What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

Answer: Option B

The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ....
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.
Sum of an AP =
$$\left(\frac{\text{First Term}+\text{Last Term}}{2}\right)\times$$     $$\text{Number}\text{of}\text{Terms}$$
We know that in an A.P., the nth term a_n = a₁ + (n - 1) × d
In this case, therefore, 998 = 101 + (n - 1) × 3
i.e., 897 = (n - 1) × 3
Therefore, n - 1 = 299
Or n = 300
Sum of the AP will therefore, be
$$\begin{array}{rl}& =\frac{101+998}{2}\times 300\\ & =1,64,850\end{array}$$