Lesson 1.1 multiplication

Vedic Mathematics
Composed by A V Prakasam for the benefit of his daughters -
from the book ‘Vedic Mathematics’
By Jagadguru Swami Sri Bharati Krsna Tirthaji Maharaja

It will be difficult to understand the concepts at the first reading, but if practiced with
the examples given, the beauty of the vedic mathematics can be appreciated.

1

Multiplication
• Nikhilam Sutra
• Principle : (x+a)*(x+b) = x(x+a+b)+ab
Thumb rule: -Make the number of digits in the Multiplier and
Multiplicand equal, by adding zeros to the left, if necessary, to the number with less
number of digits
- Chose the Base – multiple of ten like 10|100|1000 etc– nearest to one of
the numbers. The Base will have as many zeros to the right of 1 as the number of
digits in the number(s)
- Find the difference of each number from the base and write them after
the corresponding number with an appropriate sign (+ or -)
i. Add the Multiplier and Multiplicand and subtract the ‘Base’ from the total.
ii. Multiply the differences and place the product by the side of the sum.
iii. The number of digits in this product (of differences) should be the same
as the number of zeros in the base. Any excess digit in the product, the
left most digit should be added to the sum at (i). Any deficiency should
be made up by adding zeros to the left of the product.
iv. You got the product of the numbers

To understand better, see the worked out examples in the succeeding slides.
2

Nikhilam Sutra (Contd.)

Examples
I Case: Multiplicand and Multiplier are below multiples of 10.

a. Base 10

1. 8*5 2. 6*5
Note : In example 1, 5*2=10, but
8 -2 6 -4
because the Base is 10, the
5 -5 5 -5 number after The ‘|’ should be
8+5-10 5*2 6+5-10 4*5 one. Hence the 1 ( shown as a
3 10 1 20 subscript for clarity) is a overflow
And so carried over to the left of
= 40 = 30 the ‘|’ and is added to the sum on
the left hand side.
b. Base 100 Similar is the case with example 2

3. 91*91 4. 88*91 5. 99*97
91 -9 88 -12 99 -1
91 -9 91 -9 97 -3
91+91-100 9*9 179-100 108 99+97-100 1*3
82 81 79 108 96 03
= 8281 = 8008 = 9603 3


c. Base 1000

1. 888*998 2. 112*998
888 -112 112 -888
998 -002 998 -002
888+998-1000 112*002 112+998-1000 888*2
886 224 110 1776
= 886224 = 111776

d. Base 100000

1. 99979*99999
99979 -00021
99999 -00001
99979+99999-100000 00021*1
99978 00021
= 9997821

4


II. Multiplicand or Multiplier slightly above or below Multiple of 10

1. 12*11 2. 18*12 3. 111*109
12 +2 18 +8 111 +11
11 +1 12 +2 109 +09
12+11-10 2*1 18+12-10 16 111+109-100 11*9
13 2 = 132 21 6 = 216 120 99
= 12099
4. 12*8 5. 108*97
12 +2 108 +8
97 -3 Note : In Example 4, 2*- 2=-4, (-4 is
8 -2 shown as 4 with a bar on top,
12+8-10 2*-2 108+97-100 8*-3 known as ‘vinkulum’, meaning that
10 4 = 96 105 24 the digit under the vinkulum only is
104 76 negative) The figure in the units
place is -4 and hence 1, i.e. 10, is
= 10476
borrowed from the left hand side
6. 10006*9999 number 10, leaving 9 behind, and 4
10006 +6 deducted from it giving 6.
9999 -1 Similar is the case with examples 5
and 6.
10006+9999-10000 6*-1
10005 0006
10004|9994 = 100049994 5


III. Multiples & sub-Multiples.
Upa-Sutra : ‘Anurupena’
Multiplication of two numbers neither of which is near a convenient Base

Anurupena means proportionately. If there is a rational ratio-wise relationship, the
ratio should be taken into account and should lead to a proportionate
multiplication or division, as the case may be.

Example: 41*41
- 41 is nearer to 50 than to 100 and 50 is half of 100
- So workout with 50 as in the previous examples and Divide the
sum of the numbers by 2.

