SAMC

(a) We are looking for integers $m$ such that $$10^m-1=13n+8\text{ for some positive integer }n$$ The last relation is equivalent to $10^m-9\equiv 0(\mathrm{mod}13)$. We will prove that ${\mathrm{ord}}_{13}(10)=6$, that is, the smallest positive integer $s$ such that $10^s\equiv 1(\mathrm{mod}13)$ is $s=6$. Indeed, we have $$10^6=(13-3{)}^6\equiv 3^6=(3^3{)}^2=(26+1{)}^2\equiv 1(\mathrm{mod}13).$$ Moreover, $10^2\equiv 9(\mathrm{mod}13)$ and $10^3\equiv -1(\mathrm{mod}13)$, hence $${\mathrm{ord}}_{13}(10)=6.$$ From the previous property, it follows that for any integer $k\ge 0$, we have $10^{6k}\equiv 1(\mathrm{mod}13)$. Therefore, $$10^{6k+2}=10^{6k}\cdot 10^2\equiv 9(\mathrm{mod}13),$$ hence the integers $$10^{6k+2}-1=\underset{6k+2}{\underbrace{99\dots 9}},k=0,1,\dots ,$$ belong to the sequence, and all integers in the sequence containing only digit $9$ are of this form.

(b) It is clear that we have to find the greatest integer $k$ such that $10^{6k+2}-1<2010^{2010}$, that is $10^{6k+2}-1<10^{6k+2}<2010^{2010}$. It follows $6k+2<2010\log 2010$, hence $$\begin{array}{c}3k+1<1005\log 201\cdot 10=1005+1005\log 201\\ =1005+2010+1005\log \left(2+\frac{1}{100}\right)\\ =3015+1005\log \left(2+\frac{1}{100}\right)\end{array}$$ hence $$k=1004+\left[\frac{2}{3}+335\log \left(2+\frac{1}{100}\right)\right]=1105.$$ The desired number is $1106$.