Prove that in every sequence of 79 consecutive positive numbers written in decimal system there is a number whose sum of the digits is divisible by 13 
Prove that in every sequence of $79$ consecutive positive numbers written in decimal notation there is a number the sum of whose digits is divisible by $13$.

I tried to take one by one sets of $79$ consecutive positive numbers. Then I tried to solve with sets,relation,function. But I am not getting any idea how to start solving the question.

For $x\in{\mathbb N}$ denote by $r(x)$ the remainder modulo $13$ of the decimal representation of $x$. If $x$ is not divisible by $10$ then $r(x)=r(x-1)+1$. If $x$ is divisible by $10$, but not by $100$, then $$r(x)=r(x-1)-9+1=r(x-1)+5\ .\tag{1}$$ If $x$ is divisible by $100$, things are more complicated; see below.
Consider a run $R:=[a\ ..\ a+78]$ of  $79$ consecutive natural numbers, and assume that none of these numbers is divisible by $100$. There is a smallest number $c\leq a+9$ in this run which is divisible by $10$. Assume that $r(c)=1$. Then $(1)$ implies that the $r$-values in the interval $[c\ ..\ c+39]\subset R$ are given by
$$[1\ ..\ 10],\quad[2\ ..\ 11],\quad [3\ ..\ 12],\quad[4\ ..\ 13]\tag{2}$$
and cover all remainders modulo $13$. It follows that  $R$ covers all remainders modulo $13$, even if $r(c)\ne1$. Note that we would see $r(x)=0$ for an $x<c+39$ if we had not insisted in $r(c)=1$.
The run $R$ may contain at most one number $c$ divisible by $100$. Assume that $r(c-1)=12$ and $r(c)=1$. Then the $r$-values in the interval $[c\ ..\ c+39]$ are still given by $(2)$. The $r$-values in the intervall $[c-40\ ..\ c-1]$ are given by
$$[0\ ..\ 9],\quad [1\ ..\ 10],\quad [2\ ..\ 11],\quad[3\ ..\ 12]\ ,$$
and cover all remainders modulo $13$. Note that we would see $r(x)=0$ for an  $x>c-40$ if we had not insisted in $r(c-1)=12$.
It follows that in the worst possible case we don't see the $r$-value $0$ for all $x$ in the interval $[c-39\ ..\ c+38]$ containing $78$ integers. In order to realize this worst possible case we need $c$ divisible by $100$, $r(c)=1$, and $r(c-1)=12$. Assume that $c-1$ has $k\geq2$ trailing nines in its decimal expansion. Then the stated conditions imply
$$2=r(c)-r(c-1)=-9k+1\qquad({\rm mod}\>13)\ .$$
The smallest $k\geq2$ fulfilling this is $k=10$. The number $c-1=10^{10}-1=9999999999$ already has $r(c-1)=12$, and it is obvious that $r(c)=1$ in this case.  
To sum it all up: Any run of $79$ consecutive positive integers contains an $x$ with $r(x)=0$. The first run of $78$ consecutive integers containing no $x$ with $r(x)=0$ is given by $[10^{10}-39\ ..\ 10^{10}+38]$.
