2003 AIME I Problems/Problem 1
Revision as of 19:57, 4 July 2013 by Nathan wailes (talk | contribs)
Problem
Given that
![$\frac{((3!)!)!}{3!} = k \cdot n!,$](http://latex.artofproblemsolving.com/d/2/9/d29e6d13e0d8b213ee70519a8250268a6a19a4c1.png)
where and
are positive integers and
is as large as possible, find
Solution
We use the definition of a factorial to get
![$\frac{((3!)!)!}{3!} = \frac{(6!)!}{3!} = \frac{720!}{3!} = \frac{720!}{6} = \frac{720 \cdot 719!}{6} = 120 \cdot 719! = k \cdot n!$](http://latex.artofproblemsolving.com/1/f/c/1fceb00dfff187998b6486b198e21a28c7d216a3.png)
We certainly can't make any larger if
is going to stay an integer, so the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.