2012 AIME II 第 10 题

Fix $\lfloor x \rfloor = a;$ as $x$ runs from $a$ to $a + 1,$ the product $x\lfloor x \rfloor$ sweeps through the integers from $a^2$ to $a^2 + a - 1$

Each $a$ contributes exactly $a$ values of $n;$ sum over all $a$ whose values stay below $1000$

Fix $\lfloor x \rfloor = a \ge 1.$ For $a \le x \lt a + 1$ the product $n = ax$ ranges over $[a^2, a^2 + a),$ and every integer $n$ in that interval is achieved by $x = \frac{n}{a}$ (which indeed has floor $a$). So each $a$ contributes exactly $a$ values of $n,$ namely $a^2, a^2 + 1, \ldots, a^2 + a - 1.$

For $a = 31$ the largest value is $31^2 + 30 = 991 \lt 1000,$ so all values through $a = 31$ qualify, while $a = 32$ already starts at $1024.$ The count is $1 + 2 + \cdots + 31 = \frac{31 \cdot 32}{2} = 496.$