Advantages of Vedic Mathematics

Contributed by:

Harshdeep Singh Wed, Apr 06, 2022 02:53 PM UTC

This PDF contains Basics, Addition, Subtraction, Multiplication, Multiplication, Division, Squares, and Square roots, Cubes and cube roots, Reciprocals, Arithmetic, Algebra, Geometry, Trigonometry, and Calculus.

1. Vedic Mathematics
By
Prof M. Basanna
*****
Introduction
Vedic Mathematics is a great boon to the modern world from our ancient sages. It is an ancient
technique, revived by His Holiness Jagadguru Bharathi Krishna Tirthaji. Vedic Mathematics is based on
16 Sutras (Formulas) and 16 Upa-Sutras (Sub-Formulas). These are in words and are very much useful in
solving the problems. The Vedic methods are simple, short and speedy. The answers can be worked out in
2 or 3 lines. This saves a lot of time, space and energy. Answers can be computed mentally without using
pen and paper.
Vedic mathematics has many advantages:
• Addition, subtraction, multiplication, etc. can be done from left to right as against traditional right
to left.
• The answers can be verified easily and quickly.
• Multiplication, division, squares and square roots, cubes and cube roots, reciprocals etc. can be
done easily and accurately.
• Accuracy will be far better than calculators. Calculators have their own limitations.
• Alternate methods are possible to solve a problem.
• It works well with arithmetic, algebra, geometry, trigonometry, and calculus.
Vedic mathematics is very much useful to the students studying in schools, colleges and every person
in general. It is also helpful for those who are preparing for competitive and entrance examinations. It is
well tested by scientists and engineers at NASA and taught in UK and other European countries.
Basics
Mathematics is a game of numbers. Numbers contain digits 1 – 9 and 0. To learn and understand
mathematics we must know the meaning of the terms used and the rules of various operations.
Complement
If sum of two digits is equal to 10, then each digit is said to be the complement of the other from 10.
e.g., 4 + 6 = 10. Here, 4 is the complement of 6 from 10 and vice versa.
Complements of digits from 10 and 9 are shown below. These must be studied carefully and
remembered. This will help in understanding Vedic Mathematics clearly and properly.

2. Complements from 10
Digit 1 2 3 4 5 6 7 8 9
Complement 9 8 7 6 5 4 3 2 1
Complements from 9
Digit 0 1 2 3 4 5 6 7 8
Complement 9 8 7 6 5 4 3 2 1
Addition of two digits
Sum of two positive digits is equal to the sum of the digits with positive sign.
E.g., 2 + 3 = 5, 6 + 3 = 9, etc.
Sum of two negative digits is equal to the sum of the digits with negative sign.
E.g.,(–2) + (–3) = –5, (–6) + (–3) = –9, etc.
Sum of a positive digit and a negative digit is equal to the difference between the digits with the sign
of larger digit. E.g., 2 + (–3) = –1, (–6) + 3 = –3, etc.
Subtraction of two digits
Subtraction is nothing but negative addition. To subtract one digit from another digit, change the sign
of the digit to be subtracted and add the two digits.
E.g., 2 – 3 = 2 + (–3) = –1, 6 – 3 = 6 + (–3) = 3, etc.
Multiplication of two digits
Product of two positive digits is equal to the product of the digits with positive sign.
E.g., 2 x 3 = 6, 6 x 3 = 18, etc.
Product of two negative digits is equal to the product of the digits with positive sign.
E.g., (–2) x (–3) = 6, (–6) x (–3) = 18, etc.
Product of a positive digit and a negative digit is equal to the product of the digits with the negative
E.g., 2 x (–3) = –6, (–6) x 3 = –18, etc.
Subtraction from 100, 1000, 10000, …
In Vedic mathematics, powers of 10, viz., 10, 100, 1000, … are used as base. When a number is
subtracted from a base we get the complement of that number from that base.
E.g., 100 – 36 = 64. Here, 64 is the complement of 36 from base 100.
1000 – 36 = 964. Here, 964 is the complement of 36 from 1000.

3. To find the complement of a number from a base we need not subtract the number from the base
every time. We can use the Vedic Sutra "All from 9 and the last from 10", known as Nikhilam Sutra. It is
also called "All from 9", in short. The following examples explain the use of this Sutra.
1. Consider the number 853 and base 1000.
Applying All from 9, 8 from 9 is 9 – 8 = 1
5 from 9 is 9 – 5 = 4
And Last from 10, 3 from 10 is 10 – 3 =7
Thus, 147 is the complement of 853 from base 1000. Note that 147 + 853 = 1000.
2. Consider the number 52 and base 1000. Here, 1000 has 3 zeros. So, we rewrite 52 as 052 and
apply Nikhilam.
All from 9, 0 from 9 is 9 – 0 = 9
5 from 9 is 9 – 5 = 4
Last from 10, 2 from 10 is 10 – 2 =8
Thus, 948 is the complement of 52 from base 1000. Again, note that 948 + 52 = 1000.
Similarly, the complement of any number can be computed. The calculations shown above are only
guidelines. We need not write all the steps. The calculations can be done mentally after a little practice. Try
some more cases by yourselves.
Addition
In Mathematics addition and subtraction are simple and easy operations. But, addition is easier than
subtraction. These topics are covered here for the reason that Vedic Mathematics allows us to add, subtract
and multiply numbers from left to right or right to left. This is the greatness of Vedic Mathematics.
Left to right
We write the numbers from left to right. Also, we read numbers from left to right. But, in traditional
mathematics addition, subtraction and multiplication are done from right to left. Would it not be fine if we
can do these operations from left to right? In some problems we need to know first 2 or 3 digits. Here we
shall see how this could be done!
Addition
As said earlier addition is the simplest and easiest operation. We shall see here how addition could be
done from left to right. The method is given below.
Method
• Write the numbers one below the other.
• Add the digits column by column from left to right or right to left.

4. • If the sum exceeds 9, retain last digit and carry over other digits to left.
• Compute the final answer.
Example 1: Add 548 + 236 + 198.
548 548 548
2 36 235 236
+198 +198 +19 8
8 86 8 62
1 1 2
982
Working from left to right (Mental work)
Step 1: 1 + 2 + 5 = 8. Write as shown
Step 2: 9 + 3 + 4 = 16. –"–
Step 3: 8 + 6 + 8= 22. –"–
Add the two rows to get the final answer.
Example 2: Add 5678 + 3728 + 4379.
5678 5678 5678 5678
3728 3728 3728 3728
+4379 +4379 +4379 +4379
26 266 266 2665
11 111 111 1112
1 3785
Working from left to right (Mental work)
Step 1: 4 + 3 + 5 = 12. Write as shown
Step 2: 3 + 7 + 6 = 16. "
Step 3: 7 + 2 + 7 = 16. "
Step 4: 9 + 8 + 8 = 25. "
Add the two rows to get the final answer.
Subtraction
Subtraction is little harder than addition. But it is made simpler using Vedic Mathematics techniques.
Instead of subtracting numbers directly we convert the number to be subtracted into Vinculum and then
subtract. The method adopted is explained below.

5. Method:
• Write numbers one below the other.
• Check the lower digit from left to right. If it is larger than the upper digit, put a star on the its
previous digit to increase its value by 1.e.g., 1–>1+1=2,2–> 2+1=3, etc.
• If the lower digit is smaller than the upper digit write the difference between digits.
• If the lower digit is larger than the upper digit write the complement of the difference between
the digits. (Alternately, add the complement of the lower digit to the upper digit)
Example1: Subtract: 224 – 192. Working from left to right (Mental work)
Step 1: 9 > 2. A (*) is put on 1. So, 1* = 1+1=2.
2 2 4
Step 2: 2 – (1 + 1) = 0, (Write this in 1st column)
- 1* 9 2
Step 3: 2 < 9. So no (*), 2 – 9 + 10 = 3. (2nd column)
0 3 2
(Alternately, complement of 9 from 10 is 1 and 1 + 2 = 3)
The answer is = 32.
Step 4: 4 > 2. So, 4 – 2 = 2, (3rd column)
Example 2: Subtract: 322 – 197. Working from left to right (Mental work)
Step 1: 9 > 2. A (*) is put on 1. So, 1* = 1+1=2.
3 2 2
7 > 3. A *(*) is put on 9. So, 9* = 9+1=10.
- 1* 9* 7
Step 2: 3 – (1 + 1) = 1, (Write this in 1st column)
1 2 5
Step 3: 2 < 9*.So, 2 – (9 + 1) + 10 = 2. (2nd column)
The answer is = 125.
(Alternately, complement of 10 from 10 is 0 and 0 + 2 = 2)
Step 4: 2<7So, 2 – 7 + 10 = 5, (3rd column)
(Alternately, complement of 7 from 10 is 3 and 3 + 2 = 5)
Final answer is 125.
Multiplication
In mathematics multiplication is harder than addition and subtraction. In Vedic Mathematics we have
simpler methods. Before taking up them we will discuss some special cases.
Multiplication by 11
Multiplication of a number by 11 is very easy. It is as good as addition. This method is explained below.
Method
• Sandwich the given number between zeros.
• Starting from left end add the digits taking them in pairs.
• If the total exceeds 9, retain the first digit and carry over the other digits to the left.

6. Example1: Multiply 135 x 11 Working from left to right (Mental work)
Write the multiplicand as shown Addition of digits in pair is shown below.
and add the digits in pair. 0 + 1 = 1,
135 x 11 1 + 3 = 4,
0135 0 3 + 5 = 8,
1485 5+0=5
Thus, 135 x 11 = 1485
Example 2: Multiply Working from left to right (Mental work)
58403 x 11 Addition of digits in pair is shown below.
58403 x 11 0 + 5 = 5, 5 + 8 = 13, 8 + 4 = 12,
0584030 4 + 0 = 4, 0 + 3 = 3, 3 + 0 = 3.
532433
11
642433
Thus, 58403 x 11 = 642433
The above method can be extended to multiply any number by numbers like 111, 1111, 11111, etc.
Sandwich the given number between a pair of zeros and add the digits in threesome (taking three at a
Example 3: Multiply 123 x 111 Working from left to right (Mental work)
123 x 111 Addition of digits in three-some is shown below.
0012300 0 + 0 + 1 = 1, 0 + 1 + 2 = 3, 1 + 2 + 3 = 6,
13653 2 + 3 + 0 = 5, 3 + 0 + 0 = 3
Thus, 123 x 111 = 13653
Example 4: Working from left to right (Mental work)
Multiply 5786 x 111 Addition of digits in threesome is shown below.
5786 x 111 0 + 0 + 5 = 5, 0 + 5 + 7 = 12,
00578600 5 + 7 + 8 = 20, 7 + 8 + 6 = 21,
520146 8 + 6 + 0 = 14, 6 + 0 + 0 = 6.
1221
642246
Thus, 5786 x 111 = 642246

7. Multiplication by 12
This is similar to the one discussed earlier. But, there is a slight difference. This employs the Vedic
Sutra "Ultimate and twice the penultimate". According to this, we must add twice the penultimate digit to
the ultimate digit.
Consider the number 32. Here, penultimate digit is 3 and the ultimate digit is 2. By the above Sutra,
the required sum = 3 x 2 + 2 = 8.
Example 1: Multiply 123 x 12 Working from left to right (Mental work)
123 x 12 Write the multiplicand as shown and add twice the penultimate
01230 digit to ultimate digit.
1476 0 x 2 + 1 = 1, 1 x 2 + 2 = 4,
Thus, 123 x 12 = 1476 2 x 2 + 3 = 7, 3 x 2 + 0 = 6.
Example 2: Multiply 396 x 12 Working from left to right (Mental work)
396 x 12 Write the multiplicand as shown and add twice the penultimate
03960 digit to ultimate digit.
3542 0 x 2 + 3 = 3, 3 x 2 + 9 = 15,
121 9 x 2 + 6 = 24, 6 x 2 + 0 = 12.
4752
Thus, 396 x 12 = 4752
This method can be extended for multiplication with 13, 14, 15, … , 19 with little modification. Instead
of twice we have to take three times, four times, etc. Try this!
Ekadhikena Multiplication
This is another simple method. The Vedic Sutra used in this method is "One more than the previous
one" – Ekadhikena. Two different cases arise here.
Case I – Last digits adding to 10.
The numbers used in this method must obey the following conditions.
Both the numbers must have the same previous digit(s).
The sum of the last digits must be 10.
Numbers like 54 and 56, 42 and 48, 23 and 27, 34 and 36 form the examples.
Method
• Divide the answer space into LHS and RHS by Place a slash (/) or a colon (:).
• Use Ekadhikena (Previous digit +1) to the digit on the LHS.

