Steps  75 ÷ 25  
1.  Arrange the numbers in a manner that the Divisor is in the lefthand side, with the last digit slightly raised. Leave some space between the digits of the Dividend. This slightlyraised last digit of the Divisor is the Flag. 
In this case, 2^{5} ) 7 5 

2.  Check: How many times the first part of the Divisor 'goes' into the first digit of the Dividend. If it does not 'go', put a Zero, and move on to the next digit of the Dividend. Lookup: The next digit of the Dividend should not be negative. In case, it is negative, keep reducing the Quotient till it is Zero, or Positive. 
In this case, 2^{5} ) 7 5  3 

3.  Find the next digit of the Dividend by placing the Remainder of the above operation (Step 2) before the next digit of Dividend (slightly lowered) and substracting the (Quotient of the above operation, multiplied by the Flag)  In this case, 2^{5} ) 7 _{1}5 15  0  3 

4.  Repeat the above steps (from Step 2) for every digit of the Dividend. On reaching the last digit of the Dividend, if the (Last digit to be operated) is Zero, stop operating. If the (Last digit to be operated) is nonZero but perfectly divisible, execute the operation and stop operating. If the (Last digit to be operated) is nonZero but not divisible, execute the operation, and put a decimal point before the last digit of the Quetient, and stop operating. 
In this case, we get a Zero. So, we are left with 3, which is the answer! 
2^{5} ) 2 _{0}5  5  0  1 So, 1 is the answer! Note, we stop operating because the Next Digit to be operated is Zero. 
2^{5} ) 2 _{2}2 _{4}5  0 45   22 0  0 9 So, 9 is the answer! Notes: 1. 2 'goes' into 2  once. But that would mean the Next Digit as (_{0}2)(1 × 5) = 8. As per 'Lookup' in Step 2, we reduce it to Zero. 2. Moving on to the next digit, 22: 2 'goes' into 22  11 times. But that would mean the Next Digit as (_{0}5)(11 × 5), which is negative. As per 'Lookup' in Step 2, we reduce it to 10 which gives us (_{2}5)(10 × 5), which is again negative. So, we further reduce it to 9. 3. We stop operating because the Next Digit to be operated is Zero. 
2^{5} ) 2 _{2}3 _{5}5  0 45   23 10  0 9 .4 So, 9.4 is the answer! Notes: 1. Proceeding as in the above example, we get to a Last Digit of 10. 2. 2 'goes' into 10  5 times. But, considering an imaginary Zero at the end, that would mean the Next Digit as (_{0}0)(5 × 5), which is negative. As per 'Lookup' in Step 2, we reduce it to 4. 3. And, finally we place a Decimal Point, before 4. 
8^{7} ) 9 _{1}8 _{3}3 _{2}7 _{6}4 _{8}0 _{7}0 _{9}0 _{7}0 _{4}0 _{3}0 _{8}0 _{9}0 _{10}0 _{7}0 _{7}0  7  7 21  0 49 21 35 42 21 14  7 56 21 63  0                11 26 6 64 31 49 55 28 19 16 73 34 79 7 70  1 1 3 0 .7 3 5 6 3 2 1 8 3 9 0 8 So, 1130.735632183908... is the answer! Notes: 1. Note that the imaginary Zeroes are greyed out. 2. Also, we stop at the last digit voluntarily, and can be continued for more decimal digits. 
2^{1} ) 7 _{1}8 _{1}5  3  7   15 8  3 7 .3 Thus, 785 ÷ 21 = 37.3 (Ignoring more decimal places) 
2^{4} ) 8 _{2}5 _{3}4 _{2}7 12 20 24    13 14 3  3 5 6 .1 Thus, 8547 ÷ 24 = 356.1 (Ignoring more decimal places) 
10^{3} ) 2 _{2}0 _{10}1 _{8}0 _{3}4  0  3 27 15     10 98 53 19  0 1 9 5 .1 Thus, 20104 ÷ 103 = 195.1 (Ignoring more decimal places) 
12^{6} ) 8 _{8}5 _{13}7 _{5}8 _{10}1 _{5}5 _{7}4 _{14}7  0 36 48  0 48  0 30        85 101 10 101 7 74 117  0 6 8 0 8 0 5 .9 Thus, 85781547 ÷ 126 = 680805.9 (Ignoring more decimal places) 
This division method is selfexplanatory, because it is only a variation of the 'long division' method, where the Flag is used as a graphic mnemonic device. However, the point that should be worth observing is that the remainder of previous digit is added in front of the next digit. That is, (10 x Remainder) is added to the next digit. And, tactically (Quotient of the previous digit, multiplied with the Flag  which is the Unit's digit of the Divisor) is subtracted from it  to obtain the next digit to be operated upon. Not only does this tactic provide a marked advantage of dividing the digits by (n1) digits instead of ndigits  it is also a generalized tactic that will work with any ndigit Divisor. 
« Upasūtra 13  Upasūtra 15 » 