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.

Let $x$ be any number.  
Let $d_n, d_n-1, \dots, d_1, d_0$ be the digits that make up $x$ so that $x = \sum\limits_{i=0}^{n}10^i{d_i}$
Let $m = \sum\limits_{i=2}^{n}10^i{d_i}$ so that $x = m + 10d_1 + d_0$.
If $d_1 < 6$ or $(d_1=6$ and $d_0=0)$ , then the following are distinct congruence classes modulo $13$:
Note: if $d_0 = 0$, then use $m+10(d_1), m+10(d_1)+1, \dots$ instead.


*

*$m+10(d_1+1)$

*$m+10(d_1+1)+1$

*$\dots$ 

*$m+10(d_1+1)+8$

*$m+10(d_1+1)+9$ 

*$m+10(d_1+2)+9$

*$m+10(d_1+3)+9$

*$m+10(d_1+4)+9$


and since $d_1 + 1 - d_0 < 10$:
$$m+10(d_1+4)+9 - x \le 48$$ 
If $d_1 \ge 6$, then let $e_1 = d_1+3$ and substitute $e_1$ for $d_1$ which gets us $13$ distinct congruence classes with:
Note: if $d_0 = 0$, then use $m+10(e_1), m+10(e_1)+1, \dots$ instead.


*

*$m+10(e_1+1)$

*$m+10(e_1+1)+1$

*$\dots$ 

*$m+10(e_1+1)+8$

*$m+10(e_1+1)+9$ 

*$m+10(e_1+2)+9$

*$m+10(e_1+3)+9$

*$m+10(e_1+4)+9$


and:
$$m+10(e_1+4)+9 - x \le 78$$
