A)a number that filled with $9999999......$ which we can rewrite as = $(10^n-1)$ $10^n-1= 13k-5$ $k=\frac{(10^n)}{13}+\frac{4}{13}$ where k is integer $k = \frac{10^n+4}{13}$ which we can find $n={2,8,14,20,.......}$ which form a sequence $2+6m$ where m is positive integer also k is repeating $n = 2,k=8,m=0$ $n= 8 ,k=7692308,m=1$ $n = 14,k=7692307692308,m=2$ which mean $k = (769230....) m$ times and ends with 8 , $k=(769230769230.............8)$ b)$2010^{2010}=13k-5$ $2010^{2010}+5=13k$ $[\frac{2010^{2010}+5}{13}]=k$ where [x] is flooring function