8. • Write the product of last digits on the RHS.
• Take care to see that RHS has two digits, as it should be. (Digit rule)
• Remove the slash or colon.
Example 1: Example 2: Example 3:
Multiply 51 x 59 Multiply 66 x 64 Multiply 123 x 127
Same previous digit: 5, Same previous digit: 6, Same previous digits: 12,
Sum of last digits: 1+9=10. Sum of last digits: 6+4=10. Sum of last digits: 3+7=10.
51 x 59 66 x 64 123 x 127
5 x (5+1) / 1 x 9 6 x (6+1) / 6 x 4 12 x (12+1) / 3 x 7
30 / 09 42 / 24 156 / 21
Thus, 51x59=3009. Thus, 66 x 64=4224. Thus, 123 x 127=15621.
Note that 0 has been added on RHS.
Case II – Squaring of numbers ending with 5.
When the numbers are equal and end with 5, they satisfy both the conditions. Then the product of the
numbers gives the square of that number. Thus, the square of a number ending with 5 can be computed
as above. In all such cases RHS will always be 25. These computations can be done mentally.
Example 1: Example 2: Example 3:
Find the square of 35. Find the square of 75. Find the square of 155.
Previous digit = 3 Previous digit = 7 Previous digits = 15
352 752 1552
3 x (3+1) / 25 7 x (7+1) / 25 15 x (15+1) / 25
12 / 25 56 / 25 240 / 25
Thus, 352 = 1225. Thus, 752 = 5625. Thus, 1552 = 24025.
Ekanyunena Multiplication
Ekanyunena is a Vedic Sutra which states that "One less than the previous one". This Sutra is useful in
multiplying a number with multipliers having only 9's (9, 99, 999, 9999, ….). This is also a quick method
and done mentally within no time. The answer can be written in one line.
Method
• Place a slash (/) or a colon (:) to separate answer into two parts, LHS and RHS.
• Subtract 1 from the multiplicand (Ekanyunena) and write it on LHS.
• Write the complement of multiplicand (from the base of multiplier) on RHS.

9. • Remove the slash or colon.
Three different situations need our attention.
Case I: Both multiplicand and multiplier having same number of digits.
Example 1: Multiply 4 x 9 Working from left to right (Mental work)
4x9 LHS = 4 – 1 = 3 (Ekanyunena)
4–1/6 RHS = 10 – 4 = 6 (Complement of 4 from 10)
36
Thus, 4 x 9 = 36
Example 2: Multiply 76 x 99 Working from left to right (Mental work)
76 x 99 LHS = 76 – 1 = 75 (Ekanyunena)
76 – 1 / 24 RHS =100–76= 24 (Complement of 76 from 100)
7524
Thus, 76 x 99 = 7524
Example 3: Multiply 353 x 999 Working from left to right (Mental work)
353 x 999 LHS = 353 – 1 = 352 (Ekanyunena)
353 – 1 / 647 RHS=1000 – 353=647 (Complement of 353 from 1000)
352647
Thus, 353 x 999 = 352647
Case II: Multiplier having more number of digits than multiplicand.
In this case, we equalize the number of digits both in multiplicand and multiplier by prefixing required
number of zeros with the multiplicand.
Example 1: Multiply 6 x 99 Working from left to right (Mental work)
06 x 99 LHS = 6 – 1 = 5 (Ekanyunena)
6-1 / 94 RHS = 100–06 = 94 (complement of 06 from 100)
594 (We can directly write the complement of 06 = 94)
Thus, 6 x 99 = 594
Example 2: Multiply 35 x 999 Working from left to right (Mental work)
035 x 999 LHS = 35 – 1 = 34
35-1 / 965 RHS = 1000 – 035 = 965
34965 (Complement of 035 = 965)
Thus, 35 x 999 = 34965

10. Example 3: Multiply 98 x 99999 Working from left to right (Mental work)
00098 x 99999 LHS = 98 – 1 = 97
98-1 / 99902 RHS = 10000 – 00098 = 99902
9799902 (Complement of 00098 = 99902)
Thus, 98x99999=9799902 Note: Apply All from 9 last from 10 Rule in all these cases.
Case III: Multiplicand having more number of digits than multiplier.
In this case, we split the multiplicand into two parts.
i. Right Hand Part having same No.of digits as of Multiplier.
ii. Left Hand Part having remaining digits.
We subtract Left Hand Part digits also (in addition to 1 as usual) from the multiplicand and write it on
LHS and complement of Right Hand Part digits on RHS.
Example 1: Multiply 72 x 9 Working from left to right (Mental work)
7/2 x 9 LHS = 72 – 1 –7 = 64
72 – 1 – 7 / 8 RHS = 10 – 2 = 8 (complement of 2 from 10)
71 – 7 / 8
648
Thus, 72 x 9 = 648
Example 2: Multiply 123 x 99 Working from left to right (Mental work)
1/23 x 99 LHS = 123 – 1 – 1 = 121
123 – 1 – 1 / 77 RHS = 100 – 23 = 77
122 – 1 / 77
12177
Thus, 123 x 99 = 12177
Example 3: Multiply 7936 x 99 Working from left to right (Mental work)
79/36 x 99 LHS = 7936 – 1 – 79 = 7856
7936 – 1 – 79 / 64 RHS = 100 – 36 = 64
7935 – 79 / 64
785664
Thus, 7936 x 99 = 785664

11. Nikhilam Multiplication - Base System
As pointed earlier the powers of 10, i.e., 10, 100, 1000, etc. are taken as base. In Vedic Mathematics
numbers can be expressed in base system. This helps us to make various computations easier.
The difference between the number and the nearest base is called deviation. If the number is less than
the base, the deviation will be negative. On the other hand, if h number is more than the base, the
deviation will be positive. The combination of the number and its deviation forms the base system.
The following table gives us an idea of writing the number in base system.
Number Base Deviation Base System
8 10 8 – 10 = – 2 8–2
12 10 12 – 10 = +2 12 + 2
96 100 96 – 100 = – 4 96 – 4
106 100 106 – 100 = + 6 106 + 6
86 100 86 – 100 = – 14 86 – 14
112 100 112 – 100 = + 12 112 + 12
Nikhilam Multiplication
As the deviation is obtained by Nikhilam sutra we call the method as Nikhilam multiplication. This is a
special method to multiply two numbers near a base or one number near the base and the other a little
away from the base. The method of multiplication is given below.
Method
• Write the numbers one below the other in base system.
• Divide the answer space into LHS and RHS by placing a slash (/) or a colon (:).
• Add or subtract one number with the deviation of the other number and write it on the LHS. (i.e.,
cross-sum or cross-difference)
• Write the product of the deviations on RHS.
• The number of digits on RHS must be same as the number of zeros in the base. If less, prefix the
answer with the zeros. If more, transfer extra digits to the RHS. (Digit Rule)
• Take due care to the sign (+/–) while adding or multiplying the numbers.
• Remove the slash.
Three different cases are possible.
Case I: Both the numbers below the base.
Here, both deviations are negative and the product of the deviations will be positive.

12. Example 1: Multiply 6 x 8. Working from left to right (Mental work)
6–4 Base = 10, RHS Digits = 1
x8–2 LHS = 6 – 2 = 4 or 8 – 4 = 4
4/8 RHS = - 4 x – 2 = 8
48 Note: Numbers are written in Base System.
Thus, 6 x 8 = 48
Example 2: Multiply 92 x 97. Working from left to right (Mental work)
92 – 08 Base = 100, RHS Digits = 2
x 97 – 03 LHS = 92 – 03 = 89 or 97 – 08 = 89
89 / 24 RHS = - 08 x – 03 = 24
8924 Note the number of digits in deviation.
Thus, 92 x 97 = 8924
Example 3: Multiply 91 x 99. Working from left to right (Mental work)
91 – 09 Base = 100, RHS Digits = 2
x 99 – 01 LHS = 91 – 01 = 90 or 99 – 09 = 90
90 / 09 RHS = - 09 x – 01 = 09 (Digit Rule)
9009
Thus, 91 x 99 = 9009
Example 4: Working from left to right (Mental work)
Multiply 993 x 996. Base = 1000, RHS Digits = 3
993 – 007 LHS = 993 – 004 = 989 or 996 – 007 = 89
x 996 – 004 RHS = - 007 x – 004 = 028 (Digit Rule)
989 / 028 Note the number of digits in deviation.
989028
Thus, 993 x 996 = 989028
Example 5: Multiply 79 x 84.
79 – 21
Working from left to right (Mental work)
x 84 – 16
Base = 100, RHS Digits = 2
63 / 36
LHS = 79 – 16 = 63 or 84 – 21 = 63
3/
RHS = - 21 x – 16 = 336
6636
3/36 –> 3 is carried over left. (Digit Rule)
Thus, 79 x 84 = 6636

13. Example 6: Multiply 64 x 96. Working from left to right (Mental work)
64 – 36 Base = 100, RHS Digits = 2
x 96 – 04 LHS = 64 – 04 = 60 or 96 – 36 = 60
60 / 44 RHS = - 36 x – 04 = 144
1/ 1/44 –> 1 is carried over to left. (Digit Rule)
6144
Thus, 64 x 96 = 6144
Case II: Both the numbers above the base.
Here, both deviations are positive and the product of the deviations will also be positive.
Example 1: Multiply 12 x 14. Working from left to right (Mental work)
12 + 2 Base = 10, RHS Digits = 1
x 14 + 4 LHS = 12 + 4 = 16 or 14 + 2 = 16
16 / 8 RHS = 2 x 4 = 8
168
Thus, 12 x 14 = 168
Example 2: Multiply 15 x 18. Working from left to right (Mental work)
15 + 5 Base = 10, RHS Digits = 1
x 18 + 8 LHS = 15 + 8 = 23 or 18 + 5 = 23
23 / 0 RHS = 5 x 8 = 40
4/
270
Thus, 15 x 18 = 270
Example 3: Multiply 101 x 108. Working from left to right (Mental work)
101 + 01 Base = 100, RHS Digits = 2
x 108 + 08 LHS = 101 + 08 = 109 or 108 + 01 = 109
109 / 08 RHS = 01 x 08 = 08. (Digit Rule)
10908
Thus, 101 x 108 = 10908
Example 4: Multiply 111 x 123. Working from left to right (Mental work)
111 + 11 Base = 100, RHS Digits = 2
x 123 + 23 LHS = 111 + 23 = 134 or 123 + 11 = 134
134 / 53 RHS = 11 x 23 = 253.
2/

