2000 AIME II Problems/Problem 14
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
Solution 1
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)
Solution 2 (less formality)
Let . Note that since (or is significantly smaller than ), it follows that . Hence . Then , and as , it follows that . Hence , and we now need to find the factorial base expansion of
Since , we can repeat the above argument recursively to yield , and so forth down to . Now , so .
The remaining sum is now just . We can repeatedly apply the argument from the previous two paragraphs to find that , and if for some , and for all other .
Now for each , we have . Thus, our answer is .
See also
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 |
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