2000 AIME II Problems/Problem 14

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

Thus for all $m\in\mathbb{N}$ ,

$(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.$

So now,

$ $\begin{aligned} 16! - 32! + 48! - 64! + \dots + 1968! - 1984! + 2000! & = 16! + (48! - 32!) + (80! - 64!) \dots + (2000! - 1984!) \\ = 16! + \sum_{m = 1}^{62} (32 m + 16)! - (32 m)! \\ = 16! + \sum_{m = 1}^{62} \sum_{k = 32 m}^{32 m + 15} k \cdot k! \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg)

Therefore we have $f_{16} = 1$ , $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$ .

Therefore we have:

$ $\begin{aligned} f_{1} - f_{2} + f_{3} - f_{4} + \dots + (- 1)^{j + 1} f_{j} & = (- 1)^{17} \cdot 1 + \sum_{m = 1}^{62} \sum_{k = 32 m}^{32 m + 15} (- 1)^{k + 1} k \\ = - 1 + \sum_{m = 1}^{62} [\sum_{j = 16 m}^{16 m + 7} (- 1)^{2 j + 1} 2 j + \sum_{j = 16 m}^{16 m + 7} (- 1)^{2 j + 2} (2 j + 1)] \\ = - 1 + \sum_{m = 1}^{62} \sum_{j = 16 m}^{16 m + 7} [(- 1)^{2 j + 1} 2 j + (- 1)^{2 j + 2} (2 j + 1)] \\ = - 1 + \sum_{m = 1}^{62} \sum_{j = 16 m}^{16 m + 7} [- 2 j + (2 j + 1)] \\ = - 1 + \sum_{m = 1}^{62} \sum_{j = 16 m}^{16 m + 7} 1 \\ = - 1 + \sum_{m = 1}^{62} 8 \\ = - 1 + 8 \cdot 62 \\ = 495 \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg)

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2000 AIME II Problems/Problem 14

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