Algebra

We will tackle the general case first. What is the sum of all positive integers less than 1000 ending in a digit n?

The numbers must be of the form $10m + n$ where $0 \leq m \leq 99$.Therefore, the answer will be $\displaystyle \sum_{m = 0}^{99} (10m + n) = 100n + 49500$.

Hence, the answer to your original question will be $4(49500) + 100(3 + 4 + 6 + 9) = 200200$.