Difference between revisions of "2003 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
| + | Given that | ||
| − | == Solution == | + | <center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center> |
| + | |||
| + | 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 == | ||
| + | Note that<cmath>{{\left((3!)!\right)!}\over{3!}}= | ||
| + | {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.</cmath>Because <math>120\cdot719!<720!</math>, we can conclude that <math>n < 720</math>. Thus, the maximum value of <math>n</math> is <math>719</math>. The requested value of <math>k+n</math> is therefore <math>120+719=\boxed{839}</math>. | ||
| + | |||
| + | ~yofro | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2003|n=I|before=First Question|num-a=2}} | |
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:27, 21 November 2023
Problem
Given that
where 
and 
are positive integers and is as large as possible, find 
Solution
Note that![\[{{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.\]](http://latex.artofproblemsolving.com/8/6/8/868eddc32a5e4c6b77ab9b1510ff91597ff430c9.png)
Because 
, we can conclude that 
. Thus, the maximum value of is 
. The requested value of 
is therefore 
.
~yofro
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. 
