2003 AMC 10A Problem 25

Write $n = 100q + r = (q + r) + 99q.$ Since $99q$ is a multiple of $11,$ $q + r$ is divisible by $11$ if and only if $n$ is. The $5$-digit multiples of $11$ satisfy $10000 \le n \le 99999,$ and there are $$\begin{aligned} &\left\lfloor\dfrac{99999}{11}\right\rfloor - \left\lfloor\dfrac{9999}{11}\right\rfloor \\ &= 9090 - 909 \\ &= 8181. \end{aligned}$$ Thus, the correct answer is B.