Wrap Math :: Division Tricks Summary-(1)

1. Write the highest digit number right below. - ①
2. Sum ① with the 2nd digit number and write it right below the 2nd digit number. - ②
3. Sum ② with the 3rd digit number and write it right below the 3rd digit number. - ③
   ···················································
   Cumulative Sum shall be put in each digit
   You have to calculate the cumulative sum with full number, if any Cumulative Sum ≥ 10, when you calculate the sum for next digit. → This explanation would be more clear with handling problem 2.
4. The answer is the number arrayed, ①②③········
5. The last cumulative sum shall be the remainder.
   
   
   Let's follow this process.

→ The highest digit # is 1 in 121011. So, write 1 right below 1.
→ The 2nd digit # is 2 in 121011. Sum up 2 with 1, which results in 3.
→ The 3rd digit # is 1 in 121011. Sum up 3 with 1, which results in 4.
→ The 4th digit # is 0 in 121011. Sum up 4 with 0, which results in 4.
→ The 5th digit # is 1 in 121011. Sum up 4 with 1, which results in 5 :::::: Answer = 13445
→ The 6th digit # is 1 in 121011. Sum up 5 with 1, which results in 6 :::::: Remainder = 6

         121011
     ÷            9
 ------------------------------------
  Q = 13445             R = 6

How about 231062?

         231062              ÷ 9
 ------------------------------------
  Q = 25673              R = 5