Almost Divisible Three Threes

The sum

$\begin{aligned} S & = 101+104+107+...+998 \\ & = \dfrac {101+998}{2} \left( \dfrac {998-101}{3} + 1\right) \\ & = \dfrac {1099}{2} \times 300 \\ & = \boxed{164850} \end{aligned}$

First we see for the numbers which leave remainder $0$ . Clearly these are $102,105,108......,996,999$ So the numbers that leave remainder 2 are $101,104,107.........,995,998$ This is an AP with a=101,d=3 & l=998 You can also find that $n=300$ . Hence required sum = $\frac{n}{2}(a+l)$ = $\boxed{164850}$