What is the sum of all remainders obtained when the first 130 natural numbers are divided by 9?
Choices:
A) 397
B) 401
C) 403
D) 412
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=412.   Answer: 412.
<result>D</result>
What is the sum of all remainders obtained when the first 130 natural numbers are divided by 9?