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

Revision as of 18:03, 23 May 2008

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 $k + n = 120 + 719 = \boxed{839}$ .

@@ 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>
-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>.
+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>.
 == See also ==

2002 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