Difference between revisions of "2003 AIME I Problems/Problem 1"
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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>. | 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>. | ||
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| + | ==Solution 2 (Alcumus)== | ||
| + | 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>, conclude that <math>n</math> must be less than 720, so the maximum value of <math>n</math> is 719. The requested value of <math>k+n</math> is therefore <math>120+719=\boxed{839}</math>. | ||
| + | |||
| + | ~yofro | ||
== See also == | == See also == | ||
Revision as of 14:36, 9 November 2020
Problem
Given that
where 
and 
are positive integers and is as large as possible, find 
Solution
We use the definition of a factorial to get
We certainly can't make any larger if is going to stay an integer, so the answer is 
.
Solution 2 (Alcumus)
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 
, conclude that must be less than 720, so the maximum value of is 719. 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. 
