n</math> is 719. The requested value of <math>k+n</math> is therefore <math>120+719=\boxed{839}</math>.
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~yofro
  
 
== See also ==
 
== See also ==
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{{AIME box|year=2002|n=I|before=First Question|num-a=2}}
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{{AIME box|year=2003|n=I|before=First Question|num-a=2}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Revision as of 15:36, 9 November 2020

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

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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