The Essentials of Vedic Mathematics by Rajesh Thakur 260820173117
The Essentials of Vedic Mathematics by Rajesh Thakur 260820173117
The Essentials of Vedic Mathematics by Rajesh Thakur 260820173117
Vedic
Mathematics
Vedic
Mathematics
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Allahabad Bengaluru Chennai
Hyderabad Jaipur Kathmandu
Kolkata Mumbai
The views and opinions expressed in this book are the author’s
own and the facts are as reported by him/her which have been
verified to the extent possible, and the publishers are not in any
way liable for the same.
ISBN: 978-81-291-2374-9
10 9 8 7 6 5 4 3 2 1
Foreword xi-xii
Preface xiii-xiv
What is Vedic Mathematics? xv–xvii
Why is Vedic Mathematics Essential? xix–xxi
Vedic Sutras xxiii–xxiv
Vedic Sub-sutra xxv–xxvi
viii Contents
13. Fourth Root of a Number: Introduction, Vedic 196–202
sutra –Vilokanam, its meaning and application.
14. Simultaneous Equation: Introduction, Vedic 203–211
Method – Paravartya Yojayet, Anurupye Sunyam
anyat, Sankalana- Vyavkalana bhyam, theirts
meanings and applications.
15. Cubic Factorization: Introduction, Vedic method 212–217
– Gunit Samucchaye Samuccaye Gunita, their
meanings and applications.
16. Quadratic Equation: Introduction, different 218–227
Vedic sutras- Vilokanam, Sunyam Samya
samuccaya, Anurupye and Sunyam anyat –
their meanings and applications.
17. Casting Out Nines: Introduction, Fundamental 228–239
Rule of Navasesh, Vedic method of checking
the accuracy of addition, subtraction,
multiplication, division, square, and cube.
18. Trigonometry: Introduction, How to compute 240–249
Pythagorean Triplets, Computing trigonometric
ratio of Sin A, Sin 2A, Sin 3A, Sin A/2 Sum
and Difference of Compound Angles, etc.
19. Questions for Practice 250–260
20. Feedback from Students 261–263
21. Bibliography 265–667
Contents ix
Foreword
Dr J.J. Rawal
President
Indian Planetary Society, Mumbai
xii Foreword
Preface
xiv Preface
What is Vedic Mathematics?
• Sankalana-Vyavakalanabhyam (ladyu&O;odyukH;ke~) – By
addition and subtraction.
Addition
Introduction
iwj.kk iwj.kkH;ke~
(Puranapuranabhyam)
ladyu O;odyukH;ke~
(Sankalan Vyavkalanabhyam)
,dkf/kdsu iwosZ.k
(Ekadhikena Purvena)
Meaning of Vedic Sutra:
Rule:
Solution:
26 + 59 + 394 + 66 + 11 + 14
Step 2: Rearrange the pair and add as per the pairing done above.
= (40 + 460) + 70
= 500 + 70
= 570
Addition 3
Example 2: Add: 456 + 361 + 244 + 119 + 11 = ?
Solution:
Step 1: Here the unit digit of 456 and 244 makes a rounded result.
The same complement rule applies for 361 and 119. Arrange
these pairs.
Example: Add: 36 + 5 + 23 + 2 + 14
36 + (5 + 23 + 2) + 14
= 50 + 30
= 80
a) 24 = 20 + 4
b) 39 = 40 – 1
c) 543 = 550 – 7 = 500 + 40 +3
d) 793 = 700+ 90 +3 = 800 – 7
The above splitting is done by choice and you are the best judge
to decide what splitting will work for you. The splitting may be
breaking a number as the sum of two/more numbers or difference
of two numbers.
Example: Add 74 + 69
Solution: 74 + 69
= 70 + 4 + 70 – 1
= (70 + 70) + (4 – 1)
= 140 + 3
= 143
Addition 5
Example: Add 596 + 498 + 345 + 765
Solution:
4 8 6
– – –
6 5 4
–
+ 9 8 7
Running total 9 1 7
Explanation:
Running total 9 1 7
Dots 1 2 1
Sum 2 1 2 7
Solution:
6 4 8 9
– – – –
5 6 4 2
+
3 2 4 1
4 2 6 2
Explanation:
Addition 7
Now count the number of dots in each column and place it down
to the number next to the unit place as done in the first example
and add the two to get the final result.
Running total 4 2 6 2
Dots 1 1 1 1
Sum 1 5 3 7 2
Subtraction
Introduction
Example:
9 2 7 8 Minuend
– 3 0 4 1 Subtrahend
6 2 3 7 Remainder
Solution: 7 13
3 12
2 16
8 4 3 6
– 4 7 6 8
3 6 6 8
The general look of a copy will be the same. The number of steps
involved in this operation obviously makes it a time consuming
process. The operation can be simplified by either of the following
techniques:
Decomposition Method:
Solution:
Example:
Rule:
Subtraction 11
Example: Subtract 10000 – 462
Here, the extreme left digit i.e. 4, will get diminished by 1, and
all the zeros thereafter will change into 9, except the last one.
The last zero on the extreme right will be changed to 10.
Number 0 1 2 3 4 5 6 7 8 9
Complement 10 9 8 7 6 5 4 3 2 1
Rule:
• When the digit at minuend (upper digit) > subtrahend
digit (lower digit), normal subtraction is done.
• In case the upper digit < lower digit, we take the
complement of the difference as shown in the complement
Solution: From 10
8 5 4
– 5 6 9
5
Step 1: Since 4< 9, we take the complement of the difference of
the digits. This complement will be taken from 10. The difference
of 9 and 4 is 5. From the complement table it is evident that the
complement of 5 is 5, so write 5 at the unit place.
From 9
8 5 4
– 5 6 9
8 5
Step 3: The difference of the digits at the hundred’s place can
easily be carried out as 8 > 5. So we don’t need to take the
complement for the third column. As we are now out of the
complement, subtract 1 more in this column. Hence, instead of
subtracting 8 − 5, we subtract 8 − 5 − 1 = 2.
Subtraction 13
8 5 4
– 5 6 9
2 8 5
Example: 8745 – 4599 = ?
Solution: From 10
8 7 4 5
– 4 5 9 9
6
Step 1: Since, 5 < 9, take the complement of the difference.
9− 5 = 4. The complement of 4 is 6.
From 9
8 7 4 5
– 4 5 9 9
4 6
Step 3: Now, 7 – 5 is easy to carry out and since we applied the
complement in the next digit so subtract 1 more in this column.
Hence, the digit at the hundred’s place will be 7 − 5−1 = 1
8 7 4 5
− 4 5 9 9
1 4 6
Step 4: Here, 8 – 4 = 4
8 7 4 5
− 4 5 9 9
4 1 4 6
From 10
8 7 6 5 2
– 4 0 2 6 9
8 3
Step 3: In the third column, we are out of complement as 6 > 2,
so we subtract 1 more in that column and now the digit at the
hundred’s place = 6 – 2 – 1 = 3
8 7 6 5 2
– 4 0 2 6 9
3 8 3
Step 4: Since 7 > 0, write 7 – 0 = 7 in the fourth column.
Moreover, 8 > 4, so write 8 – 4 = 4 in the fifth column.
8 7 6 5 2
– 4 0 2 6 9
4 7 3 8 3
Example: 459876 – 389924 = ?
Solution:
4 5 9 8 7 6
– 3 8 9 9 2 4
5 2
Subtraction 15
Step 1: 6 > 4 and 7 > 2, so normal subtraction will take place
here. Hence,
Unit digit = 6 – 4 = 2 and
Ten’s digit = 7 – 2 = 5
Step 2: Here, 8 < 9, so the Vedic sutra “All from 9 and last
from 10” will be applied here. Now take the complement of the
difference of the two numbers. The complement will be taken
from 10.
From 10
4 5 9 8 7 6
– 3 8 9 9 2 4
9 5 2
Step 3: Here, in the fourth column, 9 – 9 = 0; subtract the
difference from 9.
4 5 9 8 7 6
– 3 8 9 9 2 4
9 9 5 2
4 5 9 8 7 6
– 3 8 9 9 2 4
6 9 9 5 2
i.e. 4 – 3 – 1 = 0
• Subtract the given digit from 10 and place a bar over it.
• Apply Ekanyunena Purvena and add 1 to the digit on
the left.
–
Example: 1. 438, on conversion to mishrank, will become 4 4 2.
––
2. 4213883, on conversion, will become 42141 23.
Subtraction 17
Explanation: In the first example, the mishrank digit 2 will be
subtracted from 10 (10 – 2 = 8)and its previous digit, 4, will
get reduced by 1.
5 6 7
– – –
+ 3 8 9
– –
2 2 2
– –
Here, 7 + 9 = 7 – 9 = 2
– –
6 + 8 = 6 – 8 = 2
– –
5 + 3 = 5 – 3 = 2
– –
Now, convert 2 2 2 = 1 7 8
Apply, fuf[kya uor’pjea n’kr% (All from 9 and last from 10).
Here the mishrank digit 2 at the unit place will get subtracted
–
from 10 i.e. 10 – 2 = 8 and the 2 on the immediate left will
be subtracted from 9 i.e. 9 – 2 = 7. The digit at the hundred’s
place will get reduced by 1 making the 2 at hundred’s place 1.
4 5 6 8 5 6 8
– – – – – – –
+ 3 4 7 8 9 8 9
– – – –
Mishrank digit 1 1 1 0 4 2 1
Original digit 1 0 8 9 5 7 9
Subtraction 19
Digit Separator Method
The digit separator method is effective and works better for all
types of subtraction problems. Before I take a few examples, let
me introduce the modus operandi of this method.
Rule:
Solution:
1 –1 9 –1 6
= 086
Example: Subtract 5428 – 3765
Solution:
• Draw as many vertical digit separator lines as the number
of digits.
1st 2nd 3rd 4rd
5 4 2 8
– 3 7 6 5
Subtraction 21
• Whenever the minuend at the top is less than the
subtrahend at the bottom, put (–1) in the remainder
column of each digit separator.
Solution:
–1 –1 –1 –1
5 -5 -2 -5 -1 1 ------ Remainder
Subtraction 23
• Write the complement of each digit with negative sign.
Here the complement of -5, -2, -5, and -1 in the 2nd, 3rd,
4th and 5th column are respectively 5, 8, 5 and 9.
–1 –1 –1 –1
5 5 8 5 9 1 ------ Remainder
= 5 –1 5 –1 8 –1 5 –1 9 1
= 5-1 / 5-1 / 8-1 / 5-1 / 9 / 1
= 4 4 7 4 9 1
Multiplication
Introduction
4 2 3 5 Multiplicand
X 2 5 Multiplier
2 1 1 7 5
+ 8 4 7 0 X
1 0 5 8 7 5 Product
This method is good, but there is no room for experimentation.
