### Vedic Math - Divide bigger numbers by more than two digit number near base

In the last article, we learn to divide three and four digit numbers with two digit numbers near base. In this article, we shall continue to discuss it further to divide bigger numbers by more than two digit number near base.

Procedure is the same as mentioned in the previous article. Please check that for more details about formula.

In below examples, we have deliberately taken 'divisior' of big digits to understand the concept.

12345 divided by 8897
Here p = 10000-8897 = 1103
1 | 2345
| 1103        (p x a)
1 | 3448
Quotient = 1 and Remainder = 3448

51235 divided by 7999
Here p = 10000-7999 = 2001
5 | 1235
|10005        (p x a)
5 |11240 (Here, remainder is greater than the divisor. Add 1 to quotient i.e. 5+1=6 and
subtract divisor from remainder i.e.(11240-7999=3241))
6 | 3241
Quotient = 6 and Remainder = 3241

12345 divided by 882
Here p = 1000-882 = 112, and a = 1
12 | 345
1 | 12      (p x a)
| 336    (p x (b+e))
13 | 801
Quotient = 13, Remainder = 801

2002002 divided by 89998
Here p = 1000000-89998 = 10002
20 | 02002
2 | 0004     (p x a)
| 20004   (p x (b+e))
22 | 22046
Quotient = 22 and Remainder = 22046

11001 divided by 88
Here p = 100-88 = 12
110 | 01
12 |         (p x a = 12 x 1 = 12)
2 | 4      (p x (b+e) = 12 x (1+1) = 24)
| 48    (p x (c+f+g) = 12 x (0+2+2) = 48)
124 | 89    (Remainder greater than divisor)
125 | 1
Quotient = 125 and Remainder = 1

11111111 divided by 99979
Here p = 1000000-99979 = 00021
111 | 11111
00 | 021
0 | 0021
| 00021
111 | 13442
Quotient = 111, Remainder = 13442

### Vedic Math - Dividing Three and Four Digit Numbers by any Two Digit Numbers Near Base

In last article, we discussed about division of numbers by any one digit number (near base). In this article, we shall discuss about division of three and four digit numbers by any two digit numbers (near base).

If we are dividing by 2 digit number than we need to have 2 places in the right column.

Dividing three digit numbers by any two digit numbers

To divide abc by mn, we apply the same rule as before:
1) Put a, b and c in the first row (as shown below)
2) List p x a in the second column of the second row. (p = base- mn)
a | b c
| p x a
----------------
a |(p x a) + bc

Quotient: a, Remainder: (p x a) + bc

Example:
To divide 102 by 75, the nearest base is 100, so p = 100-75 = 25

102 divided by 75
1 | 02
| 25
--------------
1 | 27
Quotient= 1, Remainder= 27

234 divided by 73
p = 100 - 73 = 27
2 | 34
| 54
-------------
2 | 88    (Remainder(88) is greater than 73, Add 1 to quotient i.e. 2+1=3 and subtract divisor(73)                        from remainder i.e.(88-73=15))
3 | 15
New Quotient = 3, New remainder = 15

Dividing four digit numbers by any two digit numbers

abcd divided by mn

This gets a little more complicated and one must be very careful with the places in the columns.
p = base - mn = 100 - mn

### Vedic Math - Dividing numbers by any one digit numbers (DIVISION BY 8 ETC.)

In the last article, we discussed division by 9. And in this article, we discuss divisors below base value 10, 100 etc. In these, we do the pairing from right.

Easy way to divide by 9 can be extended for 8, 7 etc.

ab divided by m
The rule we use is:
1) Put a and b in the first row
2) Write a as quotient in third row
3) List p x a (a the quotient) in the second column of the second row. (p= 10 - m)
a | b
| p x a
----------
a | b +(p x a)
Quotient: a, remainder: b + (p x a)
Clearly this is easiest for 9, p= 10 - 9= 1

8) 3 | 1               7) 2 | 0                6) 1 | 1
| 6                       | 6                       | 4
---------              ----------              ----------
3 | 7                    2 | 6                    1 | 5

In the above examples with divisor 8, 7, and 6, we get quotient 3 reminder 7 , quotient 2 reminder 6 and quotient 1 reminder 5 respectively.
We have to multiply the quotient by 2 in the case of 8, by 3 in the case of 7, by 4 in the case of 6 and so on. What this means, that we have to multiply the quotient-digit by the Divisor's complement from 10.

For example, Divide 31 by 8
Step 1: We bring the first digit 3 into the answer as quotient.
Step 2: Then instead of adding this to the 1 as we do when dividing by 9, we add double of 3
(i.e. 3 * (10-8)) to the 1 and get 7 for the remainder.
We double the 3 because 8 is 2 below 10.

Lets extend it for three digits Dividend

abc divide by m, p=10-m
a | b          | c
| p x a    | p x ((p x a) + b)
--------------------------------------
Quotient: a / (p x a) + b, Remainder: (p x ((p x a) + b)) + c

Similarly, for 311 divided by 8

8) 3  | 1 |  1
| 6
--------------
3  | 7 | 14
--------------
37   |  15
---------------
38   |   7
• We bring down the first digit i.e. 3,
• Add 2 x 3 (i.e. p x a) to 1 (i.e. b) in the next column and put down 7 (i.e.(p x a) + b),
• Then add 2 x 7 (i.e. p x ((p x a) + b)) to the 1 in the last column and put down 15 as the remainder.
• Since this remainder is bigger than 8, hence we compute answer further.  As 15 > 8, so add 1 to quotient and subtract 8 from remainder. Final answer is quotient = 38 and reminder = 7