Finding the remainder when a large number is divided by 13

Brute force isn't demanding so much effort, actually a handful of two-digits subtractions, using the table $13,26,39,52,65,78,91,104,117$.

$$\color{blue}{13}5792468135792468\\\color{blue}{57}92468135792468\\\color{blue}{59}2468135792468\\\color{blue}{72}468135792468\\\color{blue}{74}68135792468\\\color{blue}{96}8135792468\\\color{blue}{58}135792468\\\color{blue}{61}35792468\\\color{blue}{93}5792468\\\color{blue}{25}792468\\\color{blue}{127}92468\\\color{blue}{109}2468\\\color{blue}{52}468\\\color{blue}{46}8\\\color{blue}{78}\\\color{blue}0.$$

$135-792+468-135+792-468=0\implies$

• $7$ divides $135792468135792468$ without remainder
• $11$ divides $135792468135792468$ without remainder
• $13$ divides $135792468135792468$ without remainder

This trick is applicable since each one of them divides $1001$ without remainder.

A little digression to speculate on the source of the problem.

The test for divisibility by $9$ is well known: the remainder is the sum of the digits (mod $9$).

The test for divisibility by $11$ is a little less well known: look at the alternating sum of the digits. That works because odd powers of $10$ are $-1 \pmod{11}$ while even powers are $1$.

Now note the lovely fact that $7 \times 11 \times 13 = 1001$. That means you can find the remainder mod $1001$ and hence mod $7$, $11$ and $13$ by alternately adding and subtracting "digits" in groups of three - essentially thinking of the number as written in base $1000$.

(Written while @Barakmanos was posting essentially the same argument.)