We are bound to do the same in all types of multiplication. The
process involves multiple stages and is error prone. A simple
mistake is sufficient to make the calculation faulty. This is not
all; a multiplication of 5 x 5 digits is bound to take at least
3-4 minutes. The best advantage of the Vedic Method over the
present day calculation is that Vedic mathematics has ample fruit
in its basket. You have different choices and you can choose
the best possible method in the best situation. There are many
special sutras that help you to find the answer of a special type
of multiplication even in seconds and the Urdhva Tiryagbhyam
method helps you to encounter all types of multiplication. So
friends, jump into the ocean of Vedic sutras and gather the pearls
of your own choice
Vedic Sutras for Multiplication
vUR;;ksnZ’kds·fi (Antyayordashakepi)
vuw:I;s.k (Anurupyen)
vUR;;ks’krds·fi (Antyayoshatakepi)
Multiplication 27
unit digits are the same. This is also not a Vedic Sutra
and is not taken from the Original Vedic Text.
8. m/oZfr;ZxH;ke~ (Urdhava Tiryagbhyam): It is a general
formulae applicable to all cases of multiplication. It is
a process of vertical and cross-wise multiplication. This
method has been further simplified and dealt with Dot
and Cross method in this book. A better understanding
of this formula will also help you in multiplying two
numbers with the other formulae mentioned above.
Vedic Multiplication
vUR;;ksnZ’kds·fi (Antyayordashakepi)
Rule:
Example 1: Multiply 24 by 26
Solution:
2 4
x 2 6 } Here 4 + 6 = 10
Hence, 24 x 26 = 624
Example 2: Multiply 62 by 68
Example 3: Multiply 93 by 97
This sutra works better when both the multiplicand and multiplier
are very close to the base. The base should be in the form of
10n, where n is a natural number.
Rule:
Multiplication 29
a) The left hand part will be obtained by cross operation
of two numbers written diagonally.
b) The right side of the answer will be obtained by
multiplying the deviations.
• The number of digits in the right hand part will be in
accordance to the number of zeros in the base number.
In simple words, if the base is 100, the right hand part
will have two digits and if the base is 1000, the right
hand part will have three digits.
• In case there is lesser number of digits in the right side,
accommodate as many zeros before the right hand part
so that the total number of digits in that part is equal
to the number of zeros in the base.
• Here is the table that will guide you in deciding the
number of digits to be placed on the right hand side.
Example 4: Multiply 8 by 7
Solution:
Example 5: Multiply 95 by 91
Solution:
95
x 91
Multiplication 31
95 – 5
x 91 – 9
d) Write the left hand digit by cross operation of any of
the two diagonals. Here 95 – 9 = 86 or 91 – 5 = 86 is
written in the left hand part.
95 – 5
x 91 – 9
86 /
e) The right hand digit will be the multiplication of the
deviation.
95 – 5
x 91 – 9
86 /45
Example 6: Multiply 15 by 11
Solution:
Solution:
Example 8: Multiply 12 by 8
Multiplication 33
Solution:
Solution:
Multiplication 35
Case 4: Adjustment of right side digit of the product.
Sub case (a): When the number of digits on the right hand side
is more than the permissible limit.
Solution:
16 + 6
x 15 + 5
21 / 30
+
= 240
Solution:
Multiplication 37
13 + 3
18 + 8
21 / 24
Since, the base is 10, the number placed at the right side should
be of one digit, so transfer the extreme left digit of RHS to the
LHS and add them.
13 + 3
x 18 + 8
21 / 24
+
= 234
Sub case (b): When the number of digits on the right hand side
is less than the permissible limit.
Solution:
Since, the base is 100, the number placed at the right side should
consist of two digits. But there is a single digit in the RHS. In
such a case, we place the zero to the left in RHS so that the
total number of digits in RHS is equal to the permissible number
of digits. See Table 1 for better understanding.
96 − 4
98 − 2
94 / 08
Solution:
Multiplication 39
e) The right hand digit will be the multiplication of the
deviation.
989 − 11
x 995 − 5
984 / 55
Since the base is 1000, the number placed at the right side should
consist of three digits. So in order to meet the requirement of
permissible digit in RHS, we place the zero to the left in the RHS
Till now, we have seen examples in which both the numbers were
closer to the base. Now let us consider a case where the two
numbers are nearer to a different base. Hey, are you worried?
Don’t panic, the problem will be solved in a similar fashion with
a slight change in the LHS.
Deviation Base
107 +
7 (100)
x
1008 +
8 (1000)
Solution:
Multiplication 41
Deviation of 48 from the working base = 48 – 50 = – 2
Deviation of 42 from the working base = 42 – 50 = – 8
48 – 2
x 42 – 8
The working procedure is almost the same except for the fact
that the left hand side figure will be divided by 2 as our working
base is half of the theoretical base. As far as the right hand
figure is concerned, it will remain unaffected.
48 – 2
x 42 – 8
40 | 16
= ½ x 40 | 16
= 2016
Solution:
494 – 6
x 488 – 12
482 | 72
494 – 6
x 488 – 12
482 | 072
42 The Essentials of Vedic Mathematics
Now, divide the LHS by 2
494 – 6
x 488 – 12
2 ) 482 | 072
= 241 / 072
Solution:
Step 1: Write the deviation (–2) and (–1) against the number 48
and 49 taken from its working base.
48 – 50 = –2 and 49 – 50 = – 1.
48 – 2
x 49 – 1
48 – 2
x 49 – 1
47 | 02
Multiplication 43
48 – 2
x 49 – 1
2 ) 47 | 02
= 23 ½ | 02
= 23 | 50+ 2
= 2352
Solution:
251 + 2
x 252 + 1
52 + 2
x 48 – 2
52 + 2
x 48 – 2
50 | –04
52 +2
x 48 –2
) 50 | –04
2
= 25 | –04
24 | 96
Multiplication 45
Example 20: Multiply 512 by 494
512 + 12
x 494 – 6
512 + 12
x 494 – 6
506 | –072
512 +12
x 494 −6
2 ) 506 | − 072
= 253 | − 072
= 252 | 1000 – 72
= 252 | 928
This sutra works equally well when the base is the multiple of
10 (20, 30, 40, 60, 70...), multiple of 100 (200, 300, 400,) etc.
The modus operandi of such multiplication is slightly different.
Instead of dividing the LHS figure by 2, 4, 6, or 8, multiply the
LHS by the sub-base number. If you are taking the base = 20
= 2 x 10; then base = 10 and sub- base = 2.
Solution:
Solution:
252 – 4 8
x 299 – 0 1 (Working base: 100 X 3 =300)
251 | 48
x 3
= 753 | 48
= 75348
Solution:
687 – 013
x 695 – 005 (Working base: 100 X 7 =700)
682 | 65
x 7
= 4774 | 65
= 477465
Solution:
889 – 011
x 895 – 005
(Working Base = 900= 9 x 100)
884 | 055
x 9
7956 | 055
= 7956055
Multiplication 47
Ekanyuena Purvena
569876943 x 999999999 = ?
Rule:
Solution:
LHS = 89654876 – 1 = 89654875
RHS = 99999999 – 89654875 = 10345124
Hence, 89654876 x 99999999 = 8965487510345124
Solution:
LHS = 83465087629 – 1 = 83465087628
RHS = 99999999999 – 83465087628 = 16534912371
Hence, 83465087629 x 99999999999 =
8346508762816534912371
Solution:
LHS = 45682 – 1 = 45681
Multiplication 49
RHS = 99999 – 45681 = 54318
Hence, 45682 x 99999 = 4568154318
Solution:
LHS = 456 – 1 = 455
RHS = 9999 – 455 = 9544
Hence, 456 x 9999 = 4559544
Solution:
LHS = 56892 – 1 = 56891
RHS = 9999999 – 56891 = 9943108
Hence, 56892 x 9999999 = 568919943108
Solution:
LHS = 13324 – 1 = 13323
RHS = 99999999 – 13323 = 99986676
Hence, 13324 x 99999999 = 1332399986676
This case is a little bit different from the last two cases discussed
so far under Ekanyena Purvena. In order to get the result you
have to –
50 The Essentials of Vedic Mathematics
a) Add as many zero as the numbers of 9s to the
multiplicand.
b) Subtract the original multiplicand from the figure
obtained in Step 1.
Solution:
The multiplicand 1564 has 4 digits, whereas there are two 9s
in the multiplier.
1 5 6 4
X 9 9
1) Since there are two 9s, put two zeros at the end of 1564,
making it 156400.
2) Subtract (1564 original multiplicand) from 156400
i.e. 1 5 6 4 0 0
–1 5 6 4
1 5 4 8 3 6
Hence, 1564 x 99 = 154836
Solution:
1) Since there are four 9s, put four zeros at the end of
783459, making it 7834590000
2) Subtract the original number 783459 from 7834590000.
7 8 3 4 5 9 0 0 0 0
– 7 8 3 4 5 9
7 8 3 3 8 0 6 5 4 1
Hence, 783459 x 9999 = 7833806541
Solution:
1) Since there are three 9s, put three zeros at the end of
45678, making it 45678000.
2) Subtract the original number from 45678000.
Multiplication 51
4 5 6 7 8 0 0 0
– 4 5 6 7 7 8
45632322
Hence, 45678 x 999 = 45632322
This can be done by the Eknyuenane Purvena method effortlessly
after a little practice. Let’s see how it works. In this method:
a) Subtract 1 from the original number and place it in LHS.
b) Write as many digits from right to left equal to the number
of 9s that of multiplicand in RHS. Suppose you have to
multiply 147 by 99 so RHS part of answer should contain
2 digits 47 taken from right, equal to the number of 9
in the multiplier.
c) The remaining digits in the original number, after
removing the digits from the right to left, placed in the
RHS should be subtracted in the LHS.
d) Write the complement of the digits placed in the RHS
by applying the Nikhilam sutra.
Example: Multiply 147 by 99
Solution: Since there are two 9s in the multiplier, two digits
from right to left of the multiplicand will be placed in the RHS.
In the LHS, subtract 1 from the original number.
LHS = 147 – 1 = 146
RHS = Complement of 47
Now subtract the remaining digits i.e. 1 from LHS and write the
complement of 47 in RHS.
LHS = 146 – 1 = 145
RHS = 100 – 47 = 53
Hence, 147 x 99 = 14553
Example: Multiply 259648 by 9999.
Solution: Since there are four 9s in the multiplier, four digits
from right to left of multiplicand i.e. 9648 will be placed in the
RHS. In LHS, subtract 1 from the original number.
Now subtract the remaining digits i.e. 25 from LHS and write
the complement of 9648 in RHS.
Now subtract the remaining digits i.e. 528 from LHS and write
the complement of 76 in RHS.
vUR;;ks’krds·fi (Antyayoshatakepi):
Rule:
• This sutra is applicable when the sum of the last two
digits (unit and ten’s) in the multiplicand and multiplier
is 100 and the rest of the digits are the same.
• Multiply the last two digits and write the product in RHS
• Multiply the remaining digits that are the same, with
the next digit i.e digit + 1 and write the product in
the LHS. In simple word, if the digit at the hundred’s
place is 7, multiply it with the next digit 8 and write
the product in LHS.