Base : 100|2 = 50, so the proportion is 2
41 -9
41 -9
(41+41-50) divided by 2 |81
16|81 = 1681
if the Base is 40, then 40 = 10*4 and the propotion is 4. The result is
41 +1
41 +1 = (41+41-40) Multiplied by 4 |01 = 1681 6


Examples:

1. 49*49
Working Base 100|2 = 50 Working Base 10*5 = 50
49 -1 49 -1
49 -1 49 -1
(49+49-50)|2 |01 (49+49-50)*5 |1
24|01 = 2401 240|1 = 2401

2. 46*44
Working Base 10*5=50 Working Base 100|2 = 50
46 -4 46 -4
44 -6 44 -6
(46+46-50)*5 | 24 (46+44-50)|2 |24
20|24 = 2024
200+2|4 = 2024

7


3. 54*46
Working Base 10*5 = 50 Working Base 100/2=50
54 +4 54 +4
46 -4 46 -4
(54+46-50)*5 | 16 (54+46-50)/2 | 24
250-1 | 6 =249-1 | 4 = 2484 25 | 16 = 2484

Note : 4*- 4=-16, which is written with a vinkulum as
16, but because the Base is 10, the number after The
‘|’ should be units. Hence the -1 is a overflow And so
carried over to the left of the ‘|’ and because it is a
negative number, is deducted from the sum on the
left hand side. Again, the remaining figure in the units
place is -6 and hence 1, i.e. 10, is borrowed from the
left hand side number and 6 deducted from it giving
4.
4. 19*19
Working Base 10*2 = 20
19 -1
19 -1
(19+19-20)*2 | 1 = 36|1 = 361 8


5. 48*49 6. 249*246
Working Base 100/2 = 50 Working Base is 1000/4 = 250
48 -2 249 -1
49 -1 246 -4
(48+49-50)/2 | 2 (249+246-250)/4 | 0004
47/2 | 2 = 23 2 | 2 = 23 / (50+2) = 2352
1/
245/4 | 0004 =61 |250 + 4
= 61254
Note : Here division of 47 by 2 gives us a fraction.
The original base 100 is multiplied by this fraction Note : Here division of 245 by 4 gives us a
and the result added to the number on the right fraction. The original base 1000 is
hand side. multiplied by this fraction and the result
added to the number on the right hand side

9


IV. Practice Session

1. 87965*99998 879624070
2. 48*97 4656
3. 889*9998 8888222
4. 77*9988 769076
5. 532*472 251104
6. 389*516 200724

10


V. Corollaries
1. First Corollary: To find the Square of a number
Principle: Find a base (power of 10) nearest to the number and find the difference
between the number and the base. The difference may be less or more than the base.
Whatever is the extent of its deficiency (surplus), lessen (add ) it still
further to that very extent; and also set up the square of the deficiency.
Algebraical Expression: (a+or-b)2 = a2 +or- 2ab +b2
or a2 = (a+b)(a-b)+ b2
Thumb rule: to find the square of 9:
i. Take the power of 10 nearest to the given number, 10 in this case
ii. Find the difference between the nearest power and the given number,
it is 1 in this case. ( called the deficiency. If the number is 12 instead of 9, the suplus
will be 2 which should be added to 12 )
iii. Deduct this deficiency from the given number ( lessen it still further to that
extent), in this case the result is 8. This is the left hand side portion of
the answer.
iv. Put down the square of the deficiency, i.e. square of 1 in this example,
on the right hand side.
v. you got the answer (92 = 8 | 1)
11


Examples:
i. With Base multiples of 10
1. 72 = (7-3) | 32 = 4 | 9 = 49
2. 112 = (11+1) | 12 = 12 | 1 = 121
3. 1082 = (108+8) | 82 = 116 | 64 = 11664
4. 9892 = (989 – 11) | 112 = 978 | 121 =
978121

ii. With base other than multiples of 10
1. 412
Base 40 = 10*4
( 41+1)*4 | 12
168 | 1 = 1681
2. 592
Base 50 = 100/2
(59+9)/2 | 92 Note: In example 3, the right hand side figure
comes to 625, but because the base is a
68/2 | 81 = 3481 Mutile of 100, the right hand side contains only 2
3. 7752 Digits. Consequently 6, shown as subscript,
is carry forward and added to 6000.
Base 800 = 100*8
(775-25)*8 | 252
750*8 | 625 = 6000 | 625 = 600625 12


2.1 Second corollary: Squaring of numbes ending in 5

This is a special case of the previous corollary, applicable to numbers
ending in 5. The principle says ‘ By one more than the previous one’.

