Vedic Mathematics: Quicker Division by 9

Step 1: Take a 2 digit number, say 26 and write it with each digit separated, 2/6.
Step 2: Now leave a blank under the number and then draw a horizontal line under like this,

2/6

____
0/0

Step 2: Put the first digit under the last digit of the number like this,

2/6

0/2
_____

Step 3: Add the two numbers, each digit separately,

2/6

0/2
____
2/8

Step 4: The first digit before “/” is your quotient and that comes after”/” is the remainder.
Here, 2 is the quotient and 8 is the remainder.

Now for a 3 digit number say, 153 lets proceed in similar way.
Step 1: Rewriting the number, last digit separated by a “/”
1/5/6

Step 2: Putting a blank space and drawing a horizontal line under the blank line,

1/5/6

____

Step 3: Add the first two digits which will give you 6. Put that as the last digit in the blank line.

1/5/6

0/0/6
____

Step 4: Now there is an extra step, again separate the left two digit by a “/”

1/5/6

0/0/6
_____

Step 5: Put the first digit in the next blank space that is, before the “/”

1/5/6

0/1/6
_____

Step 6: Now add the numbers as they are positioned,
1/5/6

0/1/6
_____
1/6/12

Step 7: Naturally the number that comes last is the remainder. But in this case, the remainder 12 is greater than 9. So repeat the same process for 2 digit numbers. This will give,

1/2

0/1
___
1/3
Step 8: Put 3 in the original calculation and carry 1 to the next position that is, to the quotient portion.

1/5/6

0/1/6
_____
1/7/3

Your answer is quotient = 17 and remainder is 3. Verify the result with the help of traditional method. You will get the same result.

Now let us discuss the same procedure as applied generally to numbers having more than one digit.
Take a number 7321
Rewrite it by separating each digit,
7/3/2/1
Now what will come under 1?
Obviously the sum of the previous digits. Right?
In this case it is 12, again greater than 9 itself. So what we actually consider instead of a 12 is,

1/2

0/1
___
1 /3

Thus we put 3 and carry 1 to the next position.

7/3/2/1

0/0/0/3
______

Again what should we write under 2? The sum of the previous digits brings us, 7 +3 =10, Adding carry 1 , we get, 11. We put 1 and carry 1 to the next position.

7/3/2/1

0/0/1/3
_______

Write 7 + 1 = 8 Under 3 as there is only one “previous digits” left. Then we add the numbers the same way mentioned earlier.

7/3/2/1

0/8/1/3
_______
8/1/3/4
This gives 813 as the quotient and 4 as the remainder. Verify this too with the traditional process. This gonna be fun.

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