Multiplication 53
Example 35: Multiply 782 by 718
Solution: Here the sum of the last two digit of the multiplicand
and multiplier is 100 (82 + 18 = 100) and the digit at the
hundred’s place in both multiplicand and multiplier are the same.
7 82
x 7 18
LHS = 7 x 8 = 56
RHS = 82 x 18 = 1476
Hence, 782 x 718 = 561476
(NB:– Multiplication of right hand digits 82 x 18 can easily be
done by the Vedic sutra, Urdhva Tiryagbhyam.
Example 36: Multiply 978 by 922
Solution: Here the sum of the last two digits of the multiplicand
and multiplier is 100 (78 + 22 = 100) and the digit at the
hundred’s place in both multiplicand and multiplier are same.
9 78
x 9 22
LHS = 9 x 10 = 90
RHS = 78 x 22 = 1716
Hence, 978 x 922 = 901716
(NB:– Multiplication of right hand digits 78 x 22 can easily be
done by the Vedic sutra, Urdhva Tiryagbhyam.
LHS = 8 x 9 = 72
RHS = 11 x 89 = 979
Solution: Here the sum of the last two digit of the multiplicand
and multiplier is 100 (91 + 09 = 100) and the digit at the
hundred’s place in both multiplicand and multiplier are the same.
8 91
x 8 09
LHS = 8 x 9 = 72
RHS = 91 x 9 = 819
The RHS should consist of 4 digits, so put one zero before
819, making it 0819
Hence, 891 x 809 = 720819
This sutra is not taken from the Vedic Mathematics book written
by Jagad guru Bharti Krishna Tirtha ji Maharaj and the original
source of this sutra is not known to me. While writing this
book, I had gone through more than 50 books and hundreds
of websites and I noticed that this sutra serves the purpose of
faster calculation and I could not stop myself from introducing
this sutra for my avid readers.
This sutra is applicable when the sum of digits placed at the
tens’ place in the multiplicand and multiplier is 10 and the unit
digit of both multiplicand and multiplier is the same.
Multiplication 55
Rule:
6 8
x 4 8
9 7
x 1 7
Like the previous sutra, this sutra is not taken from the original
Vedic Mathematics book available in the market. The origin of this
sutra is not known to me, but again it is one of the interesting
sutras that will make our calculations easier and that is my
purpose of writing this book.
This sutra is applicable when the sum of the left two digits,
other than the unit, is a multiple of ten and the unit digits
are the same.
Multiplication 57
Rule:
20 = 2 x 10
Sub-base Base
13 6
x 7 6
45 9
x 15 9
Multiplication 59
means “Vertically and Cross-wise”. Once you get mastery over
this method you can multiply a 5 digit multiplication of any
number in 15 seconds. Initially this method will seem tough to
work out, but believe me; I have seen the change in calculating
power of students after learning this sutra. The best feature of
this method is:
a) With practice, you can multiply any digits of number and
obtain the result in one line.
b) You are free to multiply from both ends due to the
flexibility of Vedic Sutra..
A further simplification can be understood by the Dot and
Stick Method..
(1) (2) (1)
(1) (2) (3) (2) (1)
(1) (2) (3) (4) (3) (2) (1)
(1) (2) (3) (4) (5) (4) (3) (2) (1)
11 X 11 = 121
111 X 111 = 12321
1111 X 1111 = 1234321
11111 X 11111 = 123454321
--------------------------------------
--------------------------------------
Looking upon the multiplication of 11 x 11, 111 x 111 …and
the dot multiplication shown pictorially, you may find a very
special clue. Did you notice the clue?
If you are multiplying a two-digit number, see the product
of 11 X 11 i.e. 121 and now notice the number written below
the dots of 2 digit multiplication.
Did you find any similarity?
Yes, the same 121 is written there. This process can be
summed up in three points.
• First multiply vertically the number placed at unit digits.
• Find the sum of the cross product of the two figures.
• Finally, multiply vertically the remaining digits.
Similarly, if you are multiplying a three-digit number, look
at the product of 111 X111 = 12321, this is the same number
Multiplication 61
written at the bottom of the dot multiplication.
Let us take a few examples to clarify the dot and cross
product technique.
Solution:
7 6 7 6 7 6
4 2 4 2 4 2
28 14 + 24 12
Arranging the number and adding them from right to left, taking
only one digit at a time, we get the final result.
= 28 | 38 | 12
+ +
= 3192
Example 47: 92 X 18 = ?
9 2 9 2 9 2
1 8 1 8 1 8
9 72 + 2 16
Arranging the number and adding them from right to left, taking
only one digit at a time, we get the final result.
= 9 | 74 | 16
+ +
= 1656
5 6 5 6 5 6
3 4 3 4 3 4
15 20 + 18 24
Arranging the number and adding them from right to left, taking
only one digit at a time, we get the final result.
= 15 | 38 | 24
+ +
= 1904
Once the concept is clear, the whole process can be done mentally
in one line.
Multiplication 63
B. Multiplication of three- digit numbers
Solution:
5 6 6 5 6 6 5 6 6 5 6 6 5 6 6
2 8 1 2 8 1 2 8 1 2 8 1 2 8 1
10 40 +12 5 + 48 +12 6 + 48 6
10 | 5 2 | 6 5 | 5 4 | 6
+ + +
= 159046
8 9 8 8 9 8 8 9 8 8 9 8 8 9 8
48 54+40 48+72+45 40+81 72
= 4 8 | 9 4 | 1 6 5 | 1 2 1 | 7 2
+
= 48 | 94 | 165 | 121 + 7 | 2
= 48 | 94 | 165 | 12 8 | 2
+
The above steps are written for the sake of readers to understand
the concept more vividly, though it is unnecessary to write all
these steps. Keep only one thing in your mind that after one
stage of operation is over, keep a single digit in each block and
move the remaining to the next. Readers are expected to do these
operations involving the addition of two or three number mentally.
In the very beginning, I had mentioned the fact that once the
multiplication of every digit is completed, you have to add the
digits from right to left, taking only one digit in each separator.
Here the extra digits, leaving the unit digit in each separator, have
been written in sub-script so that further addition becomes easy.
Multiplication 65
Example 53: Multiply 467 by 598.
Solution:
4 6 7
X 5 9 8
2
0 | 66 | 121 | 1 11 | 56
0 6 1 1 6
+ 2 6 12 11 5
2 7 9 2 6 6
5 2 6 5 2 6 5 2 6 5 2 6 5 2 6
0 4 3 0 4 3 0 4 3 0 4 3 0 4 3
0 20+0 15+8+8 6+24 18
0 | 20 | 23 | 30 | 18
= 22618
1 2 3 5 1 2 3 5 1 2 3 5 1 2 3 5 1 2 3 5 1 2 3 5 1 2 3 5
6 14 25 46 27 27 20
6 | 1 4 | 2 5 | 4 6 | 2 7 | 2 7 | 2 0
+ + + + + +
= 7698990
Solution:
8 9 8 9
x 8 8 9 2
64 | 1 3 6 | 2 0 8 | 2 3 3 | 1 6 2 | 9 7 | 1 8
+ + + + + +
= 6 4 6 8 3 2 7 8
+ 13 20 23 16 9 1
7 9 9 3 0 1 8 8
Solution:
2 1 3 4
x 3 2 6 1
6 | 7 | 23 | 26 | 27 | 4
= 6958974
Multiplication 67
4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2
0 2 3 5 0 2 3 5 0 2 3 5 0 2 3 5 0 2 3 5 0 2 3 5 0 2 3 5
6 8 18 45 43 46 10
= 0 | 8 | 18 | 45 | 43 | 46 | 10
+ + + + +
= 1029770
4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2 4 3 8 2
0 0 3 5 0 0 3 5 0 0 3 5 0 0 3 5 0 0 3 5 0 0 3 5 0 0 3 5
0 0 12 29 39 46 10
= 0 | 0 | 12 | 29 | 39 | 46 | 10
+ + + + +
=153370
Multiplication of 5 digits
Solution:
3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7 3 4 5 6 7
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
= 3 | 10 | 22 | 40 | 65 | 72 | 70 | 58 | 35
+ + + + + + + +
= 426729615
Solution:
Step 1: Write the deviation of each number from its base. Place
it against each number. For the above multiplication, the working
base = 10, and their deviations from the base are 2, 3 and 5
respectively.
Number Deviation from the Base
12 + 2
13 + 3
15 + 5
Step 2: Add all the deviations to the working base.
10 + 2 + 3+ 5 = 20
Step 3: Make all possible permutations of the deviation, taking two
at a time and add them. For example, the possible permutation
of abc is ab, bc and ac.
2 x 3 + 2 x 5 + 3 x 5 = 31
Step 4: Multiply the deviations
2 x 3 x 5 = 30
Step 5: Arrange the result obtained in above steps as shown here.
20 | 31 | 30
Multiplication 69
Adding the result from right to left in the direction of the arrows
we can find the result.
20 | 31 | 30
+ +
= 2340
Step 1: Find the deviation of each number from its base. Write
it against each number.
Number Deviation
105 + 5
104 + 4
109 + 9
Step 2: Find the sum of base and deviations.
Base + deviations = 100 + 5 + 4 + 9 = 118
Step 3: Multiply the deviations in pairs of two and sum up the
results so obtained
5 x 4 + 4 x 9 + 5 x 9 = 101
Step 4: Multiply the deviations.
5 x 4 x 9 = 180
Step 5: Arrange the result of all the above steps in a vertical
separator and add them up, from right to left as done previously.
But never forget to keep two digits in each separator as constant
because the base taken here is 100.
118 | 101 | 180
+ +
= 1190280
Step 1: Find the deviation of each number from its base. Write
it against each number.
Solution:
Step 1: Write the deviation of each number from its base. Place
it against each number.
Multiplication 71
Number Deviation
54 + 4
56 + 6
51 + 1
Step 2: Find the sum of base and deviations.
50 + 4 + 6 + 1 = 61
Step 3: Multiply the deviations in pairs of two and sum up the
results so obtained
4 x 6 + 4 x 1 + 6 x 1 = 34
Step 4: Multiply the deviations
4 x 6 x 1 = 24
Arrange the result obtained above in a vertical separator.
61 | 34 | 24
Make three columns and write the result accordingly, as shown here.
Step 1: Find the deviation of each number from its base. Write
it against each number.
Number Deviation
54 +4
48 –2
61 +11
Multiplication 73
Suppose you have to multiply a x b x c x d if the
deviation of these numbers from its base is d1 d2 d3 and
d4, then we will write–
a d1
b d2
c d3
d d4
Solution: Here all the numbers are near the base 100 and their
deviations are respectively 2, 3, 4 and 5.
102 + 2
103 + 3
104 + 4
105 + 5
= 114 | 72 | 55 | 20 = 114725520
Since the base is 100, each part will contain a maximum of two
digits and the excess digits will be transferred to the next part.
