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

Explanatory Answer
Step 1: Identify the series
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next couple of numbers that will leave a remainder of 2 when divided by 3 are 104 and 107.
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
It is evident that any number in the sequence will be a 3 digit positive integer of the form (3n + 2).
So, the given numbers are in an Arithmetic sequence with the first term being 101 and the last term being 998 and the common difference being 3.
Step 2: Compute the sum
Sum of an Arithmetic Progression (AP) = [first term + last term/2]n
We know the first term: 101
We know the last term: 998.
The only unknown is the number of terms, n.
In an A.P., the nth term an = a1 + (n - 1)*d
In this case, therefore, 998 = 101 + (n - 1)* 3
Or 897 = (n - 1) * 3
(n - 1) = 299 or n = 300.
Sum of the AP will therefore, be [101+998/2]∗300 = 164,850
Choice B is the correct answer.