14. 2/ 2/53 –> 2 is carried over to left. (Digit Rule)
13653
Thus, 111 x 123 = 13653
Example 5: Working from left to right (Mental work)
Multiply 1007x1012. Base = 1000, RHS Digits = 3
1007 + 007 LHS = 1007 + 012 = 1019 or 1012 + 007 = 1019
x 1012 + 012 RHS = 007 x 012 = 084. (Digit Rule)
1019 / 084
1019084
Thus, 1007x1012=1019084
Example 6: Working from left to right (Mental work)
Multiply 1003 x 1854. Base = 1000, RHS Digits = 3
1003 + 003 LHS = 1003 + 854 = 1857 or 1854 + 003 = 1857
x 1854 + 854 RHS = 003 x 854 = 2562.
1857 / 562 2/562 –> 2 is carried over to left. (Digit Rule)
2/
1859562
Case III: One number below the base and one number above the base.
Here, one deviation is negative, another deviation is positive; product of deviations will be negative.
Example 1: Multiply 8 x 12. Working from left to right (Mental work)
8–2 Base = 10, RHS Digits = 1
x 12 + 2 LHS = 8 + 2 = 10 or 12 – 2 = 10
10 / –4 RHS = – 2 x + 2 = – 4
10–1 / 10 –4 Transfer 1x10 from LHS to RHS. (10 is the base)
96 Then we have (10 – 1)/(10 – 4) = 96
Thus, 8 x 12 = 96 (Reduce LHS by 1 and write the complement of RHS from base 10).

15. Example 2: Multiply 94 x 102. Working from left to right (Mental work)
94 – 06 Base = 100, RHS Digits = 2
x 102 + 02 LHS = 94 + 02 = 96 or 102 – 06 = 96
96 / –12 RHS = – 06 x + 02 = – 12
96–1/100–12 Transfer 1x100 from LHS to RHS. (100 is the base)
9598 Then we have (96 – 1)/(100 – 12) = 9598
Thus, 94 x 102 = 9598
Example 3:
Multiply 984 x 1008 Working from left to right (Mental work)
984 – 016 Base = 1000, RHS Digits = 3
x 1008 + 008 LHS = 984 + 008 = 992 or 1008 – 016 = 992
992 / –128 RHS = – 016 x + 008 = –128
992–1/1000–128 Transfer 1x1000 from LHS to RHS. (1000 is the base)
991872 Then we have (992 – 1)/(1000 – 128) = 991872
Thus, 984 x 1008 = 991872
Example 4: Multiply 7 x 17. Working from left to right (Mental work)
7–3 Base = 10, RHS Digits = 1
x 17 + 7 LHS = 7 + 7 = 14 or 17 – 3 = 14
14 / –21 RHS = – 3 x + 7 = – 21
14–3/30–21 Transfer 3x10 from LHS to RHS. (10 is the base)
119 Then we have (14 – 3)/(30 – 21) = 119
Thus, 7 x 17 = 119
Example 5: Multiply 94 x 124. Working from left to right (Mental work)
94 – 06 Base = 100, RHS Digits = 2
x 124 + 24 LHS = 94 + 24 = 118 or 124 – 06 = 118
118 / –144 RHS = – 06 x + 24 = – 144
118–2/200–144 Transfer 2x100 from LHS to RHS. (100 is the base)
11656 Then we have (118 – 2)/(200 – 144) = 11656
Thus, 94 x 102 = 9598

16. Example 6: Working from left to right (Mental work)
Multiply 734x1006. Base = 1000, RHS Digits = 3
734 – 266 LHS = 734 + 006 = 740 or 1006 – 266 = 740
x 1006 + 006 RHS = – 266 x + 006 = –1596
740 /–1596 Transfer 2x1000 from LHS to RHS. (1000 is the base)
740–2 / 2000–1596 Then we have (740 – 2)/(2000 – 1596) = 738404
738404
Thus, 734 x 1006 = 738404
Working with common base
Uphill now, we studied the multiplication of numbers near a base. We shall consider the case of
multiplication of numbers having a common working base (WB). We define the base factor (BF) as,
WorkingBase
BF =
NormalBase
Consider the multiplication of 43 x 46. Here, working base is 40 = 4 x 10, where 4 is BF, such that, 43
= 4 x 10 + 3 and 46 = 4 x 10 + 6. We proceed as usual. But, we have to multiply the LHS by BF, i.e., 4
Example 1: Example 2: Example 3:
Multiply 43 x 46. Multiply 496 x 468. Multiply 3988 x 4213.
WB = 40, BF = 4. WB = 500. BF = 5. WB = 4000, BF = 4.
43 + 3 496 – 04 3988 – 012
x 46 + 6 x 468 – 32 x 4213 + 213
4 x 49 / 18 5 x 464 / 128 4 x 4201 / –2556
196 / 8 2320 / 28 16804 / –2556
1/ 1/ 16804–3/3000–2556
1978 232128
43x46=1978 496x468=232128 16801444
Observe that the LHS is multiplied by 3988x4213=16801444
the BF before applying digit rule. Transfer 3x1000 from LHS to RHS.

17. General Multiplication
So far we have discussed some special cases of multiplication. Vedic Mathematics gives a hint to
multiply any two given numbers. The Vedic Sutra "Vertically and Crosswise" – Urdhva Tiryak – helps us to
this goal. In this method multiplication can be done from left to right or from right to left.
An (n x n) digit multiplication gives 2n or (2n – 1) digits. The format and the methods are given
Multiplication of 2 x 2 digit numbers.
Method
• Write the numbers one below the other.
• a b
xc d
• Divide the answer space into three parts using slash (/) or colon (:).
• Step 1: Find (a x c) – Multiplying vertically on left side.
• Step 2: Find (a x d + b x c) – Multiplying crosswise and adding.
• Step 3: Find (b x d) – Multiplying vertically on right side.
• Write the respective products at appropriate places in the answer space.
The method can be remembered easily with the help of the following diagrams.
Each dot denotes a digit and the lines represent the multiplication of digit pairs.
The following examples will illustrate the method.
Example 1: Multiply 12 x 13 Working from left to right (Mental work)
12 Step 1: 1 x 1 = 1 – vertically left
x 13 Step 2: 1 x 3 + 2 x 1 = 3 + 2 = 5 – crosswise
1:5:6 Step 3: 2 x 3 = 6 – vertically right
156
Thus, 12 x 13 = 156
Example 2: Multiply 34 x 72 Working from left to right (Mental work)
34 Step 1: 3 x 7 = 21 - vertically left
x 72 Step 2: 3 x 2 + 4 x 7 = 6 + 28 = 34 - crosswise
21 : 34 : 8 Step 3: 4 x 2 = 8 - vertically right
Note: Number of digits in unit, ten, … places must be only one, except the
.

18. 2148 highest place. Extra digits may be transferred to next higher places.
3
2448
Thus, 34 x 72 = 2448
Example 3: Multiply 68 x 56 Working from left to right (Mental work)
68 Step 1: 6 x 5 = 30 - vertically left
x 56 Step 2: 6 x 6 + 8 x 5 = 36 + 40 = 76 - crosswise
30 : 76 : 48 Step 3: 8 x 6 = 48 - vertically right
3068
74
3808
Thus, 68 x 56 = 3808
Example 4: Multiply 84 x 92 Working from left to right (Mental work)
84 Step 1: 8 x 9 = 72 - vertically left
x 92 Step 2: 8 x 2 + 4 x 9 = 16 + 36 = 52 - crosswise
72 : 52 : 8 Step 3: 4 x 2 = 8 - vertically right
7228
5
7728
Thus, 84 x 92 = 7728
Multiplication of 3 x 3 digit numbers.
• Write the numbers one below the other.
• a b c
xd e f
• Divide the answer space into 5 parts using slash (/) or colon (:).
• Step 1: Find (a x d)
• Step 2: Find (a x e + b x d)
• Step 3: Find (a x f + b x e + c x d).
• Step 4: Find (b x f + c x e)
• Step 5: find (c x f)
• Write the respective products at appropriate places in the answer space.

19. The method can be remembered easily with the help of the following diagrams.
Each dot denotes a digit and the lines represent the multiplication of digit pairs.
The following examples will illustrate the method.
Example 1: Multiply 236 x 482 Working from left to right (Mental work)
236 Step 1: 2x4 = 8
x 482 Step 2: 2x8 + 3x4 = 16+12 = 28
8 : 28 : 52 : 54 : 12 Step 3: 2x2 + 3x8 + 6x4 = 4+24+24=52
88242 Step 4: 3x2 + 6x8 = 6 + 48 = 54
2551 Step 5: 6x2 = 12
113752
Thus, 236 x 482 = 113752
Example 2: Multiply 738 x 659 Working from left to right (Mental work)
738 Step 1: 7x6 = 42
x 659 Step 2: 7x5 + 3x6 = 35+18 = 53
42 : 53 : 126 : 67 : 72 Step 3: 7x9 + 3x5 + 8x6 = 63+15+48=126
423672 Step 4: 3x9 + 8x5 = 27 + 40 = 67
5267 Step 5: 8x9 = 72
1
486342
Thus, 738 x 659 = 486342
Example 3: Multiply 574 x 836 Working from left to right (Mental work)
574 Step 1: 5x8 = 40
x 836 Step 2: 5x3 + 7x8 = 15+56 = 71
40 : 71 : 83 : 54 : 24 Step 3: 5x6 + 7x3 + 4x8 = 30+21+32=83
401344 Step 4: 7x6 + 4x3 = 42 + 12 = 54
7852 Step 5: 4x6 = 24
479864
Thus, 574 x 836 = 479864
Example 4: Multiply 972 x 638 Working from left to right (Mental work)
972 Step 1: 9x6 = 54
x 638 Step 2: 9x3 + 7x6 = 27+42 = 69

