Difference between revisions of "2000 AIME II Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | Note that <math>1+\sum_{k=1}^{n} {k\cdot k!} = 1+\sum_{k=1}^{n} {(k+1)\cdot k!-\cdot k!} = 1+\sum_{k=1}^{n} {\cdot (k+1)!-\cdot k!} = (k+1)! | + | Note that <math>1+\sum_{k=1}^{n} {k\cdot k!} = 1+\sum_{k=1}^{n} {(k+1)\cdot k!-\cdot k!} = 1+\sum_{k=1}^{n} {\cdot (k+1)!-\cdot k!} = (k+1)!</math> |
| − | The answer is < | + | The answer is <math>\boxed{495}</math>. |
{{incomplete|solution}} | {{incomplete|solution}} | ||
{{AIME box|year=2000|n=II|num-b=13|num-a=15}} | {{AIME box|year=2000|n=II|num-b=13|num-a=15}} | ||
Revision as of 14:21, 29 March 2008
Problem
Every positive integer 
has a unique factorial base expansion 
, meaning that 
, where each 
is an integer, 
, and 
. Given that 
is the factorial base expansion of 
, find the value of 
.
Solution
Note that 
The answer is 
.
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
