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

We apply the same rule as before:
1) Write a, b, c and d in the first row, columns 1 to 4
2) Split the numbers into 2 sections (ab | cd)
3) Multiply a by p (p = 100 - mn)
4) List the result of p x a in the second and third column of the second row
5) Calculate the total of column 2 i.e.(b+e), and
6) Multiply this total by p and put the answer in column 3 and 4
7) Add up the left side to get the quotient and the right side to get the remainder

a b | c d
e  | f       (ef = p x a)
| g h   (gh = p x (b+e))
ab + e | cd + f + gh
Quotient: ab +e, Remainder: cd + f + gh

Important Note: Place 'f' on tens place in second section. So, when we add 'f', its becomes f0. Final addition is cd+f0+gh (see below example)

Following examples will help to understand it more:

1111 divide by 73
Here p = 100-73 = 27
11 | 11
2 | 7         (27 x 1)
| 81       (27 x (1 + 2) = 81)
13 | 162     (11+70+81 = 162)
15 | 16       (162 -73 -73 = 16) or (162 = (2 x 73) + 16)
Carry-over 2 and 16 is remaining
Quotient= 15, Remainder= 16

1221 divide by 73
Here  p = 100-73 = 27

12 | 21
2 | 7       (27 x 1 = 27)
|108    (27 x (2+2) = 108 )
14 |199    (12+2 = 14 and 21+70+108 = 199)
16 | 53     (199 -73-73 = 53) or (99 = (2 x 73) + 53)

Carry-over 2 and 53 is remaining
New Quotient = 16, Remainder = 53

Rule, discussed in this article, can be applied to any division. It may be a little difficult to do it fast in one go for larger numbers, however, it will become easier with practice and is still quicker than conventional method.

Hope it helps. Share your experience with these techniques.