20. 54 : 69 : 105 : 62 : 16 Step 3: 9x8 + 7x3 + 2x6 = 72+21+12=105
549526 Step 4: 7x8 + 2x3 = 56 + 6 = 62
6061 Step 5: 2x8 = 16
1
620136
Thus, 972 x 638 = 620136
Multiplication of 4 x 4 digit numbers.
• Write the numbers one below the other.
a b c d
xe f g h
• Divide the answer space into 7 parts using slash (/) or colon (:).
• Step 1: Find (a x e)
• Step 2: Find (a x f + b x e)
• Step 3: Find (a x g + b x f + c x e).
• Step 4: Find (a x h + d x e + b x g + c x f)
• Step 5: Find (b x h + c x g + d x f)
• Step 6: Find (c x h + d x g)
• Step 7: Find ( d x h)
• Write the respective products at appropriate places in the answer space.
The method can be remembered easily with the help of the following diagrams.
Each dot denotes a digit and the lines represent the multiplication of digit pairs.
The following examples will illustrate the method.
Example 1: Multiply 2463 x 3728 Working from left to right (Mental work)
2463 Step 1: 2x3=6
x 3728 Step 2: 2x7+4x3 =14+12=26
6:26:50:75:65:54:24 Step 3: 2x2+4x7+6x3=4+28+18=50
6605544 Step 4: 2x8+3x3+4x2+6x7=16+9+8+42=75
257652 Step 5: 4x8+6x2+3x7=32+12+21=65
9182064 Step 6: 6x8+3x2=48+6=54
Thus, 2463 x 3728 = 9182064 Step 7: 3x8=24

21. Example 2: Multiply 6378 x 7596 Working from left to right (Mental work)
6378 Step 1: 6x7=42
x 7596 Step 2: 6x5+3x7 =30+21=51
42:51:118:154:121:114:48 Step 3: 6x9+3x5+7x7=54+15+49=118
42184148 Step 4: 6x6+8x7+3x9+7x5=36+56+27+35=154
515214 Step 5: 3x6+7x9+8x5=18+63+40=121
1111 Step 6: 7x6+8x9=42+72=114
48447288 Step 7: 8x6=48
Thus, 6378 x 7596 = 48447288
Example 3: Multiply 5743 x 6859 Working from left to right (Mental work)
5743 Step 1: 5x6=30
x 6859 Step 2: 5x8+7x6 =40+42=82
30:82:105:130:107:51:27 Step 3: 5x5+7x8+4x6=25+56+24=105
30250717 Step 4: 5x9+3x6+7x5+4x8=45+18+35+32=130
803052 Step 5: 7x9+4x5+3x8=63+20+24=107
111 Step 6: 4x9+3x5=36+15=51
39391237 Step 7: 3x9=27
Thus, 5743 x 6859 = 39391237
Summary
So far we studied various methods of multiplication. But, one may like to know which method suits
best? The choice is personal. All methods give the same result as will be seen below. We must select the
simplest and easiest method.
Let us consider a problem worked out by different methods. Suppose we want to multiply –>
95 x 95.

22. By Ekadhika Multiplication. By Nikhilam Multiplication. By Urdhva Tiryak Method.
Multiply 95 x 95. Multiply 95 x 95. Multiply 95 x 95
9 is common. 5 + 5 = 10. 95 is nearer to Base=100. 95
Ekadhika is applicable. Deviation = 95–100=–05 x 95
End digit = 5, Nikhilam is applicable 81 : 90 : 25
Previous digit = 9 95 – 05 9025
952 x 95 - 05 Therefore, 95 x 95 = 9025
9 x (9+1) / 25 95-05 / 25
90 / 25 9025
Therefore, 952 = 9025. Therefore, 95 x 95 = 9025
So, which method has to be preferred? Probably, Ekadhika method is best suited in this case. Is it not?
Remember to work out the problem mentally and write the answer in one or line.
Division
Division is the hardest of all the arithmetical operations. The traditional method of division is always the
same irrespective of the divisor. But in Vedic Mathematics there are different methods depending on the
nature of the divisor.
The format for division in Vedic Mathematics is shown below.
Divisor Dividend
Quotient / Remainder
We shall discuss various methods of division in Vedic Mathematics.
Division by 9.
Division by 9 is the simplest one in Vedic Mathematics. We divide the dividend by the devisor to get
quotient and remainder. When the remainder is equal to or greater than 9 we re-divide the remainder by 9,
carry over this quotient to the quotient side and retain the final remainder in the remainder side. We shall
study the method by some illustrations.
Example 1: Divide 12 ÷ 9. Working (Mental)
Dividend = 12, Step 1: Write the devisor and dividend as shown.
Devisor = 9. Step 2: Set off last digit for the remainder by a slash (/) or colon (:).
9 1/2 Step 3: Enter 0 under 1st digit and add. This gives quotient, Q = 1.
0 1 Step 4: Write the result (i.e., 1) under 2nd digit and add to get
1/3 2 + 1 = 3. This gives the Remainder, R= 3.Thus, 12 ÷ 9 –> Q = 1, R = 3.

23. Example 2: Working (Mental)
Divide 123 ÷ 9. Step 1: Write the devisor and dividend as shown.
Dividend = 123, Step 2: Set off last digit for the remainder by a slash (/) or colon (:).
Devisor = 9. Step 3: Enter 0 under 1st digit and add. This gives quotient, Q = 1.
Step 4: Write the result (i.e., 1) under 2nd digit and add to get 2+1=3.
9 12/3 Division is over and Q = 13.
01 3 Step 5: Write this result (i.e., 3) under 3rd digit and add 3+3=6.This gives
13/6 the Remainder, R= 6.
Thus, 12 ÷ 9 –> Q = 13, R = 6.
Example 3: Working (Mental)
Divide 2340 ÷ 9. Step 1: Write the devisor and dividend as shown.
Dividend = 2340, Step 2: Set off last digit for the remainder by a slash (/) or colon (:).
Devisor = 9. Step 3: Enter 0 under 1st digit and add. This gives quotient, Q = 2.
Step 4: Write the result (i.e., 2) under 2nd digit and add to get 3+2=5.
9 234/0 Step 5: Write this result (i.e., 5) under 3rd digit and add to get 4+5=9.
025 9 Step 6: Write this result (i.e., 9) under the 4th digit and add to get 0+
259/9 9 =9. Remainder, R= 9.
1/0 As the remainder is equal to 9, we re-divide it by 9 and compute new
260/0 quotient and remainder.
Thus we get, 2340 ÷ 9 –> Q = 260, R = 0.
Example 4: Working (Mental)
Divide 3509 ÷ 9. Step 1: Write the devisor and dividend as shown.
Dividend = 3509, Step 2: Set off last digit for the remainder by a slash (/) or colon (:).
Devisor = 9. Step 3: Enter 0 under 1st digit and add. This gives quotient, Q = 3.
Step 4: Write the result (i.e., 3) under 2nd digit and add to get 5+3=8.
9 350/9 Step 5: Write this result (i.e., 8) under 3rd digit and add to get 0+8=8.
038/8 Step 6: Write this result (i.e., 8) under the 4th digit and add to get 9+
3 8 8 / 17 8 =17. Remainder, R= 17.
1/8 As the remainder is greater than 9, we re-divide it by 9 and compute new
38 9/8 quotient and remainder.
Thus we get, 3509 ÷ 9 –> Q = 389, R = 8.
Division By 8.
This method is similar to the one with division by 9. The only difference is, we write 2, the complement
of 8, below 8. Every time we multiply the quotient digit by 2 and add it to the next digit. When the

24. remainder is equal to or greater than 8 we re-divide the remainder by 8, carry over this quotient to the
quotient side and retain the final remainder on the remainder side.
Example 1: Working (Mental)
Divide 13 ÷ 8. Step 1: Write the devisor and dividend as shown.
Dividend = 13, Step 2: Set off last digit for the remainder by a slash (/) or colon (:).
Devisor = 8. Step 3: Enter 0 under 1st digit and add. This gives quotient, Q = 1.
8 1/3 Step 4: Now, 1 x 2 = 2, 3 + 2 = 5. Remainder, R= 5.
2 0 2 Thus, 13 ÷ 8 → Q = 1, R = 5.
1/5
Example 2: Working (Mental)
Divide 257 ÷ 8. Step 1: Write the devisor and dividend as shown.
Dividend = 257, Step 2: Set off last digit for the remainder by a slash (/) or colon (:).
Devisor = 8. Step 3: Enter 0 under 1st digit and add. This gives quotient, Q = 2.
8 25/7 Step 4: Now, 2 x 2 = 4, 5 + 4 = 9.
2 0 4 18 Step 5: Again, 9 x 2 = 18, 7 + 18 = 25.
2 9 / 25 As the remainder is greater than 8, we re-divide it by 8 and compute
3/1 new quotient and remainder.
32/1 Thus we get, 257 ÷ 8 –> Q = 32, R = 1.
We can extend this method for division with 7, 6, . . .
Next we shall consider the division with divisors having more than one digit, and when the divisors are
slightly smaller or greater than the base.
Nikhilam Division
This method is useful for division with the devisors nearer and below the base. The method consists in
finding the complement of the devisor using Nikhilam Sutra "All from 9". Hence, the method is called
Nikhilam division.
Method
• Write the problem in the usual format.
• Find the complement of the devisor and write it below the devisor.
• Divide the dividend into quotient part and remainder part by a slash (/) or colon (:).
• The number of digits in the remainder part must be equal to the number of zeros in the base.

25. • Leave sufficient space between the dividend and quotient lines. The number of lines must be
equal to the number of digits in the quotient part of the dividend.
• Bring down the 1st digit of dividend as the 1st digit of the quotient.
• Multiply the 1st quotient digit with the complement of the devisor and write the product starting
under 2nd column.
• Add the digits in 2nd column to get 2nd quotient digit.
• Multiply the 2nd quotient digit again with the complement of the devisor and write the product
starting under 3rd column.
• Add the digits in 3rd column to get 3rd quotient digit.
• Repeat this as many times as the digits in the quotient part of dividend.
• Add all digits under remainder part to get the final remainder. If this exceeds the devisor, re-
divide.
• If the sum exceeds 9, carry over the excess to the left.
Example 1: Divide 1318 ÷ 88. Working (Mental)
88 13:1 8 Step 1: Complement of devisor 88 is 12.
12 1 2 Step 2: Split dividend as 13:18. Bring down 1.
4 8 Step 3: 1x12=12. Write it as shown.
14:8 6 Step 4: 3+1=4 and 4x12=48. Division over.
Step 5: Remainder = 18+20+48=86.
Hence, 1318 ÷ 88 –> Q = 14, R = 86.
Example 2: Divide 2316 ÷ 97. Working (Mental)
97 23:16 Step 1: Complement of devisor 97 is 03.
03 0 6 Step 2: Split dividend as 23:16. Bring down 2.
09 Step 3: 2x03=06. Write it as shown.
23:85 Step 4: 3+0=3 and 3x03=09. Division over.
Step 5: Remainder = 16+60+09=85.
Hence, 2316 ÷ 97 –> Q = 23, R = 85.
Example 3: Divide 38445 ÷ 896. Working (Mental)
896 3 8: 4 4 5 Step 1: Complement of devisor 896 is 104.
104 3 1 2 Step 2: Split dividend as 38:445. Bring down 3.
11 0 44 Step 3: 3x104=312. Write it as shown.
3 11: 16 6 49 Step 4: 3+8=11 and 11x104=11 0 44.
4 1 : 1709 Step 5: Remainder = 1600+60+49=1709.
4 2 : 813 1709 ÷ 896 = Q = 1, R = 813.

