What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?  Choose the correct choice from the following answers.
A) 897  B) 164,850  C) 164,749  D) 149,700  E) 156,720
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 164,850 is the correct answer.
<result>B</result>
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?