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 $a_1<a_2<\cdots<a_{78}<a_{79}$ be the $79$ consecutive numbers.
Case $1$: The digit in the hundreds place is the same for all numbers, i.e., $\left\lfloor \dfrac{a_i}{100}\right\rfloor$ is the same for all numbers. Consider the smallest of the lot, which has $0$ as its last digit, say $a_k$, where $k \in \{1,2,\ldots,9\}$. Note that $k+39<79$.


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*If $\text{sum}(a_k) \pmod{13} = 0$, then we are done.

*If $\text{sum}(a_k) \pmod{13} = b \in \{4,5,\ldots,12\} $, then $\text{sum}(a_{k+13-b}) \pmod{13} = 0$.

*If $\text{sum}(a_k) \pmod{13} = 3$, then $a_{k+19} \pmod{13} = 0$

*If $\text{sum}(a_k) \pmod{13} = 2$, then $a_{k+29} \pmod{13} = 0$

*If $\text{sum}(a_k) \pmod{13} = 1$, then $a_{k+39} \pmod{13} = 0$


Case $2$: There is a number $a_m$ such that $100$ divides $a_m$, i.e., if $k<m$, then $a_k$'s have same digit in hundreds place and if $k \geq m$, then $a_k$'s have same digit in hundreds place. Now repeat the same argument as above by considering either $a_0$ to $a_{m-1}$ or $a_m$ to $a_{79}$, whichever has $40$ numbers.
