2018 AMC 10B 第 25 题

Writing $\{x\} = x - \lfloor x \rfloor,$ the equation is $x^2 = 10{,}000\{x\},$ so $0 \le x^2 < 10{,}000$

On each interval $[a, a+1)$ there is a solution exactly when $(a + 1)^2 \lt 10{,}000$

Let $a = \lfloor x \rfloor.$ The equation reads $x^2 = 10{,}000(x - a)$ $= 10{,}000\{x\},$ and since $0 \le \{x\} < 1,$ this forces $0 \le x^2 < 10{,}000,$ so $-100 < x < 100.$ On each interval $[a, a + 1)$ the quantity $10{,}000x - x^2$ increases from $10{,}000a - a^2$ and approaches, but does not reach, $10{,}000(a+1) - (a+1)^2.$ It hits $10{,}000a$ exactly once precisely when $(a + 1)^2 < 10{,}000.$ That holds for the integers $-100 \le a \le 98,$ which is $199$ solutions. Thus, C is the correct answer.