Wrap Math :: Division Tricks Summary-(1)
- Write the highest digit number right below. - ①
- Sum ① with the 2nd digit number and write it right below the 2nd digit number. - ②
- 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. - The answer is the number arrayed, ①②③········
- 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
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