Solution: Here all numbers are near the base 100 and their
deviations are respectively 2, 3, 4 and 5.
995 −5
996 −4
997 −3
998 −2
Multiplication 75
• 4th part = Product of all deviations
= − 2 x − 3 x− 4 x− 5= 120
Here, the base is taken as 1000, so each part will consist of three
digits. In the 2nd part, we have a two-digit number, so put a zero
in front of it, making it 071. In the 3rd part, we have a negative
number, so subtract 1 from the 2nd part and the negative number
of part 3 from the base 1000. Our answer will now look like–
Solution:
506 + 6
507 + 7
508 + 8
509 + 9
Since the base = 100, each part will contain maximum two digits
and the rest will be transferred to the next part. The excess digit
is written in subscript.
506 x 507 x 508 x 509 = 66250 | 8375 | 8250 | 3024
= 66334578024
Multiplication 77
4
Multiplication through
Observation
Introduction
Mental Multiplication
A: Multiplication by 11
Multiplication of any number with 11 can be done orally in a
single line. Once the technique for multiplication of a number
with 11 is mastered, it can be further extended for a number
such as 22, 33, 44 etc by simply splitting the multiplicand as
11 x 2, 11 x 3 or 11 x 4. In mensuration, you need to calculate
the volume and surface area of three-dimensional objects such
as cylinder, sphere, cone, pyramid, frustum etc and there you
need to multiply the number by 22 (π =22/7). This method will
help you immensely here.
Rule:
• Place the number to be multiplied by 11 in a bracket
and put zeros on either side.
• Start adding the two numbers at a time from right to
left. If the sum of two numbers in any case exceeds 10,
the digit at the tenth place shall be carried over to the
next sum, as is usually done in simple addition.
Solution:
0 3 2 5 1 0
Solution:
0 4 8 7 6 2 5 4 0
= 4 | 12 | 15 | 13 | 8 | 7 | 9 | 4
= 4 | 12 | 16 | 3 | 8 | 7 | 9 | 4
= 4 | 13 | 6 | 3 | 8 | 7 | 9 | 4
= 5 3 6 3 8 7 9 4
0 3 8 4 0
= 3 | 11 | 12 | 4
= 3 | 12 | 2 | 4
= 4 2 2 4
B: Multiplication by 111
0 3 4 0 0
0
Keep adding three digits from the right at a time as shown above.
= 0 + 0 + 3 | 0 + 3 + 4 | 3 + 4 +0 | 4 + 0 + 0
= 3774
0 0 4 9 7 0 0
Keep adding from right to left, taking the sum of three digits
at a time.
0
0 4 9 7 0 0
= 4 | 13 | 20 | 16 | 7
= 4 | 13 | 21 | 6 | 7
= 4 | 15 | 1 | 6 | 7
= 5 5 1 6 7
0 0 2 1 7 2 0 0 0
0
Step 1: 0 0 0 2 1 7 2 0 0 0
0 + 0 + 0 + 2 = 2
Step 2: 0 0 0 2 1 7 2 0 0 0
0 + 0 + 2 + 7 = 9
Step 3: 0 0 0 2 1 7 2 0 0 0
0 + 2 + 7 + 1 = 10
Step 4 : 0 0 0 2 1 7 2 0 0 0
2 + 7 + 1 + 2 = 12
Step 5: 0 0 0 2 1 7 2 0 0 0
0 + 2 + 1 + 7 = 10
Step 6: 0 0 0 2 1 7 2 0 0 0
0 + 0 + 2 + 1 = 3
Step 7: 0 0 0 2 1 7 2 0 0 0
0 + 0 + 0 + 2 = 2
= 2 | 3 | 10 | 12 | 10 | 9 | 2
= 2| 3 | 10 | 13 | 0 | 9 | 2
= 2 | 3 | 11 | 13| 0 | 9 | 2
= 2 | 4 | 1 | 3| 0 | 9 | 2
Hence, 2172 x 1111 = 2413092
Example 1: Multiply 16 by 25
Example 2: Multiply 98 by 25
E: Multiplication by 125
Rule: Put three zeros to the right of the multiplicand and divide
it by 8.
Rule: Put 4 zeros (0000) to the right of the number and divide
it by 16.
Example 1: Multiply 42 by 5
Solution: Place 1 zero after 42, making it 420
Example 1: Multiply 47 by 50
I: When the sum of the unit’s place digit is 10 and the rest
of the digits are the same
Rule:
4 6
X 4 4
4 6
x 4 4
/ 24
(4+1)= 5 6
x 4 4
20/ 24
Hence, 46 x 44 = 2024
11 3
x 11 7
/ 21
(11+1) = 12 3
x 11 7
132/ 21
16 8
x 16 2
Multiply the encircled digit and write it to the right side.
16 8
x 16 2
/16
Increase the multiplicand by 1 and then multiply it with the
original multiplier.
(16+1) = 17 8
x 16 2
272/ 16
H T
O
(1 + 1)=
2 0
3
X 1 9
7
2 /0291
Solution:
Multiply the number (unit and ten’s place digit) of the multiplicand
and the multiplier whose sum is 100.
H T
O
4 2
5
X 4 7 5
/1875
(Here the sum of the encircled number is 100 (25 +75 =100).
For multiplication of 25 x 75, see Rule D.
Solution: 42
x 51
• Place 42 at the right side
42
x 51
/42
• Place half of the multiplicand to the left, i.e. 42/2 = 21
42
x 51
21 /42
Hence, 42 x 51 = 2142
Solution: 124
x 51
• Place 124 at the right side
124
x 51
/124
• Place half of the multiplicand to the left, i.e. 124/2 = 62.
The excess digit from the right side will get transferred
to the left side.
124
x 51
62/124
= 63/24
Hence, 124 x 51 = 6324
Example: Multiply 41 by 51
Solution: 41
x 51
Example: Multiply 23 by 51
Solution: 23
x 51
• Place 23 at the right side
23
x 51
/23
• Place half of the multiplicand to the left, i.e. 23/2 = 11½.
23
x 51
11½/23
= 11 / 50 + 23= 73
Since, the left most part is fractional, add 50 to the right side
and remove the fractional part from the left.
Hence, 23 x 51 = 1173
Multiplication in Algebra
Introduction
The above arrows will help you to recall what you have done
previously and ease the understanding process even in Algebraic
Multiplication.
Case 1: Multiplication of Binomial equations.
Example: Multiply x + 2y and 3x + 4y
Vedic Method
Before I proceed with the Vedic way, let me explain to you the
steps involved in solving.
Step 1: First, write the two variables on the top and put the
coefficients from each equation below. Apply the Urdhya
tiryagbhyam Vedic sutra. Remember, the carry-over to the
preceding column as done in multiplication will not be
executed here.
x y
1 2
3 4
3 | 4+6 | 8
Vertical | sum of cross wise | vertical
1 x 3 | 1 x 4 + 2 x 3 | 2 x 4
3 | 10 | 8
Multiplication in Algebra 93
Step 2: Starting from the right, add the variables to these
coefficients in the following manner.
Solution:
Step 1: Write the two variables on top and their coefficients along
their respective signs below them.
x y
8 –3
2 4
a) 4 x – 3 = – 12 (Vertical product)
b) 8 x 4 + 2 x ( – 3 ) = 32 – 6 = 26 (sum of cross wise
multiplication)
c) 8 x 2 = 16 (Vertical product of extreme left)
x y
8 –3
2 4
16 | 26 | –12
Place y2 to – 12
xy to 26
and x2 to 16.
Example 3: Multiply 3x – 7 y by 2 x − 5 y
Step 1: Write down the variable on the top and their respective
coefficient below, with the proper sign.
x y
3 –7
2 –5
(3x – 7 y ) x ( 2 x −5 y ) = 6 x2 – 29 xy + 35 y2
Let me remind you the rule before writing the one line answer
to the next few examples that can be done with a little practice.
Rule:
Multiplication in Algebra 95
Let me take you through a interesting tour of algebraic
multiplication on a ride on the very interesting Vedic Chariot
called – m/oZfr;ZxH;ke~ (Urdhva Tiryagbhyam).
Solution:
x y
2 3
11 5
22 |43 | 15
= 22 x2 + 43 xy + 15 y2
Solution:
x y
2 –5
3 –7
6 | −29 | 35
= 6 x 2 – 29 xy + 35 y2
Solution:
y2 z
2 4
3 –7
6|−14 +12|−28
= 6 y4 −2y2 z −28 z2
Since y2 x y2 = y4, 6 is obviously the coefficient of y4. Moreover,
the crosswise multiplication gives the coefficient – 2 and it is the
coefficient of y2z. The extreme right vertical multiplication – 28
is the coefficient of z x z = z2.
What is a polynomial?
The general equation for a polynomial in the variable x is
a0xn + a1xn–1 + a2xn–2 +… where a0, a1, etc., are constants and
n is the highest power of x, called the degree of the polynomial.
The multiplication of Algebriac polynomial by the traditional
method taught in our classrooms is a little bit confusing and
one needs to club variables of the same degree before coming
to an answer. Let us take an example and try to solve it by the
traditional method.
Multiply x2 + 2x + 3 by 3 x2 + 2x + 4
Solution: (x2 + 2x + 3) x (3 x2 + 2x + 4)
= x2 (3 x2 + 2x + 4) + 2x (3 x2 + 2x + 4)+ 3 (3 x2 + 2x + 4)
[Distributive Property]
= 3 x4 + 2 x3 + 4 x2 + 6 x3 + 4 x2 + 8x + 9 x2 + 6 x + 12
= 3 x4 +8 x3 +17 x2 + 14 x + 12
[on clubbing the power of like terms]
Rule:
Multiplication in Algebra 97
the coefficient below the variable shown here.
x2 x x0
1 2 3
3 2 4
5 4 8 5 4 8 5 4 8 5 4 8 5 4 8
1 5 9 1 5 9 1 5 9 1 5 9 1 5 9
x2 x x0
1 2 3
3 2 4
a) 3 x 4 = 12
b) 2 x 4 + 2 x 3 = 14
c) 1 x 4 + 3 x 3 + 2 x 2 = 17
d) 1 x 2 + 3 x 2 = 8
e) 1 x 3 = 3
x2 x x0
1 2 3
3 2 4
3 | 8 | 17 | 14 | 12
Solution:
Step 1: Put the variable x2 , x and x0 at the top and place the
coefficient of the polynomials below it along with the respective
sign.
x2 x x0
2 −4 6
3 −7 −2
x2 x x0
2 –4 6
3 –7 –2
a) 6 x – 2 = −12
b) −4 x −2 + 6 x −7 = −34
c) 2 x −2 + 3 x 6 + (−4)x (−7) = 42
d) 2 x −7 + 3 x −4 = −26
e) 2 x 3 = 6
Multiplication in Algebra 99
Step 3: Place the coefficient obtained in step 2 below in the
answer line. Starting from the extreme right i.e (–12), place the
variable to each successive coefficient in increasing order.
x2 x x0
2 –4 6
3 –7 –2
6 | –26 | 42 | –34 | –12
x0 or no variable to – 12
x to – 34
x2 to 42
x3 to – 26
x4 to 6
Hence ( 2x2 – 4x + 6 ) x ( 3 x2 – 7 x – 2) = 6 x4 – 26 x3 +
42 x 2 – 34 x – 12
Since the second equation involves only two terms i.e. x and 9
and the first equation has three terms, namely 7 x2, 6x and 5,
this gives us a clue that the x2 term is missing in the second
equation. Hence we shall write the second equation as –
x + 9 = 0 x2 + x + 9
x2 x x0
7 6 5
0 1 9
x2 x x0
7 6 5
0 1 9
0 | 7 | 69 | 59 | 45
i.e. x0 to 45,
x1 to 59,
x2 to 69
and x3 to 7.
