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 $1+\sum_{k=1}^{n} {k\cdot k!} = 1+\sum_{k=1}^{n} {(k+1)\cdot k!- k!} = 1+\sum_{k=1}^{n} {(k+1)!- k!} = (k+1)!$ The answer is $\boxed{495}$ .

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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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