Difference between revisions of "2003 AIME I Problems/Problem 1"
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Given that | Given that | ||
| − | <math> \frac{((3!)!)!}{3!} = k \cdot n!, </math> | + | <center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center> |
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> | 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> | ||
== Solution == | == Solution == | ||
| − | <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> | + | We use the definition of a [[factorial]] to get |
| + | |||
| + | <center><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></center> | ||
Therefore: <math> k + n = 120 + 719 = 839 </math> | Therefore: <math> k + n = 120 + 719 = 839 </math> | ||
Revision as of 22:09, 15 July 2006
Problem
Given that
where 
and 
are positive integers and is as large as possible, find 
Solution
We use the definition of a factorial to get
Therefore: 
