Difference between revisions of "2003 AIME I Problems/Problem 1"
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<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> | <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 == | == See also == |
Revision as of 19:03, 23 May 2008
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
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 |