Question

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 $0 \lt f_m.$ Given that $(f_1, f_2, f_3, \ldots, f_j)$ is the factorial base expansion of $\begin{aligned} &16! - 32! + 48! \\ &\quad {}- 64! + \cdots + 1968! \\ &\quad {}- 1984! + 2000!, \end{aligned}$ find the value of $f_1 - f_2 + f_3$ $- f_4 + \cdots + (-1)^{j+1} f_j.$

Answer

Answer: 495

Concepts:number base factorial telescoping

Difficulty rating: 3060

Solution:

Since $(i+1)! - i! = i \cdot i!,$ telescoping gives $a! - b! = \sum_{i=b}^{a-1} i \cdot i!$ for $a \gt b.$ Group the given number as $\begin{aligned} &16! + (48! - 32!) + (80! - 64!) \\ &\quad {}+ \cdots + (2000! - 1984!), \end{aligned}$ with $62$ parenthesized groups $(32j + 16)! - (32j)!$ for $j = 1, \ldots, 62.$

The group for $j$ contributes factorial-base digits $f_i = i$ for $32j \le i \le 32j + 15,$ and the lone $16!$ contributes $f_{16} = 1;$ all other digits are $0.$ Every digit satisfies $0 \le f_i \le i,$ so by uniqueness this is the factorial base expansion.

In the alternating sum, $f_{16} = 1$ sits at an even index and contributes $-1.$ Each group's range starts at an even index and has length $16,$ so it splits into $8$ consecutive pairs, each contributing $-i + (i + 1) = 1,$ for $+8$ per group. The total is $62 \cdot 8 - 1 = 495.$

2000 AIME II Problem 14

Answer: 495

Difficulty rating: 3060

Solution:

Problem 14 in Other Years