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2000 AIME II Problems/Problem 14

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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.

Contents

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...

So now,

\begin{align*}16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\&=16! +\sum_{m=1...

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\...

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}.

See also

2000 AIME II (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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