26. Final quotient = 3x10 + 11 + 1 = 42.
Hence, 38445 ÷ 896 –> Q = 42, R = 813.
Paravartya Division
This method is useful for division with the devisors nearer and above the base. The method is similar
to Nikhilam division, but slightly different. It uses the Vedic Sutra "Transpose and apply" known as
Paravartya. We leave the first digit, and change the sign of other digit or digits with minus (-) sign and
place them below the divisor. This is the meaning of ‘ transpose and apply ’.
Method
• Write the problem in the usual format.
• Leaving the first digit, write other digit or digits using minus (-) sign and place them below the
divisor.
• Divide the dividend into quotient part and remainder part by a slash (/) or colon (:).
• The number of digits in the remainder part must be equal to the number of zeros in the base.
• Leave sufficient space between the dividend and quotient lines. The number of lines must be
equal to the number of digits in the quotient part of the dividend.
• Bring down the 1st digit of dividend as the 1st digit of the quotient.
• Multiply the 1st quotient digit with the complement of the devisor and write the product starting
under 2nd column.
• Add the digits in 2nd column to get 2nd quotient digit.
• Multiply the 2nd quotient digit again with the complement of the devisor and write the product
starting under 3rd column.
• Add the digits in 3rd column to get 3rd quotient digit.
• Repeat this as many times as the digits in the quotient part of dividend.
• Add all digits under remainder part to get the final remainder. If this exceeds the devisor, re-
divide.
• If the sum exceeds 9, carry over the excess to the left.
Example 1: Divide 358 ÷ 12. Working (Mental)
12 35:8 Step 1: Drop 1 of devisor 12. New devisor is –2
–2 –6 Step 2: Split dividend as 35:8. Bring down 3.
2 Step 3: 3x–2= –6 . Write it as shown.
3 –1: 10 Step 4: 5+(–6)=–1 and–1x–2 = 2. Division over.
2 9 : 10 Step 5: Remainder =8+2=10.
Hence, 358 ÷ 12 –> Q =3x10–1=29, R = 10.

27. Example 2: Divide 24688 ÷ 102. Working (Mental)
102 2 4 6:8 8 Step 1: Drop 1 of devisor 122. New devisor is 0 –2
0 –2 0 –4 Step 2: Split dividend as 246:88. Bring down 2.
0 –8 Step 3: (0 –2)x2= 0 –4. Write it as shown.
0 –4 Step 4: 4+0=4 and (0 –2)x4= 0 –8. Division over.
2 4 2: 0 4 Finally, 6+(–4)=2 and (0 –2)x2= 0 –4.
Step 5: Remainder =88–80–04= 04.
Hence, 358 ÷ 12 –> Q =242, R = 4.
Example 3: Divide 36478 ÷ 1012. Working (Mental)
1012 3 6: 4 7 8 Step 1: Drop 1 of devisor 1012. New devisor 0 –1 –2
0 –1 –2 0 –3 –6 Step 2: Split dividend as 36:478. Bring down 3.
0 –6 –12 Step 3: (0 –1 –2)x3 =0 –3 –6. Write it as shown.
36: 46 Step 4: 6+0 =6 and (0 –1 –2)x6 =0 –6 –12.
Step 5: Remainder =478–360–60–12 = 46
Hence, 36478 ÷ 1012 –> Q =36, R = 46.
Straight Division
This is the general method of division applicable to all types of division. This uses the combination of
Vedic Sutras: "Dhvajanka and Urdhva Tiryak". The meaning of the word Dhvajanka is "flag digit", since it is
written as a flag little above the operator digit. The method is given below.
Method
• Separate the first digit of the devisor from other digits. The first digit is called operator and other
digits are called flag digits.
• Flag digits are placed at a higher level above the operator.
• Divide the dividend into two parts – quotient part and remainder part – using a slash (/) or a
colon (:).
• The number of digits in the remainder part must be same as the number of flag digits.
• Divide the first digit (or first two digits) by the operator to get a partial quotient and a remainder.
• This remainder with the next digit of the quotient part of dividend forms the gross dividend (GD).
• Find the product of flag digits and partial quotient to get the flag dividend (FD).
• FD is calculated using Urdhva Tiryak. Note the method of calculation carefully.
• The difference between GD and FD gives net dividend ND.
• Divide ND further by the operator to get the next partial quotient and remainder.
• Repeat the process with the remaining digits of the quotient part.

28. • Finally, evaluate the remainder (R=ND–FD).
The above steps are illustrated in the example below.
Division by devisors with 1 flag digit
Example 1: Divide 5284 ÷ 32
2 5 2 8: 4
3 2 2 :1
1 6 5:4
Here, 3 –> operator, 2 –> flag digit.
Note that the remainder part has only one digit. (Guess)
Working (Mental)
Step 1: 5 ÷ 3 = 1 R 2
(Read Q=1, R=2). Written as above.
GD = 22, FD = 2 x 1 = 2
ND = 22 – 2 = 20 ÷ 3 = 6 R 2, written as above.
Step 2: GD = 28, FD =2 x 6 = 12
ND = 28 – 12 = 16 ÷ 3 = 5 R1, written as above.
Here, division is over. We shall calculate the remainder as below.
Step 3: GD = 14, FD = 2x5=10.
Remainder = 14 – 10 = 4.
Hence, 5284 ÷ 32 --> Q = 165, R = 4.
Example 2: Divide 3275 ÷ 23
3 3 2 7: 5
2 1 1 :1
1 4 2 :9
Here, 2 –> operator, 3 –> flag digit.
Working (Mental)
Step 1: 3 ÷ 2 = 1 R 1, written as above.
GD = 12, FD = 3 x 1 = 3, ND = 12 – 3 = 9 ÷ 2 = 4 R 1, written as above.
Step 2: GD = 17, FD = 3 x 4 = 12
ND = 17 – 12 = 5 ÷ 2 = 2 R 1, written as above.
Here, division is over. We shall calculate the remainder as below.
Step 3: GD = 15, FD = 3x2= 6.
Remainder = 15 – 6 = 9.

29. Hence, 3275 ÷ 23 –> Q = 142, R = 9.
Example 3: Divide 5678 ÷ 43
3 5 6 7: 8
4 1 1 :0
1 3 2 :2
Here, 4 –> operator, 3 –> flag digit.
Working (Mental)
Step 1: 5 ÷ 4 = 1 R 1, written as above.
GD = 16, FD = 3 x 1 = 3, ND = 16 – 3 = 13 ÷ 4 = 3 R 1, written as above.
Step 2: GD = 17, FD = 3 x 3 = 9
ND = 17 – 9 = 8 ÷ 4 = 2 R 0, written as above.
Here, division is over.
Step 3: GD = 8, FD = 3x2=6.
Remainder = 8 – 6 = 2.
Hence, 5678 ÷ 43 –> Q = 132, R = 2.
Example 4: Divide 27543 ÷ 64
4 27 5 4: 3
6 3 1 :2
4 3 0: 23
Here, 6 –> operator, 4 –> flag digit.
Working (Mental)
Step 1: 27 ÷ 6 = 4 R 3, written as above.
GD = 35, FD = 4 x 4 = 16
ND = 35 – 16 = 19 ÷ 6 = 3 R 1, written as above.
Step 2: GD = 14, FD = 4 x 3 = 12
ND = 14 – 12 = 2 ÷ 6 = 0 R 2, written as above.
Here, division is over.
Step 3: GD = 23, FD = 4x0=0.
Remainder = 23 – 0 = 23.
Hence, 27543 ÷ 64 → Q = 430, R = 23.

30. Division by Devisors with 2 flag digits
Example 1: Divide 237963 ÷ 524
24 23 7 9: 63
5 3 4 :3
4 5 4 : 67
Here, 5 –> operator, 24 –> flag digits.
Note that the remainder part has two digits. (Guess)
Working (Mental)
Step 1: 23 ÷ 5 = 4 R 3 , written as above.
GD = 37, FD = 2 x 4 + 4 x 0 = 8
ND = 37 – 8 = 29 ÷ 5 = 5 R 4, written as above.
Step 2: GD = 49, FD=2 x 5 + 4 x 4 = 26
ND = 49 – 26 = 23 ÷ 5 = 4 R 3, written as above.
Here, division is over. We shall calculate the remainder as below.
Step 3: GD = 363,
FD = (2x4+4x5)/(2x0+4x4) = 28/16 = 296. (Guess how? 280 + 16)
Remainder = 363 – 296 = 67.
Hence, 237963 ÷ 524 –> Q = 454, R = 67.
Example 2: Divide 2999222 ÷ 713
13 29 9 9 2: 22
7 1 1 5 :4
4 2 0 6 : 344
Here, 7 –> operator, 13 –> flag digits.
Working (Mental)
Step 1: 29 ÷ 7 = 4 R 1 , written as above.
GD = 19, FD = 1 x 4 + 3 x 0 = 4
ND = 19 – 4 = 15 ÷ 7 = 2 R 1, written as above.
Step 2: GD = 19, FD = 1 x 2 + 3 x 4 = 14
ND = 19 – 14 = 5 ÷ 7 = 0 R 5, written as above.
Step 3: GD = 52, FD = 1 x 0 + 3 x 2 = 6
ND = 52 – 6 = 46 ÷ 7 = 6 R 4, written as above.
Here, division is over. We shall calculate the remainder as below.
Step 4: GD = 422,

31. FD = (1x6+3x0) / (1x0+3x6) = 6 / 18 = 78.
Remainder = 422 – 78 = 344.
Hence, 2999222 ÷ 713 Q = 4206, R = 344.
Example 3: Divide 35769 ÷ 512
12 35 7: 69
5 5 :6
6 9 :441
Here, 5 –> operator, 12 –> flag digits.
Working (Mental)
Step 1: 35 ÷ 5 = 6 R 5 , written as above.
(With higher quotient digit remainder becomes negative. Check?)
GD = 57, FD = 1 x 6 + 2 x 0 = 6
ND = 57 – 6 = 51 ÷ 5 = 9 R 6, written as above.
We shall calculate the remainder as below.
Step 2: GD = 669,
FD = (1x9+2x6) / (1x0+2x9) = 21 / 18 = 228.
Remainder = 669 – 228 = 441.
Hence, 35769 ÷ 512 –> Q = 69, R = 441.
Division by Devisors with 3 flag digits
Example 1: Divide 163758 ÷ 5314
341 16 3: 758
5 1 :4
3 0 : 3528
Here, 5 –> operator, 341 –> flag digits.
Working (Mental)
Step 1: 16 ÷ 5 = 3 R 1 , written as above.
GD = 13, FD = 3x3+4x0+1x0=9
ND = 13 – 9 = 4 ÷ 5 = 0 R 4, written as above.
Here, division is over. We shall calculate the remainder as below.
Step 2: GD = 4758,
FD= (3x0+4x3+1x0)/(3x0+4x0+1x3)/(3x0+4x0+1x0)
= (0+12+0) / (0+0+3)/(0+0+0) : = 12 / 3 / 0 = 1230.
Remainder = 4758 – 1230 = 3528.

