I had asked a Q similar to this.
Find the value of this 3 digit number such that dividing by it leaves REM = 11
But I want to know also if instead we take sum of all those numbers. Is there a way to solve this too in a similar way to that Q?
The problem happening is how to put limits in the sum such that the numbers are in between 100 and 999.
To sum up all the $3$-digit numbers that give remainder $11$ upon dividing by $13$.
Note that they are of the form of $13k+11$.
$$100 \le 13k+11 \le 999$$
$$89 \le 13k \le 988$$
$$\lceil \frac{89}{13}\rceil \le k \le \lfloor\frac{988}{13} \rfloor$$
$$7 \le k \le 76$$
I will leave the task of evaluating
$$\sum_{k=7}^{76}(13k+11)$$
as an exercise to you.