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

Solution 1

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,

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

Therefore we have $f_{16} = 1$ , $f_k=k$ 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:

$\begin{align*} f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ &= -1 + \sum_{m=1}^{62}8\\ &= -1 + 8\cdot 62\\ &= \boxed{495} \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Solution 2 (less formality)

Let $S = 16!-32!+\cdots-1984!+2000!$ . Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \cdots$ is significantly smaller than $2000!$ ), it follows that $1999! < S < 2000!$ . Hence $f_{2000} = 0$ . Then $2000! = 2000 \cdot 1999! = 1999 \cdot 1999! + 1999!$ , and as $S - 2000! << 1999!$ , it follows that $1999 \cdot 1999! < S < 2000 \cdot 1999!$ . Hence $f_{1999} = 1999$ , and we now need to find the factorial base expansion of

$S_2 = S - 1999 \cdot 1999! = 1999! - 1984! + 1962! - 1946! + \cdots + 16!$

Since $|S_2 - 1999!| << 1999!$ , we can repeat the above argument recursively to yield $f_{1998} = 1998$ , and so forth down to $f_{1985} = 1985$ . Now $S_{16} = 1985! - 1984! + 1962! + \cdots = 1984 \cdot 1984! + 1962! + \cdots$ , so $f_{1984} = 1984$ .

The remaining sum is now just $1962! - 1946! + \cdots + 16!$ . We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$ , and $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$ .

Now for each $m$ , we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \cdots + 1 + 1$ $= 8$ . Thus, our answer is $-f_{16} + 8 \cdot 62 = \boxed{495}$ .

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
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2000 AIME II Problems/Problem 14

Problem

Contents

Solution

Solution 1

Solution 2 (less formality)

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