32. Hence, 163758 ÷ 5314 → Q = 30, R = 3528.
Example 2: Divide 563214 ÷ 4123
123 5 6 3: 214
4 1 3 :4
1 3 6 : 2486
Here, 4 –> operator, 123 –> flag digits.
Working (Mental)
Step 1: 5 ÷ 4 = 1 R 1 , written as above.
GD = 16, FD =1x1+2x0+3x0=1
ND = 16 – 1 = 15 ÷ 4 = 3 R 3, written as above.
Step 2: GD = 33, FD = 1x3+2x1+3x0=5
ND = 33 – 5 = 28 ÷ 4 = 6 R 4, written as above.
Here, division is over. We shall calculate the remainder as below.
Step 3: GD = 4214,
FD = (1x6+2x3+3x1)/(1x0+2x6+3x3)/(1x0+2x0+3x6)
= (6+6+3) / (0+12+9)/(0+0+18)
= 15 / 21 / 18 = 1728. (Guess how?)
Remainder = 4214 – 1728 = 2486.
Hence, 563214 ÷ 4123 –> Q = 136, R = 2486.
Example 3: Divide 12345678 ÷ 8234
234 12 3 4 5: 678
8 4 9 11 :9
1 4 4 9 : 2912
Here, 8 –> operator, 234 –> flag digits.
Working (Mental)
Step 1: 12 ÷ 8 = 1 R 4 , written as above.
GD = 43, FD = 2x1+3x0+4x0=2
ND = 43 – 2 = 41 ÷ 8 = 4 R 9, written as above.
Step 2: GD = 94, FD = 2x4+3x1+4x0=11
ND = 94 – 11 = 83 ÷ 8 = 9 R 11 written as above.
Step 3: GD = 115, FD = 2x9+3x4+4x1=34
ND = 115 – 34 = 81 ÷ 8 = 9 R 9 written as above.
Here, division is over. We shall calculate the remainder as below.
Step 4: GD = 9678,

33. FD = (2x9+3x9+4x4)/(2x0+3x9+4x9)/(2x0+3x0+4x9)
= (18+27+16) / (0+27+36)/(0+0+36)
= 61 / 63 / 36 = 6766. (Guess how?)
Remainder = 9678 – 6766 = 2912.
Hence, 12345678 ÷ 8234 –> Q = 1499, R = 2912.
Summary
So far we studied various methods of division. But, one may like to know which method suits best? The
choice is personal. All methods give the same result as will be seen below. We must select the simplest and
easiest method.
Let us consider a problem worked out by different methods. Suppose we want to divide 2345 by 98.
By Nikhilam Division Method. By Paravartya Method By Straight Division Method
Divide 2345 ÷ 98. Divide 2345 ÷ 98. Divide 2345 ÷ 98.
Modified divisor = 100–98 = 02 We know 98 =100+(–2). Operator = 9
98 23:45 Modified divisor = 02 Flag digit = 8
02 0 4 98 23:45 8 23 4 : 5
06 02 0 4 9 5 :11
2 3 : 91 06 2 3 :91
2 3 : 91
So, either of the first two methods is preferable in this case.
Squares and Square Roots
When a number is multiplied by itself we get the square of that number.
E.g., 2 x 2 = 4, 3 x 3 = 9. We say that:
4 is the square of 2 and 2 is the square root of 4.
9 is the square of 3 and 3 is the square root of 9.
The squares of digits 1 – 9 are given below.
Digits 1 2 3 4 5 6 7 8 9
Squares 1 4 9 16 25 36 49 64 81
We notice that the digits equidistant from 5 have same end digits. We shall study the methods of
squaring of a number.

34. Ekadhika method
We studied earlier the method of finding the square of numbers ending in 5 under Ekadhika
multiplication. E.g., 352 = 3 x 4 / 25 = 1225,
652 = 6 x 7 / 25 = 4225, etc.
If the number is large we can use Urdhva Tiryak multiplication in concurrence to achieve this. This is
illustrated in the following examples.
Example 1: Find the square of 285. Example 2: Find the square of 1235.
2852 = 28 x 29 / 25 12352 = 123 x 124 / 25
28 But, 123
x29 x124
442 14142
37 111
812 15252
Therefore, 2852 = 812/25 = 81225. Therefore, 12352 = 15252/25 = 1525225.
Yavadunam Method
The Vedic Sutra "Deviate as much as deviation and add square of the deviation". This is known as
Yavadunam Sutra. We shall see the application of this Sutra to find the square of numbers.
The answer consists of 2 portions as given here under.
• The excess (deviation) is added to the number. This forms the Left Portion of the answer.
• The square of the initial excess forms the Right Portion of the answer.
This is explained below with examples.
Example 1: Find the square of 12.
Here, Base = 10, Deviation = 12 – 10 = +2.
Hence, 122 = 12 + 2 / 22 = 144.
Example 2: Find the square of 108.
Here, Base = 100, Deviation = 108 – 100 = +08.
1082 = 108 + 08 / 082 = 11664. (Digit Rule)
Example 3: Find the square of 1015.
Base = 1000, Deviation = 1015 – 1000 = +015.
10152 = 1015 + 015 / 0152 = 1030225.
Example 4: Find the square of 8.
Here, Base = 10, Deviation = 8 – 10 = – 2.

35. Hence, 82 = 8 – 2 / (-2)2 = 64.
Example 5: Find the square of 96.
Here, Base = 100, Deviation = 96 – 100 = – 04.
Hence, 962 = 96 – 4 / (-04)2 = 9216.
Example 6: Find the square of 985.
Here, Base = 1000,
Deviation = 985 – 1000 = – 015.
Hence, 9852 = 985 – 015 / (–015)2 = 970225.
Straight Squaring (General Method)
To find the square of any number we use the Vedic Sutra Dwandva Yoga or Duplex System. Duplex of
a number is formed (i) by squaring and (ii) by cross-multiplying. If the digit is single central digit, the
Duplex is a square and for an even number of digits equidistant from the two ends the Duplex is twice the
cross - product. Duplex of numbers are defined as follows:
No Duplex Example
2
a D(a) = a D(2) = 22 =4
ab D(ab) = 2ab D(23) = 2x2x3=12
abc D(abc)= b2+2ac D(234) = 22+2x3x4=28
abcd D(abcd)=2ad+2bc D(2345) = 2x2x5+2x3x4=44
We use Duplex system to find the square of numbers. Observe that for a n-digit number, the square of
the number contains 2n or (2n-1) digits.
Squaring of 2 digit numbers
We have: (a + b)2 = a2 + 2ab + b2
Or (a + b)2 = D(a) + D(ab) + D(b).
We use this property as follows.
(ab)2 = D(a) : D(ab) : D(b)
This is clear from the following examples.
Example 1: Find the square of 36. Working (Mental)
362 Step 1: D(3) = 32 = 9
D(3):D(36):D(6) Step 2: D(36) = 2x3x6 =36
9 : 36 : 36 Step 3: D(6) = 62 =36.
966 Hence, 362 = 1296
33

36. 1296
Example 2: Find the square of 74. Working (Mental)
742 Step 1: D(7) = 72 = 49
D(7):D(74):D(4) Step 2: D(74) = 2x7x4 =56
49 : 56 : 16 Step 3: D(4) = 42 =36.
4966 Hence, 742 = 5476
51
5476
Example 3: Find the square of 85. Working (Mental)
852 Step 1: D(8) = 82 = 64
D(8):D(85):D(5) Step 2: D(85) = 2x8x5 =80
64 : 80 : 25 Step 3: D(5) = 52 =25.
6405 Hence, 852 = 7225
82
7225
Squaring of 3 digit numbers
We have: (a + b + c)2 = a2 + 2ab + (b2 +2ac) + 2bc + c2
Or (a+b+c)2 = D(a)+D(ab)+D(abc)+D(bc)+D(c).
We use this property as follows.
(abc)2 = D(a) : D(ab) : D(abc): D(bc) D(c)
This is clear from the following examples.
Example 1: Find the square of 234. Working (Mental)
2342 Step 1: D(2) = 22 = 4
D(2):D(23):D(234):D(34):D(4) Step 2: D(23) = 2x2x3 =12
4 :12 : 25 : 24 : 16 Step 3: D(234) = 32+2x2x4=25.
42546 Step 4: D(34) = 2x3x4 =24
1221 Step 5: D(4) = 42 =16.
54756 Hence, 2342 = 1296
Example 2: Find the square of 647. Working (Mental)
6472 Step 1: D(6) = 62 = 36
Step 2: D(64) = 2x6x4 =48
Step 3: D(647) = 42+2x6x7=100.
D(6):D(64):D(647):D(47):D(7)

37. 368069 Step 4: D(47) = 2x4x7 =56
4054 Step 5: D(7) = 72 =49.
1 Hence, 6472 = 418609
418609
Example 3: Find the square of 795. Working (Mental)
7952 Step 1: D(7) = 72 = 49
D(7):D(79):D(795):D(95):D(5) Step 2: D(79) = 2x7x9 =126
49 :126 : 151 : 90 : 25 Step 3: D(795) = 92+2x7x5=151.
496105 Step 4: D(95) = 2x9x5 =90
2592 Step 5: D(5) = 52 =25.
11 Hence, 7952 = 632025
632025
This method can be extended to other numbers up to any level.
Square Roots Of Perfect Squares
The square root of a perfect square can be found by division method. Before finding the square root of
a number we must know 3 things.
• Number of digits in the square root.
If the number has n-digits, the square root will have n/2 or (n+1)/2 digits.
• First digit of the square root must be known. The number is divided into groups of 2 digits from
right to left. The nearest square root of the first group gives the first square root digit.
• The devisor must also be known.
Twice the first square root digit is taken as devisor.
• While dividing we use Duplex of digits from second digit onwards.
Example 1: Find the square root of 1156.
No of square root digits = 4/2 = 2.
Grouping of digits = 11'56.
First square root digit = 3 ( 32 < 11)
Modified Devisor = 3x2 = 6.
11 5 6
6 2 1
3 4.0
Working (Mental)
Step 1: 11 – 32 = 11 – 9 = 2

38. Step 2: 25 ÷ 6 = 4 R 1
Step 3: 16 – D(4) = 16 – 42 = 0
Hence, Square Root of 1156 = 34.
Example 2: Find the square root of 6241.
No of square root digits = 4/2 = 2.
Grouping of digits = 62'41.
First square root digit = 7 ( 72 < 62)
Modified Devisor = 7x2 = 14.
62 4 1
14 13 8
7 9.0
Working (Mental)
Step 1: 62 – 72 = 62 – 49 = 13
Step 2: 134 ÷ 14 = 9 R 8
Step 3: 81 – D(9) = 81 – 92 = 0
Hence, Square Root of 6241 = 79.
Example 3: Find the square root of 80656.
No of square root digits = (5+1)/2 = 3.
Grouping of digits = 8'06'56.
First square root digit = 2.
Modified Devisor = 2x2 = 4.
8 0 6 5 6
4 4 8 6 1
284.00
Working (Mental)
Step 1: 8 – 22 = 8 – 4 = 4
Step 2: 40 ÷ 4 = 8 R 8 (Guess)
Step 3: 86 – D(8) = 86 – 82 = 22
22 ÷ 4 = 4 R 6
Step 4: 65 – D(84) = 65 – 64 = 1
I÷4=0R1
Step 5: 16 – D(4) = 16 – 16 = 0
Hence, Square Root of 80656 = 284.
Example 4: Find the square root of 119025.

