What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?	Choose the correct choice:
A) 897 B) 164,850 C) 164,749 D) 149,700 E) 156,720

Rationale: "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 ( 3 n + 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 = a 1 + ( 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 ."
Answer: B