13. Find the number of multiples of 11 in the sequence $99,100,101,102, \cdots, 20130$.
13. Answer: 1822
The the number of multiples of 11 in the sequence $1,2, \ldots, n$ is equal to $\left\lfloor\frac{n}{11}\right\rfloor$. Thus the answer to this question is $\left\lfloor\frac{20130}{11}\right\rfloor-\left\lfloor\frac{98}{11}\right\rfloor=1830-8=1822$.