what is the sum of all remainders obtained when the first 200 natural numbers are divided by 9 ?
"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 = 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 = 401 . answer : b ."
add(const_3, multiply(divide(subtract(200, reminder(200, 9)), 9), add(add(add(const_1, const_2), const_3), add(add(const_1, const_4), add(const_4, add(add(const_1, const_2), const_3))))))