when a natural number n is successively divided by 1020 , 3040 . the remainders are 5060 , 7080 . what will be the sum of the remainders if the order of the division is reversed ?
10 20 30 40 50 60 70 80 leave the top right - most number 8 start with bottom right - most number 5 80 * 30 + 70 = 2470 2470 * 20 + 60 = 49460 49460 * 10 + 50 = 494650 this is the number required now , do the successive division in the reverse order the sum of the remainders is 28 hence , the correct option is c
add(multiply(add(add(const_4, const_3), add(const_1, const_2)), const_2), multiply(floor(divide(5060, 1020)), floor(divide(7080, 3040))))