39. No of square root digits =6/2 = 3.
Grouping of digits = 11:90:25.
First square root digit = 3.
Modified Devisor = 3x2 = 6.
11 9 0 2 5
6 2 5 4 2
3 4 5 . 00
Working (Mental)
Step 1: 11 – 32 = 11 – 9 = 2
Step 2: 29 ÷ 6 = 4 R 5
Step 3: 50 – D(4) = 50 – 42 = 34
34 ÷ 6 = 5 R 4
Step 4: 42 – D(45) = 42 – 40 = 2
2÷6=0R2
Step 5: 25 – D(5) = 25 – 5 = 0
Hence, Square Root of 119025 = 345.
Summary
So far we studied various methods of squaring. But, one may like to know which method suits best?
The choice is personal. All methods give the same result as will be seen below. We must select the simplest
and easiest method.
Let us consider a problem worked out by different methods. Suppose we want to find the square of 95.
By Ekadhika Method By Yavadunam Method By Dwandva Yoga Method
Find the square of 95. Find the square of 95. Find the square of 95.
952 = 9 x 10 / 25 Here, Base = 100, 952
952 = 9025 Deviation = 95–100 = –05. D(9):D(95):D(5)
952 = (95–05) / 052 = 9025. 81 : 90 : 25
9025
Naturally, Ekadhika method is best suited in this case.
Cubes and Cube Roots
When a number is multiplied by itself three times we get the cube of that number.
E.g., 2 x 2 x 2 = 8, 3 x 3 x 3 = 27, 4 x 4 x 4 = 64, etc. Thus,
8 is the cube of 2 and 2 is the cube root of 8,

40. 27 is the cube of 3 and 3 is the cube root of 27,
64 is the cube of 4 and 4 is the cube root of 64.
The cubes of digits from 1 – 9 are given below.
Digits 1 2 3 4 5 6 7 8 9
Cubes 1 8 27 64 125 216 343 512 729
Thus, the cubes of digits have unique end digits.
We shall next study the method of cubing a number.
Yavadunam Method
This method explained earlier can be used in this case also, with modifications. The answer consists of
3 portions as given here under.
• Twice the excess (deviation) is added to the number. This forms Left Portion of the answer.
• The product of new excess and the original excess forms the Middle Portion of the answer.
• The cube of the initial excess forms the Right Portion of the answer.
This is explained below with examples.
Example 1: Find the cube of 13. Example 2: Find the cube of 106.
Base = 10. Deviation = +3. Base = 100. Deviation = +6.
New excess =3 + 6=9 New excess = 6+12=18
(Deviation + Twice deviation) Hence, (Deviation + Twice deviation) Hence,
133 1063
13+6 : 9 x 3 : 33 106+12 : 18 x 6 : 63
19 : 27 : 27 118 : 108 : 216
1977 1180816
22 1 2
2197 1191016
Therefore, 133 = 2197 Therefore, 1063 = 1191016
Example 3: Find the cube of 1003. Example 4: Find the cube of 8.
Base = 1000. Deviation = +3. Base = 10. Deviation = – 2.
New excess =3 + 6 = 9 New excess = – 2 – 4 = – 6
(Deviation + Twice deviation) Hence, (Deviation + Twice deviation) Hence,
10033 83
1003+6 : 9 x 3 : 33 8–4 : –6 x–2 : (–2)3
1009 : 027 : 027 4 : 12 : –8

41. 1009027027 4 : 11 : (10–8)
Therefore, 10033 = 1009027027 4 : 11 : 2
(Digit Rule) 412
1
512
Therefore, 83 = 512
Example 5: Find the cube of 93. Example 6: Find the cube of 995.
Base = 100. Deviation = - 7. Base = 1000. Deviation = - 5.
New excess = - 7 - 14 = - 21 Hence, New excess = - 5 – 10 = - 15 Hence,
933 9953
93 - 14 : 147 : –343 995 - 15: 075 : –125
79 : 147 : –343 985 : 075 : –125
79 : 143 : (400–343) 985 : 074 : (1000–125)
79 : 143 : 057 985074875
794357 Therefore, 9853 = 985074875 (Digit Rule)
1 0
804357
Therefore, 933 = 804357
Straight Cubing of 2 digit numbers
To find cube of any number directly we use the formula: (a + b)3 = a3 + 3a2b + 3ab2 + b3
We rewrite this as
a3 a2b ab2 b3
2a2b 2ab2
a3 3a2b 3ab2 b3
The form makes it easy to compute the cube any 2 digit number. The following examples will show
how this could be done.
Method
• Find the values of a3, a2b, ab2, b3 and write them as shown.
• Also double the vales of a2b, ab2 and write them under respective column.
• Compute the cube of the number from the result.
Example 1: Find the cube of 26. Example 2: Find the cube of 84.
Here, a = 2, b = 6. Here, a = 8, b = 4.

42. 8 24 72 216 512 256 128 64
48 144 512 256
8 72 216 216 512 768 384 64
8266 512844
711 686
22 73
17576 592704
Hence, 263 = 17576 Hence, 843 = 592704
Example 3: Find the cube of 33. Example 4: Find the cube of 47.
Here, a = 3, b = 3. Here, a = 4, b = 7.
27 27 27 27 64 112 196 343
54 54 224 392
27 81 81 27 64 336 588 343
27117 64683
882 384
35937 353
Hence, 333 = 35937 103823
Hence, 473 = 103823
Cube Roots Of Perfect Cubes
The cube root of a perfect cube can be found by division method. Before finding the cube root of a
number we must know 3 things.
• Number of digits in the cube root.
If the number has n-digits, the cube root will have n/3 or (n+1)/3 or (n+2)/3 digits.
• First digit of the cube root must be known.
The number is divided into groups of 3 digits from right to left. The nearest cube root of the first
group gives the first cube root digit.
• The devisor must also be known.
Thrice the square of first cube root digit is taken as the devisor.
Example 1: Working (Mental)
Step 1: 13 – 23 = 13 – 8 = 5. Write it as shown.
Find the cube root of 13824.
Step 2: 58 ÷ 12 = 4 R 10. Write it as shown.
No of cube root digits = 5+1/3 = 2.
Step 3: But, 3ab2 = 3x2x42 = 96
Grouping of digits = 13 ' 824
First cube root digit = 2 ( 23 < 13) 102 – 96 = 6 ÷ 12 = 0 R 6

43. Modified Devisor = 3x22 = 12. Step 4: But, b3 = 43 = 64
13 8 2 4 64 – 64 =0 ÷ 12 = 0 R 0
12 5 10 6 Hence, Cube Root of 13824 = 24.
24.00
Example 2:
Find the cube root of 195112.
Working (Mental)
No of cube root digits = 6/3 = 2.
Step 1: 195 – 53 = 195 – 125 = 70. As shown
Grouping of digits = 195 ' 112
Step 2: 701 ÷ 75 = 8 R 101. As shown
First cube root digit = 5 ( 53 < 195)
2
Step 3: But, 3ab2 = 3x5x82 = 960
Modified Devisor = 3x5 = 75.
1101 – 960 = 51 ÷ 75 = 0 R 51
195 1 1 2 Step 4: But, b3 = 83 = 512
75 70 101 51 512 – 512 = 0 ÷ 75 = 0 R 0
58.00 Hence, Cube Root of 195112 = 58.
Divisibility
If we divide one number by another number and get a whole number, we say that the first number is
divisible by the second. This property of division is called divisibility. For example,
25 is divisible by 5.; 12 is divisible by 2, 3, 4, 6.
63 is divisible by 3, 7, 9. etc.
Divisibility Criteria - Here are divisibility Criteria for a first few integers:
Divisor Criteria Examples
2 All even numbers 12; 24; 136
3 Sum of digits is divisible by 3 15; 234
4 Last two digits divisible by 4 1932; 2016
5 Last digit is 0 or 5 15; 210; 305
6 Numbers divisible by 2 and 3 36; 1236
8 Last four digits divisible by 8 451936
9 Sum of digits divisible by 9 27; 171;
Test for divisibility
Suppose we want to find whether a number is divisible by prime numbers like 7, 13, 17, 19 . . . we
have no clue in our modern mathematics. But, Vedic Mathematics gives us a nice method to do it.

44. Any number can be split into prime numbers. A number divisible by itself or 1 is called prime number.
Otherwise, it is a composite number. To find the divisibility of one number by another number, we must try
division by prime numbers. A number divisible by a composite number is also divisible by all its prime
The prime numbers within 100 are given below:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 51, 53, 57, 59, 61, 67, 71, 73, 79, 83, 87, 89, 91
and 97.
Operators
We observe that the prime numbers have 1, 3, 7,9 as end digits. We can convert such numbers either
to end with 1 or 9 using a multiplier.
Converting numbers ding with 9 Converting numbers ending with 1
End digit Multiplier Example End Multiplier Example
digit
1 9 11x9=99 1 1 31x1=31
3 3 13x3=39 3 7 13x7=91
7 7 7x7=49 7 3 17x3=51
9 1 19x1=19 9 9 19x9=171
Now, we have two choices.
Division by numbers ending in 9 and Division by numbers ending in 1.
Therefore, we select two operators – one positive operator and another negative operator.
Positive operator is used with devisors ending in 9. The Vedic Sutra Ekadhika – One more than the
previous, gets it. Here are some positive operators.
Devisor Operator
19 1+1=2
29 2+1=3
39 3+1=4
49 4+1=5
etc etc
Negative operator is used with devisors ending in 1. Just dropping end digit 1 gets it. Here are some
negative operators.
Devisor Operator
11 1
21 2

45. 31 3
41 4
etc etc
Testing Divisibility by a Positive Operator
Testing the divisibility of a number using a positive operator is very simple.
Find the operator first using Ekadhika.
Separate the last digit from the number.
Multiply this digit by the operator.
Add the result to the remaining number.
Apply this rule again and again as necessary.
If the final result is divisible then so is the first number.
Example 1: Is 171 divisible by 19?
Here, Operator = 1 + 1 = 2.
Now, 171 –> 17 + 1x2 = 17 + 2 = 19.
Hence, given number is divisible.
Example 2: Is 348 divisible by 29?
Here, Operator = 2 + 1 = 3.
Now, 348 –>34 + 8x3 = 34 + 24 = 58
58 –> 5 + 8x3 = 5 + 24 = 29.
Hence, given number is divisible.
Example 3: Is 1764 divisible by 49?
Here, Operator = 4 + 1 = 5.
Now, 1764 –> 176 + 4x5 = 176 + 20 = 196
196 –> 19 + 6x5 = 19 + 30 = 49.
Hence, the result.
Example 4: Test the divisibility of 3614 by 13.
Here, devisor ends with 3. Therefore, 13x3=39.
Operator = 3 + 1 = 4
Now, 3614 –> 361 +4x4 = 361 + 16 = 377.
377 –> 37 + 7x4 37 + 28 = 65.
65 –> 6 + 5x4 = 6 + 20 = 26.
26 –> 2 + 6x4 = 2 + 24 = 26. Repeated.
Hence, 3614 is divisible.

