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!- k!} = 1+\sum_{k=1}^{n} {(k+1)!- k!} = (k+1)!</math>
+
Note that <math>1+\sum_{k=1}^{n} {k\cdot k!} = 1+\sum_{k=1}^{n} {(k+1)\cdot k!- k!} = 1+\sum_{k=1}^{n} {(k+1)!- k!} = (n+1)!</math>
  
 
The answer is <math>\boxed{495}</math>.
 
The answer is <math>\boxed{495}</math>.

Revision as of 14:22, 29 March 2008

Problem

Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$, meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$, where each $f_i$ is an integer, $0\le f_i\le i$, and $0<f_m$. Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$, find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$.

Solution

Note that $1+\sum_{k=1}^{n} {k\cdot k!} = 1+\sum_{k=1}^{n} {(k+1)\cdot k!- k!} = 1+\sum_{k=1}^{n} {(k+1)!- k!} = (n+1)!$

The answer is $\boxed{495}$.

Template:Incomplete

2000 AIME II (ProblemsAnswer KeyResources)
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