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

Revision as of 11:48, 25 October 2006

Given that

$\frac{((3!)!)!}{3!} = k \cdot n!,$

where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$

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 $k + n = 120 + 719 = 839$ .

@@ Line 4: / Line 4: @@
 <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 integer]]s and <math> n </math> is as large as possible, find <math> k + n. </math>
 == Solution ==
@@ Line 11: / Line 11: @@
 <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>
+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 = 839 </math>.
 == See also ==
+* [[2003 AIME I Problems/Problem 2 | Next problem]]
 * [[2003 AIME I Problems]]
+[[Category:Introductory Number Theory Problems]]