Hence, (7x2 + 6 x + 5) x (x + 9) = 7 x3 + 69 x2 + 59 x + 45
Example: Multiply 2 x2 – 7 by 3 x2 + 4 x
x2 x x0
2 0 –7
3 4 0
x2 x x0
2 0 –7
3 4 0
6 | 8 | –21 | –28 | 0
Multiplication in Algebra 101
Step 3: Place the variables in ascending order from the right to left.
x2 x x0
2 0 –7
3 4 0
6 x4 + 8 x3 − 21 x2 −28 x
Hence, (2 x2 – 7) x (3 x2 + 4 x) = 6 x4 + 8 x3 − 21 x2 −28 x
Division
Introduction
1st stage = 12 – 3 = 9
2nd stage = 9 – 3 = 6
3rd stage = 6 – 3 = 3
4th stage = 3 – 3 = 0
Since at the 4th stage, we get a remainder of 0, 12 ÷ 3 = 4.
The conventional method is lengthy and time-consuming as it
is based on a hit and trial method. This method is effective as
long as the divisor is a single digit, but when the divisor is a
bigger number we are at loss because we keep multiplying the
dividend by different numbers from 1 to 9. On the other hand, in
the Vedic system, we get the quotient and remainder in one line.
Conventional Method
54 4 3 8 5 4 8 1 2
–4 3 2 Rough area
6 5 54 x 6 = 324
–5 4 54 x 7 = 378
1 1
4 54 x 8 = 432
– 1 0 8
6
Vedic one-line Dhwajak method
54 4 3 3
8 1
5 1
4
8 1 2 6
Working Rule
Division 105
Navataha Charmam Dastah method as explained in the
Subtraction chapter. Write the complement of the divisor
below it.
2. Separate the extreme right digit of the dividend by
drawing a slash equal to the number of digits in the
divisor. This block is known as the remainder block and
the left block is known as the quotient block.
3. The number of digits to be placed in the remainder column
should be equal to the number of zeros in the base.
4. Carry down the first digit of divisor in the first nd column.
This gives you the first digit of the quotient. Multiply
the quotient digit by the complement, and place it in the
dividend column; next to the first digit of the dividend.
5. Write mechanically the sum of the digits of the second
column to get the second digit of the quotient.
6. Repeat the process until you get a number in the
remainder column. If the remainder is greater than the
divisor, continue the same process in the remainder block
until the digit in the reminder column is less than that
of the original divisor.
Case 1: When the remainder is less than the divisor.
Example 1: Divide 22 by 8
Solution:
Divisor 8) 2 / 2
Complement 2 4 (Quotient x Complement)
2 / 6
Here the divisor is nearer to the base 10.
Complement = 10 – 8 = 2
• Since the base has one zero, one digit will be separated by
a slash for the remainder column. In the above example,
the right-most 2 is separated by a slash.
• Carry down the first digit (2) of the dividend. This is the
first digit of the quotient.
Division 107
Example 3: Divide 10025 by 88
1 1 3 8 1
Hence,
Quotient = 113
Remainder = 81
Explanation:
1. Here Divisor = 88
Nearest base in power of 10 = 100
Complement = Base – Divisor
= 100–88 = 12
2. Arrange the digits in the columns as shown above,
separating quotient and remainder. Since there are two
zeros in the base, the remainder column will have the
two right-most digits of the dividend.
3. Carry down 1 of column 2 (the first digit of the dividend).
This is the first digit of the quotient.
4. Multiply the first digit of the quotient with the complement
and place it in the dividend column, next to the first digit
of the dividend. 12 X 1 = 12 is placed below 0.
1 1
5. Carry down the sum of the circled digits; this will give
you the second digit of the quotient. The second digit
of quotient is 0 + 1 = 1
6. Multiply the second digit of the quotient with the
complement and place it in the dividend column, next
to the second digit of the dividend. 12 x 1 = 12 is
placed below the second zero.
Division 109
9. Sum up the digits of Column 3 to get the remainder. The
above process repeats until the digit thus obtained in the
remainder column is less than the original divisor. Here
the sum of the digits of Column 3 = 81 and 81< 88
(Divisor), hence Quotient = 113 and Remainder = 81
Solution:
Explanation:
1. Here Divisor = 8988
Nearest base in power of 10 = 10000
Complement = Base – Divisor
= 10000 – 8988 = 1012
2. Arrange the digits in the columns as shown above,
separating quotient and remainder. Since there are four
zeros in the base, the remainder column will contain 4
digits.
3. Carry 1 of column 2 (the first digit of the dividend) down.
This is the first digit of the quotient.
4. Multiply the first digit of the quotient with the complement
and place it in the dividend column, next to the first digit
of the dividend. 1012 x 1 = 1012 is placed below 1.
1 2
1 2 4
Division 111
Column 1 Column 2 (Q) Column 3 (R)
Divisor 8 9 8 8 1 1 2 1 1 3 4
Complement = 1 0 1 2 1 0 1 2
2 0 2 4
4 0 4 8
1 2 4 6 6 2 2
Since the remainder 133 > 87, add the complement 13 in the
remainder column.
1 1 8 46
Since the base is 100, the number in the remainder column should
be less than 100; therefore the left-most 1 of the remainder
column will be transferred to the quotient column. Thus Quotient
= 117 and Remainder = 46.
Division 113
Column 2 (Q) Column 3 (R)
99979 1 1 1 99171
00021 0 0 0 2 1 – –
0 0 0 2 1 –
0 0 0 2 1
1 1 1 1 0 1 5 0 2
+ 0 0 0 2 1
1 1 1 1 0 1 5 2 3
Explanation:–
1. Base =100
Complement = 113 – 100 = 13
Writing each digits of complement with the changed sign
= – 1– 3.
Since the number of zeros in the base is 2, the number
of digits in the remainder column is 2.
2. Carry 1 (the fist digit of Column 2) down; this will give
you the first digit of the quotient.
Division 115
Column 1 Column 2(Q) Column 3(R)
1 1 3 1 3 5 8
Complement = 1 3 –1 –3
Revised Complement= –1 –3 + –2 –6
1 2 0 2
Quatient = 21
Remainder = 4 0 0 6
Rules:
Division 117
quotient and flagged digits. The steps written here will
help you in this regard.
• When there is a single flagged digit, we subtract the
product of the last quotient digit and the flagged digit
at each step.
• When there are two flag digits, in the first step product
of the first flagged digit and the first quotient digit is
subtracted and from the second step onwards, the cross
product of the two flagged digits and the last two quotient
digits is subtracted.
• When there are three flagged digits: – first subtract the
product of the first flagged digit and the first quotient
digit. In the second stage, the cross product of the first
two flagged digits and first two quotient digits will be
subtracted from the gross dividend. In the third stage,
the cross product of the three flagged digits and the
three quotients will be subtracted. After the third stage,
the subtraction of the cross product of the three flagged
digits and the three quotients will be continued.
• Let us take some examples to understand the modus
operandi more clearly.
Solution:
Step 1: Take 4 as the main divisor and 2 as the flag digit. Arrange
the divisor, flag digit, quotient and digit for remainder, if any,
according to the rule explained above. Since we have taken only
one digit as the flagged digit, the unit digit will be put in the
remainder column.
42 1 7 6 4
Quotient Remainder
Step 3: Divide 8 by 2
Quotient = 2
Remainder = 0
42 1 7 1
6 0
4
Solution:
32 3 8 7
Division 119
Step 2: Divide 3 by 3. Quotient = 1 and Remainder = 0
Put the quotient in thequotient column and remainder 0
before 8 in the dividend column.
32 3 0
8 7
1
Gross Dividend = 08
Net Dividend = 08 – first digit of quotient x flag digit
= 08 – 1 x 2 = 6
Step 3: Divide 6 by 3
Quotient = 2 and Remainder = 0.
0 is now placed before 7 in the dividend column, making
it 07, the next dividend.
32 3 0
8 0
7
1 2
Gross Dividend = 07
Net Dividend = 07 – second digit of quotient x flag digit
= 07 – 2 x 2 = 3
Since,our calculation has moved inare in remainder side,
we do not do the division.
32 3 0
8 0
7
1 2 3
Hence, 387 ÷32, Quotient = 12, Remainder = 3
Solution:
Step 1: Out of the divisor 73, we put down only the first digit
i.e. 7 in the divisor- column and put the other digit i.e. 3 on
the top of the flag.
73 3 8 9 8 2
Hence, a line is drawn vertically from the right, leaving one digit
at the end in the remainder column as shown above. Now the
entire division is to be carried out by 7.
Step 2: As the first digit from the left of dividend (3) is less than
7, we take 38 as our first dividend. Divide 38 by 7
Q1=5 R1=3
Q1 = The first quotient
R1 = The first remainder.
Put the quotient below the horizontal line and prefix the remainder
3 below the digit 9.
73 3 8 3
9 8 2
5
Step 3:
New gross dividend = 39
Subtract the product of dhvajanka 3 and the first quotient
(5) i.e. 3 x 5 =15 from 39.
Net divided = 39 – 3 x 5 = 24
Divide 24 by 7
Q2=3 R2=3
Q2 = The second quotient
R2 = The second Remainder
Prefix the remainder 3 below 8 above the horizontal line
73 3 8 3
9 3
8 2
Step 4:
New gross dividend =38
Net Dividend = Gross Dividend – product of 2nd quotient
and dhvajanka
= 38 – 3 x 3 =29
Division 121
Net dividend =29
Divide it by 7
Q3 =4 R3 =1
Q3 = The third digit of quotient
R3 = The third digit of remainder
Place 4and 1 as discussed above.
73 3 8 3
9 3
8 1
2
5 3 4
Step 5:
New Gross dividend =12
73 3 8 3
9 3
8 1
2
5 3 4 0
Net Dividend = Gross Dividend - product of dhvajanka
and third quotient
= 12 – 3 x 4 =0
Since, we are in remainder part so we stop division
process.
Hence, Quotient = 534 and Remainder =0
Example 12: Divide 72 38 761 by 524?