Thumb rule: to find the square of 15:
i. Here the last digit is 5 and the ‘previous one’ is 1
ii. One more than the ‘previous one’ is 1+1=2
iii. Multiply the previous one i.e. 1 by one more than the previous one i.e. 2.
This is the left hand side digit which is 1*2 = 2
iv. Put down the square of 5 on the right hand side i.e. square of = 25
v. you got the answer (152 = 2 | 25 = 225)

Examples:
1. 252 = (2*3) | 25 = 625
2. 1052 = (10*11) | 25 = 11025
3. 1852 = (18*19) | 25 = 34225

13


2.2 Second Corollary (Contd.): Multiplication of numbers whose first
digits are same and whose last digits add up to 10.
The above rule is applicable not only to the squaring of a number ending
In 5, but also to the multiplication of two numbers whose last digits
together total 10 and whose previous part is exactly the same.

Examples:
1. 27*23
2*(2+1) | 7*3 = 621
2. 96*94
9*(9+1) | 6*4 = 9024

The above rule is capable of further application for the multiplication of
numbers whose last digits in sets of 2,3 and so on together total 100, 1000
Etc.

Examples:
1. 191* 109 = 1*(1+1)*10 | 91*9 = 20 819
2. 884*816 = 8*(8+1)*10 | 84*16 = 720 | 1344 ( 1 is a carry)
= 721344 14


2.3. Third corollary: Multiplier digits are only 9s.

Principle: This relates to a very special case where the multiplier digits consists
entirely of 9s

Case 1 Multiplicand and the multiplier contain the same number of digits
i.e. case like 9*1, 99*11 etc.
In these cases, the left hand side of the result is one less than the
multiplicand and the right hand side is the 9’s complement of the left
hand side.
Examples:
1. 2*9 = 18 Explanation – left hand is one less than 2 ( the
multiplicand) and the right hand side 8 is 9’s complement of
1
2. 3*9 = 27
3. 7*9 = 63
4. 17*99 = 1683
5. 19*99 = 1881
6. 777*999 = 776223
7. 9879*9999 = 98780121 15


Case 2: The multiplicand consists of a smaller number of digits than the
multiplier
In this case, add zeros to the left hand side of the multiplier to
make the number of digits equal to those in the mutiplier.

Example:
1. 7*99 = 07*99 = 0693
2. 79*999 = 079*999 = 078921
3. 79*9999999 = 0000079*9999999 = 00000789999921

Case 3: The multiplier contains a smaller number of digits than the
multiplicand.
In this case, the following procedure applies:
i. Divide the mutiplicand off by a vertical line into a right hand
portion consisting of as many digits as the multiplier
ii. Subtract from the multiplicand one more than the whole excess
portion on the left
iii This gives the right hand side portion of the product
iv. The right hand portion of the product would be 10’s
complement of the right hand side portion of the multiplicand 16


Examples:
Note: - Refer to Example 1
1. 43*9 There is one digit in the multiplier i.e. 9.
( 43-5) | (10 – 3) - Therefore draw a vertical line after one digit
38 | 7 = 387 from the right in the multiplicand 43
-To the right of the line is 3 and to the left of the
line is 4
2. 63*9 -Add 1 to 4 giving 5
(63-7) | (10 – 3) -Subtract 5 from 43 giving 38. This is the left
56 | 3 = 567 hand portion of the product
- The 10’s complement of 3 is 7 which is the
3. 122*9 -Right hand side portion of the product.
(122-13) | (10 – 2) -The product is therefore 387
109 | 8 = 1098
4. 112*99
(112-2) | (100 – 12) = 11088
5. 11119*99
(11119-112) | (100-19) = 1100781
6. 1000001*99999
(1000001-11) | 100000 – 00001)
= 9999099999 End of Chapter 1A – Nikhilam Sutra
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Lesson 1.1 multiplication

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