2001 AIME II 第 10 题

Factor $10^j - 10^i = 10^i(10^{j-i} - 1)$ and note that $1001$ is coprime to $10^i$

$10^k - 1$ is divisible by $1001$ exactly when $k$ is a multiple of $6,$ so count pairs $0 \le i \lt j \le 99$ with $j - i$ a multiple of $6$

Factor $10^j - 10^i = 10^i(10^{j-i} - 1).$ Since $1001 = 7 \cdot 11 \cdot 13$ is coprime to $10^i,$ we need $10^{j-i} - 1$ to be divisible by $1001.$ The multiplicative order of $10$ is $6$ modulo $7,$ $2$ modulo $11,$ and $6$ modulo $13,$ so $10^k \equiv 1 \pmod{1001}$ exactly when $k$ is a multiple of $6.$ Distinct pairs $(i, j)$ give distinct values, so we just count the pairs.

For $j - i = 6d$ with $1 \le d \le 16,$ the index $i$ can be $0, 1, \ldots, 99 - 6d,$ giving $100 - 6d$ choices. The total is $$\begin{aligned} &\sum_{d=1}^{16} (100 - 6d) \\ &= 94 + 88 + \cdots + 4 \\ &= \frac{(94 + 4) \cdot 16}{2} \\ &= 784. \end{aligned}$$