2003 AMC 12B Problem 23

The intercepts occur where $\frac{1}{x} = k\pi,$ that is $x = \dfrac{1}{k\pi}$ for a nonzero integer $k.$ The condition $0.0001 \lt \dfrac{1}{k\pi} \lt 0.001$ becomes $$\frac{1000}{\pi} \lt k \lt \frac{10000}{\pi}.$$ The number of such integers is $$\begin{aligned} &\left\lfloor \frac{10000}{\pi} \right\rfloor - \left\lfloor \frac{1000}{\pi} \right\rfloor \\ &= 3183 - 318 = 2865, \end{aligned}$$ closest to $2900.$ Thus, the correct answer is A.