46. Example 5: Find if 13294 divisible by 23?
Here, devisor ends with 3. Therefore, 23x3=69.
Operator = 6 + 1 = 7.
Now, 13294 –> 1329 + 4x7 = 1329 + 28 = 1357.
1357 –> 135 + 7x7 = 135 + 49 = 184.
184 –> 18 + 4x7 = 18 + 28 = 46.
46 –> 4 + 6x7 = 4 + 42 =46. Repeated.
Hence, 13294 is divisible by 23.
Example 6: Find if 3171 divisible by 7?
Here, devisor ends with 7. Therefore, 7x7=49.
Operator = 4 + 1 = 5.
Now, 3171 –> 317 + 1x5 = 317 + 5 = 322.
322 –> 322 + 2x5 = 32 + 10 = 42.
42 –> 4 + 2x5 = 4 + 10 = 14. Divisible by 7.
Hence the result.
Testing Divisibility by Negative Operator
Testing the divisibility of a number using a positive operator is very simple.
Find the operator first using Ekadhika.
Separate the last digit from the number.
Multiply this digit by the operator.
Subtract the result from remaining number.
Apply this rule again and again as necessary.
If the final result is divisible then so is the first number.
Example 1: Find if 1331 divisible by 11?
Here, Operator = 1.( End digit is dropped.)
Now, 1331 –> 133 – 1x1 = 133 – 1 = 132.
132 –> 13 – 2x1 = 13 – 2 = 11.
11 –> 1 – 1x1 = 1 – 1 = 0.
Hence, 1331 is divisible by 11.
Example 2: Is 5376 divisible by 21?
Here, Operator = 2.( End digit is dropped.)
Now, 5376 –> 537 – 6x2 = 537 – 12 = 525.
525 –> 52 – 5x2 = 52 – 10 = 42.

47. 42 –> 4 – 2x2 4 – 4 = 0.
Hence, the result.
Example 3: Test the divisibility of 3844 by 31.
Here, Operator = 3.( End digit is dropped.)
Now, 3844 –> 384 – 4x3 = 384 – 12 = 372.
372 –> 37 – 2x3 = 37 –6 = 31.
31 –> 3 – 1x3 = 3 – 3 = 0.
Therefore, 3844 is divisible by 31.
Example 4: Find if 27249 divisible by 31?
Here, Operator = 3.( End digit is dropped.)
Now, 27249 –> 2724 – 9x3 = 2724 – 27 =2697.
2697 –> 269 – 7x3 = 269 – 21 = 248.
248 –> 24 – 8x3 = 24 – 24 = 0.
Hence, the result.
Example 5: Is 4318 divisible by 17?
Here, devisor ends with 17. Therefore, 17x3=51.
Therefore, Operator = 5.
Now, 4318 –> 431 – 8x5 = 431 – 40 = 391.
391 –> 39 – 1x5 = 39 – 5 = 34.
34 –> 3 – 4x5 = 3 – 20 = – 17, divisible by 17.
Hence, the result.
Example 6: Is 3171 divisible by 7?
Here, devisor ends with 7. Therefore, 7x3=21.
Therefore, Operator = 2.
Now, 3171 –> 317 – 1x2 = 317 – 2 = 315.
315 –> 31 – 5x2 = 31 – 10 = 21.
21 –> 2 – 1x2 = 0.
Hence, the result.
Note: Compare this problem with the same one under positive operator above. Both give the same
answer. This shows either of the method could be used. Selection of easier method is our choice.
Suppose we want to test for divisibility by 21 or 63 (21=3x7, 63=3x3x7). First we check for
divisibility by 3 (digit sum divisible by 3). If it does, then try for 7.
Now let us take an example where divisibility is not possible, to show the strength of this test.
Example: Is 3245 divisible by 7?

48. ( i ) We test this by positive operator.
Here, 7 x 7 = 49. Operator = 5.
Now, 3245 –> 324 – 5x5 =324 – 25 = 299.
299 –> 29 – 9x5 = 29 – 45 = – 16.
This is not divisible by 7.
( ii ) We now test this by negative operator.
Again, 7 x 3 = 21. Operator = 2.
3245 –> 324 – 5x2 = 324 – 10 = 314.
314 –> 31 – 4x2 = 31 – 8 = 23.
This is also not divisible by 7.
Therefore, it is clear that we can go for either of the methods.
Recurring Decimals
A decimal with a sequence of digits that repeats itself indefinitely is called recurring decimal.
The Vedic Sutra Ekadhika helps us to convert Vulgar fractions of the type 1/p9, where p = 1, 2, 3, . . .
9, into recurring decimals. The number of decimal places before repetition is the difference of numerator
and denominator. The devisor can be found using Sutra "One more than the previous one".
Conversion of a few vulgar fractions is shown below. The method is simple and does not require actual
division. It is only simple division with small figures.
Convert 1/19 into recurring decimal.
Here, number of decimal places is (19 – 1) = 18.
The denominator is 19, and the previous one is 1.
Hence, Ekadhika = (1 + 1) = 2.
The method of division is as follows:
Now, 1/19 = 1/20 = 0.1/2
Divide 1 by 2, answer is 0, remainder 1
–> 0.10 (Remainder is put before the answer)
Next 10 ÷ 2 is 5 –> .05
Next 5 ÷ 2 is 2 remainder 1 –> 0.10512
Next 12 (remainder 1) ÷ 2 is 6 –> 0.10526
Next 6 ÷ 2 is 3 –> 0.051263
Next 3 ÷ 2 is 1 remainder 1 –> 0. 105126311
Next 11 ÷ 2 is 5 remainder 1 –> 0.10512631115
and so on...

49. Continue like this till the end result got is
1/19 = 0.1051263151718914713168421
Therefore, 1/19 = 0.052631578947368421
Convert 11/19 into recurring decimal.
The denominator = 19. Previous digit = 1. Ekadhika is 1+1=2.
11/19= 11/20 = 1.1/2 = 0.151718914713168421
Therefore, 11/19 = 578947368421.
Convert 1/7 into recurring decimal.
Here, number of decimal places is (7 – 1) = 6.
1/7 = 7/49. The denominator is 49.
Previous digit = 4. Ekadhika is 4+1=5.
7 / 49 = 7/50 = 0.7/5 = 0.21144228357
Therefore, 1/7 = 0.142857.
Convert 9/39 into recurring decimal.
The denominator = 39. Previous digit = 3. Ekadhika is 3+1=4.
9/39= 9/40 = 0.9/4 =0.12330769. Therefore, 9/39 = 0.230769.
Equations
Simple Equations
A simple equation consists of one variable or unknown. The variable may be present on one side or
both sides. To solve such problems the Vedic Sutra “Transpose and apply” (Paravartya Sutra) comes to our
rescue. We can transpose the terms from left to right or from right to left. In such cases, positive(+) sign
becomes negative(–) sign and vice versa. We can also change numerator to denominator or denominator to
numerator. Then multiplication(×) sign becomes division(÷) sign and vice versa.
Paravartya Method
We shall consider the following types of equation under simple equations.
Type 1: ax + b = cx + d. Solution: x = d − b
a−c
Example 1: Solve: 3x + 5 = 5x – 3.
Here, a = 3, b = 5, c = 5, d = –3.
∴ x = d −b = −3−5 = 4
a−c 3−5

50. Example 2: Solve: (2x/3) + 1 = x - 1
--> 2x + 3 = 3x – 3
Here, a = 2, b = 3, c = 3, d = –3.
∴ x=
d − b −3−3
= =6
a−c 2−3
Type 2: (x + a) (x + b) = (x + c) (x + d)
Solution: x = cd − ab . Note: If ab = cd, then x = 0.
(a + b ) − (c + d )
Example 1: Solve: (x + 1) ( x + 2) = ( x – 3) (x – 4)
Here, a = 1, b = 2, c = –3, d = –4.
∴ x= cd − ab ( −3 × −4) − (1 × 2)
= =1
( a + b) − (c + d ) (1 + 2) − ( −3 − 4)
Example 2: Solve: (x + 3) ( x + 2) = ( x – 3) (x – 2)
Here, a = 3, b = 2, c = –3, d = –2.
∴ x= cd − ab (−3 × −2) − (3 × 2)
= =0
(a + b) − (c + d) (3 + 2) − (−3 − 2)
ax + b m md − bn
Type 3: = Solution: x =
cx + d n an − mc
Example: Solve: 2x + 3 = 1
5x + 4 2
Here, a = 2, b = 3, c = 5, d = 4, m = 1, n = 2.
∴ x = md − bn = 1 × 4 − 3 × 2 = − 2 = 2
an − mc 2 × 2 − 1× 5 −1
Type 4: m n − (mb + na )
+ = 0 Solution: x =
x+a x+b (m + n )
Example: Solve: 2 3
+ =0
x+2 x+7
Here, a = 2, b = 7, m = 2, n = 3.
∴ x = − ( mb + na ) = − (2 × 7 + 3 × 2) = − 20 = −4
(m + n ) ( 2 + 3) 5
Sunyam Samya Method

51. Next we shall explore another Vedic Sutra “Common group is zero” (Sunyam Samya Sutra). The group
may be a single variable or an expression and it should be equated to zero. We shall consider the following
Type 1: (i) ax + bx = cx + dx, and (ii) m(x + a) = n(x + a).
Example 1: 2x + 3x = 5x + x.
Here, x is a common factor (group). Therefore, x = 0.
Example 2: 3(x + 2) = 5(x + 2).
Here, (x + 2) is again a common factor. Hence, x + 2 = 0 or x = -2, is the solution.
Type 2: (x + a)(x + b) = (x + c)(x + d). If a×b = c×d then x = 0.
Example: (x + 2)(x + 6) = (x – 3)(x – 4)
Here, 2 × 6 = –3 × –4 = 12. Hence, x = 0.
Type 3: N1 + N 2 = 0 . N1 and N2 are numerators, D1 and D2 are denominators.
D1 D2
If N1 = N2, D1 + D2 = 0, is the solution. From this the value of x can be found.
Example: Solve: 1 1
+ =0
2x + 3 x − 6
Here, D1 + D2 = 2x + 3 + x – 6 = 0 or 3x – 3 = 0. ∴ x = 1, is the solution.
Type 4: N1 N 2 . N1 and N2 are numerators, D1 and D2 are denominators
=
D1 D2
If N1 + N2 = D1 + D2 = k (D1 + D2), where k is a number, then
N1 + N2 = D1 + D2 = 0, is the solution. From these the value of x can be found.
Example 1: Solve: 3x + 4 = 2 x + 11
2x + 11 3x + 4
Here, N1 + N2 = 3x + 4 + 2x + 11 = 5x + 15
D1 + D2 = 2x + 11 + 3x + 4 = 5x + 15
Therefore, N1 + N2 = D1 + D2 = 0 or 5x + 15 = 0. ∴ x = –3 is the solution.
Example 2: Solve:
2x + 3 x + 1
=
4x + 5 2x + 3
Here, N1 + N2 = 2x + 3 + x + 1 = 3x + 4
D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8 = 2(3x + 4)

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