When a natural number N is successively divided by 10,20,30,40. the remainders are 50,60,70,80. What will be the sum of the remainders if the order of the division is reversed?	Options:
A) 16
B) 14
C) 28
D) 11
E) 9
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= 
<gadget id="calculator">80*30+70</gadget>
<output>2_470</output>
2470   2470*20+60= 
<gadget id="calculator">2470*20+60</gadget>
<output>49_460</output>
49460   49460*10+50= 
<gadget id="calculator">49460*10+50</gadget>
<output>494_650</output>
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 28
<result>C</result>
When a natural number N is successively divided by 10,20,30,40. the remainders are 50,60,70,80. What will be the sum of the remainders if the order of the division is reversed?