Solution: If we take two digits as dhvajanka, then the remainder
column will certainly have two digits. See the arrangement below.
524 7 2 3 8 7 6 1
Step 1: Divide 7 by 5
Q1 = 1 and R1 = 2
524 7 2
2 3 8 7 6 1
1
Step 2: Gross dividend = 22
Net dividend = Gross Dividend – Product of the first quotient
and the first flag digit
= 22 –1 x 2 = 20
Q2 = 4 and R2 = 0
Since R2 = 0, we have to take the quotient below 4 i.e. New
quotient = 4 – 1 =3. This is because we cannot consider the
remainder zero in the middle of the division. The same process
will be applicable in case the net divisor in the middle is either
zero or negative.
Hence, Revised Quotient ( Q2 ) = 3 and R2 = 5
524 7 2
2 5
3 8 7 6 1
1 3
524 7 2
2 5
3 3
8 7 6 1
1 3 8
524 7 2
2 5
3 3
8 5
7 6 1
1 3 8 1
Division 123
Step 6: Gross dividend =57
Net dividend = 57 – (sum of cross product of 24 and 81)
= 57 – (2 x 1 + 4 x 8)
= 57–34 2 4
= 23
Divide 23 by 5 8 1
Q5 = 4 and R5 = 3
Remainder = 361 – (sum of cross product of 24 and 14) x 10
– (last flag digit x last quotient)
= 361 – (8 +4) x 10 – (4 x 4) 2 4
= 361–120-16= 225
1 4
524 7 22 53 38 57 36 1
1 3 8 1 4 225
20132 9 8 6 2 1 4 5
Division 125
20132 9 8 18
6 22
2 22
1 20
4 13
5
4 8 9 8 7
9213649875 5 3 6 2 9 6 8 5 2 7
9213649875 5 3 8
6 2 9 6 8 5 2 7
0.5
Step 2: Divide 76 by 9 2 1
Q = 8 , R = 4
5 8
Gross Dividend = 42
Net Dividend = 42 – cross product of 21 and 58
Step 3: Divide 21 by 9 2 1 3
Q = 2, R = 3
5 8 2
Gross Dividend = 39
Net Dividend = 39 – cross product of 213 and 582
= 39 – 27
= 12
9213649875 5 3 8
6 4
2 3
9 6 8 5 2 7
0.5 8 2
Step 4: Divide 12 by 9
Q = 1, R = 3
9213649875 5 3 8
6 4
2 3
9 3
6 8 5 2 7
0.5 8 2 1
Division 127
7
Square
Introduction
Squaring a number means multiplication of the number by itself.
Mathematically, a x a = a2. Here is the geometrical representation
of first five square numbers.
1 2 x 2 3 x 3 4 x 4 5 x 5
The school curriculum follows only two methods for finding the
square of a number.
• By multiplying the number by itself through long
multiplication process.
Example: (12)2 = 12 x 12 = 144
• By Algebraic Expansion:
Here, we generally use either of the two formulae—
i) (a+b)2 = a2 + 2ab + b2
ii) (a− b)2 = a2 − 2ab + b2
Vedic Method
Square 129
2. Yavadunam Tavaduni kritya vargena yojayet (;konwue rkonwuh d`R;
oxsZ.k ;kst;sr): This Vedic sub-sutra is used for squaring numbers
which are closer to the base (10n). With a little practice, though,
you can extend it to numbers which are farther from the base using
the sub-base provided the sub-base is a multiple of 10n.
This Vedic-sutra simply says –
(a) Find the extent or deficiency of a number to be squared
with respect to its base. This extent or deficiency is termed
here as the deviation.
(b) Set up the square of the deviation at the end.
3. Dwanda-Yoga or Duplex Mehod (}a} ;ksx): The two Vedic methods
discussed above have limited application. So the question is – what
will you do if the number that is to be squared does not satisfy
either of the above condition?
The Duplex combination is applicable in all the cases. The term
Dwanda-Yoga is used in two different senses– The first one
is by squaring and the second one is by cross multiplication.
i) In the case of a single central digit a, the duplex is its
square. i.e. a2
ii) In the case of an even number of digits (say a and b)
equidistant from the two ends, the duplex is taken as
double the cross product of a and b (i.e.2ab)
4. Urdhva–Tiryak (m/oZ fr;Zd): Squaring a number means multiplying
a number twice by itself. This method of squaring is nothing new
but it is the same as that of the multiplication of two numbers in the
Cross and Dot method discussed in the chapter on Multiplication. It
is therefore expected from readers to do the squaring of any number
by Urdhva-Tiryak Method, which is nothing but the multiplication
of a single number twice.
RHS = Square of 5 = 25
LHS = Remaining digit x Next digit
= 8 x 9 = 72
Hence, (85)2 = 7225
Solution:
RHS = Square of 5 = 25
LHS = Remaining digit x Next digit
= 5 x 6 = 30
2
Hence, (55) = 3025
Solution:
RHS = Square of 5 = 25
LHS = Remaining digit x Next digit
= 12 x13 = 156
Multiplication of 12 x 13 can be done by the Nikhilam method.
12 + 2
13 + 3
15 / 6
Moreover, multiplication in such a case can be done by the Urdhva-
Tiryak method.
1 2
x 1 3
Square 131
1 x 1 / 1 x 3 + 1 x 2 / 2 x 3
= 1 5 6
Hence, (125)2 = 15625
Solution:
RHS = Square of 5 = 25
LHS = Remaining digit x Next digit
= 16 x17 = 156
Multiplication of 12 x 13 can be done by the Nikhilam method.
16 + 6
17 + 7
23 / 42
= 272
= 2 7 2
Solution:
RHS = Square of 5 = 25
LHS = Remaining digit x Next digit
= 24 x 25 = 600
(See special method of multiplication by 25 in Multiplication
through Observation.)
Hence, (245)2 = 60025
Case 1: When the number is near the base 10, 100, 1000...... 10n
Deviation = 13 – 10 = 3.
(13)2 = 13 + 3 / 32
= 169
Deviation = 16 – 10 = 6.
(16)2 = 16 + 6 / 62
= 22 / 36
= 256
Square 133
Example 8: Find the square of 91?
Deviation = 91 – 100 = −9
(91)2 = 91 – 9 / (−9)2
= 82/ 81
Deviation = 97 – 100 = −3
(97)2 = 97 – 3 / (−3)2
= 94/ 9
Since the Base =100, the RHS should have 2 digits, so one
additional zero will be placed before 9.
(97)2 = 9409
Case 2: When the base is not in the form of 10n, but the
multiple of 10.
If the number to be squared is near the base 20, 30, 40, ---- or
200, 300, 400, ---- or 2000, 3000, 4000, --- the Yavadunam
Tavduni sub-sutra will work with a slight change.
The answer will be arrived at in two parts.
The RHS part of the answer will be the square of the deviation
from the base. The LHS part of the answer should be written
with utmost care. LHS = (Number to be squared + Deviation)
x sub-base.
Deviation = 32 – 30 = 2
30 = 3 x 10
Sub-base = 3
Actual base =10
= 416/ 16
= 41616
= 5 x 464/ 324
= 232 0/3 24
= 232324
Square 135
Let us take the base 500 and find the square of 482 in another
way.
500 = 1000/2
Hence, Base = 1000 and sub-base = ½
Deviation = 482 −500 = −18
(482)2 = (482 −18) x ½ / (−18)2
= 232 / 324
= 232324
= 7 x 718/ 81
= 502681
= 9 x 8978/ 121
= 80802 /121
= 80802121
Square 137
Duplex of 320416 = 2 x (3 x 6 ) + 2 x (2 x 1) + 2 x
(0 x 4) = 40
Duplex of 125673 = 2 x (1 x 3 ) + 2 x (2 x 7) + 2 x
(5 x 6) = 94
( 11) 2 = 1 2 1
( 1 1 1 ) 2= 1 2 3 2 1
(1 1 1 1 )2 = 1 2 3 4 3 2 1
(1 1 1 1 1 )2 = 1 2 3 4 5 4 3 2 1
( 1 1 1 1 1 1)2 = 1 2 3 4 5 6 5 4 3 2 1
(1 1 1 1 1 1 1)2 = 1 2 3 4 5 6 7 6 5 4 3 2 1
----------------------------------------------------------------------------------
Grouping of a number
4 4 4
D(2) D(24) D(4)
4 4 4
D(2) D(24) D(245) 4 4
D(45) D(5)
1 digit 2 digit 3 digit 2 digit 1 digit
4 4 4 4 4 4 4
D(2) D(24) D(245)
D(2456)
D(456) D(56) D(6)
1 digit 2 digit 3 digit 4 digit 3 digit 2 digit 1 digit
4 4 4
D(3)
2
D(32) D(2)
= 3 | 2 x 3 x 2 | 22
= 9 | 1 2 | 4
+
= 1024
Square 139
Example 17: Find the square of 49
4 4 4
D(4) D(49) D(9)
= 42 | 2 x 4 x 9 | 92
= 16 | 7 2 | 8 1
+ +
= 16 / 80 / 1
= 2401
4 4
D(4)
2
D(46) 4
D(465)
2
4 4
D(65) D(5)
= 4 2 x 4 x 6 2 x 4 x 5 + 6 2 x 6 x 5 52
= 16 | 4 8 | 7 6 | 6 0 | 2 5
+ + + +
= 20 / 1 5/1 2/2/5
= 216225
Square 141
Duplex of 5612 = 2x 5 x 2 + 2 x 6 x 1 = 32
Duplex of 612 = 2 x 6 x 2 + 12 = 25
Duplex of 12 = 2 x 1 x 2 = 4
Duplex of 2 = 22 = 4
Square Root
Introduction
Table 1
Table 2
Vedic Method
Vedic Method of
Extracting Square
Root
Square Root 145
of 3–4 digits in 2–3 seconds, merely by observing the
above two tables. The first table will help you to find
the unit digit of the exact square root, whereas Table 2
will help you to find the ten’s digit.
2. }a}&;ksx (Duplex method): A detailed description of this
method can be found in this book itself in the chapter
on Squares. This Vedic sutra is applicable to all, whether
the given number is a perfect square root or not.
Rule:
• Make a group of two starting from the right
• Look at the unit digit of the number and observe your
answer in Table 1. This will help you to decide the digit
at the unit place
• Now move to the second group and find the ten’s digit
of your square root
Solution:
21 16
2nd group 1st group
Hence, √ 2116 = 46
Example 2: Find the square root of 5184.
Solution:
51 84
2nd group 1st group
Hence, √ 5184 = 72
Example 3: Find the square root of 9216.
Solution:
92 16
2nd group 1st group
Square Root 147
We know that 952 = 9025, since 9216 > 9025, therefore the
desired square root will be more than 95.
Solution:
06 76
2nd group 1st group
Rule:
• Make a group of two digits starting from the right. Here
we will have three groups. Denote the left digit by L, the
middle digit by M and the right digit by R
• The first (L) and third (R) group will give us the hundred’s
place digit and the unit place digit. These two can be
written only through observation, with the help of Table
1 and Table 2.
