What is the sum of all remainders obtained when the first 300 natural numbers are divided by 9?  Choose the correct choice from the following answers
A) 397 B) 401 C) 403 D) 405 E) 399
A positive integer can give only the following 9 remainders when divided by 9: 1, 2, 3, 4, 5, 6, 7, 8, and 0.   1 divided by 9 gives the remainder of 1;   2 divided by 9 gives the remainder of 2;   ...   8 divided by 9 gives the remainder of 8;   9 divided by 9 gives the remainder of 0.   We'll have 11 such blocks, since  99/9= 
<gadget id="calculator"> 99/9</gadget>
<output>11</output>
11. The last will be:   91 divided by 9 gives the remainder of 1;   92 divided by 9 gives the remainder of 2;   ...   98 divided by 9 gives the remainder of 8;   99 divided by 9 gives the remainder of 0.   The last number, 100, gives the remainder of 1 when divided by 9, thus the sum of all remainders will be:   11(1+2+3+4+5+6+7+8+0)+1=405.   Answer: 405.
<result>D</result>
What is the sum of all remainders obtained when the first 300 natural numbers are divided by 9?