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!} = | + | Note that <math>1+\sum_{k=1}^{n-1} {k\cdot k!} = 1+\sum_{k=1}^{n-1} {(k+1)\cdot k!- k!} = 1+\sum_{k=1}^{n-1} {(k+1)!- k!} = n!</math> |
| − | + | Thus for all <math>m\in\mathbb{N}</math>, | |
| − | {{ | + | <math>(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.</math> |
| + | |||
| + | So now, | ||
| + | |||
| + | <math> | ||
| + | \begin{align*} | ||
| + | 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ | ||
| + | &=16! +\sum_{m=1}^{62}(32m+16)!-(32m)!\\ | ||
| + | &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! | ||
| + | \end{align*} | ||
| + | </math> | ||
| + | |||
| + | Therefore we have <math>f_{16} = 1</math>, <math>f_k=k</math> if <math>32m\le k \le 32m+15</math> for some <math>m=1,2,\ldots,62</math>, and <math>f_k = 0</math> for all other <math>k</math>. | ||
| + | |||
| + | Therefore we have: | ||
| + | |||
| + | <math> | ||
| + | \begin{align*} | ||
| + | f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ | ||
| + | &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ | ||
| + | &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ | ||
| + | &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ | ||
| + | &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ | ||
| + | &= -1 + \sum_{m=1}^{62}8\\ | ||
| + | &= -1 + 8\cdot 62\\ | ||
| + | &= \boxed{495} | ||
| + | \end{align*} | ||
| + | </math> | ||
{{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 23:43, 5 May 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 
Thus for all 
,
So now,
$\begin{align*} 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ &=16! +\sum_{m=1}^{62}(32m+16)!-(32m)!\\ &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Therefore we have 
, 
if 
for some 
, and 
for all other .
Therefore we have:
$\begin{align*} f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ &= -1 + \sum_{m=1}^{62}8\\ &= -1 + 8\cdot 62\\ &= \boxed{495} \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
| 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 | ||