• Subtract L2 from the 1st pair and carry down the next
digit from the dividend, as done in simple division.
Solution:
69 22 24
L M R
69 22 24
– 82 carry down the next digit from the dividend
5 2 new dividend
Square Root 149
Digit sum of 69 22 24 Digit sum Digit sum
of (832)2 of (838)2
6 + 9 + 2 + 2 + 2 + 4 = 7 7 1
Hence, √ 69 22 24 = 832
Example 1: Find the square root of 103041.
Solution:
10 30 41
L M R
10 30 41
– 32 carry down the next digit from the dividend
1 3 new dividend
Rule:
• The given number is first arranged in two-digit groups
from right to left, and a single digit if any left over at
the left hand end, is counted as a simple group by itself.
• The number of digits in the square root will be the same
as the number of digit-groups in the given number itself
including a single digit if any such there is.
• If a square root contains n digits, the square must consist
of 2n or 2n–1 digits.
• And conversely, if the given number has n digits, the
square root will contain n| 2 or (n+1)|2 digits.
• Group the number by placing a bar and put them in
between the horizontal and vertical line bar, as shown
in the given examples.
Working procedure of the Duplex Method
The working of the Duplex method is as simple as straight division.
Let us take a few examples to understand the modus operandi
of this method.
Example: Find the square root of 529 by using the Duplex method.
Solution:
– —
• Group the number 5 29 by placing a bar over it.
• Put a horizontal and vertical line as shown below.
5 29
Q
Square Root 151
• Since 22 < 5 < 32, the first digit of the square root in
the quotient column is 2. Double the quotient and set
this down as divisor
4 5 29
Q 2
Remainder = 5 – 22 = 1
4 5 1
29
Q 2 3
Solution:
— —
Group the number 42 25 by placing a bar over it.
Put a horizontal and vertical line as shown below.
42 25
Q
• Since 62 < 12 < 72, the first digit of the square root in
the quotient column is 6. Double the quotient and set
this down as divisor.
Remainder = 42 – 62 = 6
12 42 6
25
Q 6
12 42 6
2 2
5
Q 6 5
• Next dividend = 25 Subtract the square of the
quotient from the next dividend.
Net dividend = 25 – 52 = 0
Since no more digits are left, the square root of 4225 is
65.
Example: Find the square root of 20736
Solution:
• Group the number 2 07 36, by placing a bar over it.
• Put horizontal and vertical line as shown below.
2 07 36
Q
• Since 12 < 2 < 22 , the first digit of the square root in
the quotient column is 1. Double the quotient and set
this down as divisor.
2 2 07 36
Q 1 4
Remainder = 2 – 12 = 1.
Square Root 153
• Put this remainder below the next dividend digit. Hence
the next gross dividend = 10
2 2 1
07 36
Q 1
• Divide 10 by 2
10÷2 = 5, Q = 5 and R = 0
2 2 1
0 27 36
Q 1 4
• Next dividend = 27
Subtract the square of quotient from the next dividend.
Net dividend = 27 – 42 = 11
Divide 11 by 2 and write the quotient ( Q = 5) and
remainder (R = 1) at their respective places.
2 2 1
0 27 1
36
Q 1 4 5
• Gross dividend = 13
Net dividend = 13 – (Duplex of 45)
= 13 − 2 x 4 x 5
= 13 – 40 = − 27 < 0
(Since Net dividend is less than zero, we can’t take the
quotient (Q= 5) as taken above. Now for 11÷ 2, Revised
Q = 4 and Revised R = 3.)
2 2 1
0 27 3
36
Q 1 4 4
2 2 1
0 27 1
3 1
6
Q 1 4 4.0
10 25 7
4 45 54 57 1 6
Q 5 0 7 4.000
A step by step illustration has been done here for your convenient.
Square Root 155
• Since we have so far got 4 digits in the quotient column,
the perfect square root is obtained. The next digit in the
quotient column will give the remainder, if any.
• Remainder= 54 – 2 x 0 x 4 − 72 = 5
• Next dividend = 57 and remainder
= 57 – 2 x 4 x 7 =1
• Next dividend =16 and the last remainder
= 16 – 42 = 0
• I think you have understood the Duplex method, therefore
the next two examples given below are without much
detail, though a brief description is provided here for
your convenience.
Solution:
12 45 9
3 91 69 38 12 4
Q 6 7 3 2 .000
Solution:
14 52 3
4 64 43 13
9 70 7
Q 7 2 4 1. 8
Square Root 157
9
Introduction
Vedic Method
Deviation = 79 – 81 = – 2
79 = 81 − 2
2 x 81
Deviation = 34 –36 = −2
34 = 36 − 2
2 x 36
= 6 − 1/6
= 6 – 0.166 = 5.844
Example 7: Find the square root of 204
Solution: Perfect square approaching 204 is 196.
Deviation = 204 – 196 = 8
204 = 196 + 8
2 x 196
= 14+ 8/28
= 14 + 0.285
= 14. 285
Cube
Introduction
Number 1 2 3 4 5 6 7 8 9 10
Cube 1 8 27 64 125 216 343 512 729 1000
;konwue
(Yavadunam)
Vedic Method
For Cubing a
Number
vuq:I;s.k fuf[kye~
(Anurupyen) (Nikhilam)
1. Yavadunam (;kcnwue)
This formula works better when the number to be cubed is near
the base. The base should be in the form of 10n, where n is a
natural number. This formula has limited application.
Working Rule
• Check whether the number is near the base 10, 100,
1000... or not
• Find the excess or deficit number from the base.
• The whole operation is to be performed in three parts. In
the 1st part, add twice the excess/ deficit to the original
number.
• 2nd part = New Excess (Number obtained in 1st part –
base) x original excess/deficit
• 3rd part = cube of excess.
Cube 163
Mathematically, if a = original number and d = deviation (Excess/
Deficit from the Base) then the whole operation can be summed
up as —
a3 = a+2d / [ (a + 2d ) – base ] x d / d3
let us take some example.
1st part = a + 2d = 12 + 2 x 2 = 16
2nd part = (16 – 10) x 2 = 12
3rd part = 23
Combining all the three parts we get,
(12)3 = 16 | 12 | 8
= 17 | 2 | 8
= 1728
1st part = a + 2d = 96 − 2 x 4 = 88
2nd part = (88 – 100) x −4 = 48
3rd part = (–4)3 = – 64
Combining all three parts we get,
(96)3 = 8 8 | 4 8 | −64
= 88 |47 + 1 | −64
= 88 | 47 | 100 – 64
= 88 | 47 | 36
rd
(3 part is negative, so add 100 to the negative part to make it
100 – 64 = 36. Subtract 1 from the previous part, thus making
48 to 47.)
2. Anurupyen (vuq:I;s.k):
Cube 165
• Double the second and the third number and put it down,
under the second and third numbers. Finally add up the
two rows
Solution: (12)3 = ?
Important points:
1. If you start with the cube of the first digit and multiply
with the geometric ratio up to the next three numbers,
Solution: ( 15 )3 = ?
(15)3 = 1 5 25 125
+ 10 50
1 1 5 7 5 1 2 5
+ + +
= 3 3 75
The excess digit, leaving the unit digit from each column (from
right to left) is transferred to the next column. In the above
example, the excess digit is underlined.
Solution: ( 19 )3 = ?
Cube 167
Example: Find the cube of 32
Solution: (32)3 = ?
Solution: ( 46 )3 = ?
64 96 144 216
192 288
64 288 432 216
= 64 288 432 216
= 64 288 453 6
= 64 333 3 6
= 97336
Solution: (105)3 = ?
Cube 169
the other hand, if the base = 40, then the sub- base =
4, because 40 = 4 x 10,
• The whole cubing process then involves 3 steps.
A) (Number to be cubed + 2 x deviation from the base)
x (sub-base)2
B) {3 x (deviation)2} x sub-base
C) (Deviation)3
• If there is no sub-base, then the calculation becomes
very easy.
st nd
1 term 2 term 3rd term
= 1850 | 960 | 512
= 1850 | 1011 | 2
= 1951 | 1 | 2
Hence (58)3 = 195112
= 112 | 48 | 64
= 1124864
(Since the base =100, there should be 2 digits in each digit
separator)
Example: Find the cube of 997 by using Nikhilam Sutra
Solution: Working base = 1000
Deviation = 997 − 1000 = − 3
(997)
3
: : :
= 997 + (−3) x 2 | 3 x (−3)2 | (−3)3
st
1 part 2nd part 3rd part
= 991 | 27 | −27
= 991 | 026 | 1000 – 27
= 991 026 973
Since the base is 1000, there should be 3 digits in each digit
separator. Therefore, 26 in the second part should be changed to
026. In the 3rd part, there is a negative sign, so subtract it from
the base 1000. Hence in the 3rd part, we will have, 1000 – 27
= 973.This change will be adjusted by reducing 1 from the 2nd
part. Thus, 27 in the second part now become 27 – 1 = 26.
(997)3 = 991026973
Cube 171
11
Cube Root
Introduction
⇒ a = x1/3
Solution:
2 592704
2 296352
2 148176
2 74088
2 37044
2 18522
3 9261
3 3087
3 1029
7 343
7 49
7 7
1
Hence,
: : : :
592704 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 x 7 x 7
= 2 x 2 x 3 x 7
= 84
The Vedic method will instantly give you the answer 84 by mere
observation.
The great Indian Astronomer and Mathematician Aryabhatta
has also mentioned in his book, Ganita Pada, a method to extract
the cube root of any number, but the method is too complex to
understand. The fifth sloka of Aryabhatta’s book Ganita-Pada
reads as follows.
Cube Root 173
elegant method described by Aryabhatta has more symmetry
with the Vedic Method used by Swami Bharathi Krishna Tirtathji
Maharaj
Before we move further, look at the following table carefully.
This table will help you to determine the unit digit of a cube root.
Table 1
Table 2
There is another very important table that will help you to find the
cube in ambiguous cases. The details will follow the next table.
Table 3
Cube Root 175
Example: Is 1729 a perfect cube?
Vedic Method
Vilokanam Beejank
(foyksdue~) (chtkad)
Let us now start with the Vedic method of finding cube root.
For the convenience of readers, this chapter has been divided
into two segments.
Cube Root 177
unit digit of cube root and Table 2 will give you the Left
digit of the cube root.
Case 1: Cube Root of a number having less than 7 digits.
The Vedic method of extracting the cube root of a number having
less than 7 digits is done by the Vilokanam ( foyksdue~) method.
Form the group of three digits from right to left as discussed
above and find the unit digit and ten’s digit with the help of
Table 1 and Table 2.
Solution:
1. Place the bar over the number from left to right, leaving
two digits at a time.
9 4 1 1 9 2
2. Since the bar is placed on two numbers, the cube root
will contain only two digits.
3. Since the unit digit of this number is 2, the unit digit
of cube root is 8.
4. For the ten’s digit, take the least of
93 < 941 < 103
Ten’s digit = 9
3
Hence, 941192 = 98
Rule:
Cube Root 179
unit digit of the number obtained in the previous step.
In case you obtain more than one value of M, check by
theBeejank method (Refer Table 3) which value of M is
best suited in this case.
4. The cube root of such a number = L M R.
Example 4: Find the cube root of 76765625
Solution:
1. Group the number from right to left by placing the bar.
3
7 6 7 6 5 6 2 5
2. The unit digit of the cube root is 5 (Refer Table 1). Hence
R = 5.
3. Moreover 43 < 76 < 53 , so L = 4 (Refer Table 2)
4. Subtract R3 from the number and forget the last digit (0)
obtained after subtraction.
7 6 7 6 5 6 2 5
–1 2 5
7 6 7 6 5 5 0 0
8 4 6 0 4 5 1 9
–7 2 9
8 4 6 0 3 7 9 0
Solution:
1. Placing the bar over the number we find that there are
three groups of numbers, so the cube root will have three
digits.
3
2 7 9 7 2 6 2 6 4
2. From the Table 1, we can say that the unit digit of the
cube root is 4. Hence R = 4.
Cube Root 181
3. Moreover, 63 < 33 < 73 , so L = 6 (Refer Table 2)
4. Subtract R3 from the number and cancel out the last
zero.
2 7 9 7 2 6 2 6 4
– 6 4
2 7 9 7 2 6 2 0 0
2 9 7 9 1
Since there are two groups, the cube root of 29791 contains a
two digit number.
Unit digit = 1
Ten’s digit = 3
Hence, the cube root of 29791 = 31
We have,
3 3
1906624 = 8 x 8 x 29791
= 2 x 2 x 31 = 124
8 51478848
8 6434856
804357
804 357
Cube Root 183
Since there are two groups, the cube root of 804357 contains a
two digit number.
Unit digit = 3 (Refer Table 1)
Since, 93 < 804 < 103
Hence, Ten’s digit = 9 (Refer Table 2)
We have,
3 3
51478848 = 8 x 8 x 804357
= 2 x 2 x 93 = 372
Let us extract the cube root by the Beejank method.
Here the method is slightly changed. In case 2, we have subtracted
R3 from the number and neglected the last zero but here L3 is
subtracted. You may consider either of these cases.
8 7738893352
967361669
2. The unit digit of the cube root is 9 (Refer Table 1). Hence
R = 9
3. Moreover 93 < 967 < 103, so L = 9 (Refer Table 2)
4. Subtract R3 from the number and neglect the last zero.
9 6 7 3 6 1 6 6 9
– 7 2 9
9 6 7 3 6 0 9 4 0
Hence,
3
967 361 669 = 9 8 9
Therefore,
3 3
7738893352 = 8 x 967361669 = 2 x 989 = 1978
Cube Root 185
12
Introduction
Pascal Triangle:
N=1 1
N=2 1 2 1
N=3 1 3 3 1
N=4 1 4 6 4 1
N= 5 1 5 10 10 5 1
Blaise Pascal was born on 19 Jun 1823 in France. He was a
true genius. At the age of 12, he discovered that the sum of the
angles of a triangle is 180 degrees. He invented the first digital
calculator to help his father, who was a tax collector. He laid the
foundation of the theory of Probability. His work on the topic in
Tratise on the Arithmetical Traingle carried the most interesting
triangle called the Pascal Triangle.
Since we are expanding the binomial with (+) sign, put the (+)
sign in between each assigned value in the above line. Replace
the value of a0 and b0 by 1. Now the formula is–
Coefficients are– 1 4 6 4 1
Assign the value of a and b
1 a4b0 4a3b1 6a2b2 4a1b3 1a0b4
Solution: Apply the above method and write down the coefficients
and assign the value of a and b thereafter. Lastly, place the (+)
and (−) sign alternatively.
Vedic Method:
From the above explanation, you must have learnt to find the
expansion of any binomial. The above Pascal Triangle is for n
=5, but you can extend it for more values of n.
Now, let’s return to our business. The above expansion of
(a+b)4 can be re-written as-
(a+b)4
= a4 + a3b + a2b2 + ab3 + b4
3 a3b + 5a2b2 + 3ab3
1st column 2nd column 3rd column 4th column 5th column
1st column 2nd column 3rd column 4th column 5th column
1 2 4 8 16
2 x 3 =6 4 x 5 =20 8 x 3 = 24
1st column 2nd column 3rd column 4th column 5th column
1 2 4 8 16
6 20 24
1 8 24 32 16
1st column 2nd column 3rd column 4th column 5th column
1 2 4 8 16
6 20 24
1 8 24 32 16
1 8 24 33(32 +1) 6
1 8 27(24 +3) 3 6
1 10(8 +2) 7 3 6
2 (1 +1) 0 7 3 6
1st column 2nd column 3rd column 4th column 5th column
1st column 2nd column 3rd column 4th column 5th column
24 = 16 16 x 1/2 8 x 1/ 2 4 x 1/2 2 x 1/ 2
= 8 = 4 = 2 =1
1st column 2nd column 3rd column 4th column 5th column
16 8 4 2 1
8 x 3 =24 4 x 5 =20 2 x 3 = 6
1st column 2nd column 3rd column 4th column 5th column
16 8 4 2 1
24 20 6
16 32 24 8 1
1st column 2nd column 3rd column 4th column 5th column
16 8 4 2 1
24 20 6
16 34(32+2) 4 8 1
19(16+3) 4 4 8 1
Write the fourth power of a in the 1st column and the remaining
4 terms in the rest of the columns as done in the previous two
example. The table will look like
1st column 2nd column 3rd column 4th column 5th column
14 = 1 1 x 1 = 1 1 x 1 = 1 1 x 1 = 1 1 x 1 = 1
1st column 2nd column 3rd column 4th column 5th column
1 1 1 1 1
1 x3 =3 1 x 5 = 5 1 x 3 = 3
1st column 2nd column 3rd column 4th column 5th column
1 1 1 1 1
3 5 3
1 4 6 4 1
1st column 2nd column 3rd column 4th column 5th column
1 4 16 64 256
4 x 3 = 12 16 x 5 64 x 3
= 80 = 192
1st column 2nd column 3rd column 4th column 5th column
1 4 16 64 256
4 x3 =12 16 x 5 64 x 3
= 80 = 192
1 16 96 256 256
1st column 2nd column 3rd column 4th column 5th column
1 4 16 64 256
4 x 3 =12 16 x 5 64 x 3
= 80 = 192
1 16 96 256 256
1st column 2nd column 3rd column 4th column 5th column
1 4 16 64 256
12 80 192
1 16 96+ 28 1 6
= 124
1 16 +12 4 1 6
= 28
1+2 =3 8 4 1 6
1st column 2nd column 3rd column 4th column 5th column
256 320 400 500 625
960 2000 1500
1st column 2nd column 3rd column 4th column 5th column
256 320 400 500 625
960 2000 1500
256 1280 2400 2000 625
1st column 2nd column 3rd column 4th column 5th column
256 320 400 500 625
960 2000 1500
256 1280 2400 2000 625
256 1540 6 2 5
410 0 6 2 5
Introduction
If, x4 = 16 so, x = 4 16
= 4 2 x 2 x 2 x 2
= 2
The above example clearly shows that in order to get the fourth
power of a desired number, we break it into 4 equal factors and
take one factor out of four.
In general, there is no traditional method taught in our
curriculum that helps us to find the fourth root of a number,
except the labyrinth Prime Factor Method. The traditional method
is too clumsy and time-consuming. As long as the number is
2–4 digits, the prime factor method works effectively, but if the
number is more than 4 digits, this method takes more than 5
minutes to arrive at an answer.
Let us understand it with an example;-
4 4
Hence, 1679616 = 4x4x4x4 x 3x3x3x3 x 3x3x3x3
= 4 x 3 x 3 = 36
Table 1
Table 2
Solution:
676 5201
2nd Pair (B) 1st Pair (A)
A B 4 B3 A
1 5 500
3 5 1500
7 5 3500
9 5 4500
• Since our new dividend = 515 and the above table shows
that only A = 1 satisfies the result, we reject the other
option and take A = 1. This might seem lengthy to you,
but with a little practice, you will be able to guess the
result quite comfortably, without setting different values
of A. From the above table, it is clear that you could
have saved time without calculating the value of A =
3, 5 or 7.
• Hence, the fourth root of 6765201 = 51
Solution:
2541 1681
2nd Pair (B) 1st Pair (A)
2541 1 6 8 1
−74 Carry down the next digit from the number
140 1 new dividend
A B 4 B3 4 B3 A
1 7 1372 1372
3 7 ------------- -----------
7 7 ------------ ------------
9 7 ----------- -----------
Solution:
530 8416
2nd Pair (B) 1st Pair (A)
A B 4 B3 4 B3 A
2 4 256 512
4 4 256 1024
6 4 256 1536
8 4 256 2048
Simultaneous Equation
Introduction
• Method of Elimination
• Method of Comparison
• Method of Substitution
• Method of Cross Multiplication
Vedic Method
ijkoR;Z ;kst;sr
(Paravartya Yojayet)
vkuq:I;s 'kwU;etU;r~
(Anurupye Sunyamanyat)
ladyu &O;kodyukH;ke
(Sankalana -Vyavakalana-bhyam)
Paravartya Yojayet
Simultaneous Equation 205
To make the concept more clear let us put the whole thing in a
simple diagrammatical structure.
Solution:
Solution:
Solution:
Simultaneous Equation 207
Independent Term Coefficient of x
29 4
y = −3 12
Coefficient of x Coefficient of y
4 7
12 3
Hence, x = −21 −87 = −108 = −3
84 – 12 −72 2
y = 348 + 12 = 360 = 5
84 – 12 72
5x + 8y = 40
10x + 11 y = 80
12 x + 78 y = 12
16 x + 96 y = 16
Hence, y = 0
Put y = 0 in either of the two equations to get x = 1
44x + 178 y = 22
132 x + 243 y = 66
27 x + 144 y = 720
42 x + 72 y = 360
Simultaneous Equation 209
Example: Solve for x and y
23 x + 31 y = 18
31 x + 23 y = 90
45 x – 23 y = 113
23 x – 45 y = 91
Solution: We have,
Solution: We have,
23 x – 29 y = 98
29 x – 23 y = 110
Solution: We have,
23 x – 29 y = 98 ---(1)
29 x – 23 y = 110 --(2)
Simultaneous Equation 211
15
Cubic Factorization
Introduction
Solution:
Since 6 has the factor ± 1, ±2, ±3, and ±6, we invariably put
these values in the polynomial p(x) until we get the remainder
0. Let us see how it works.
P(x) = x3 + 6 x2 + 11x + 6.
P(1) = (1)3 + 6. (1)2 + 11(1) + 6 ≠ 0
P(–1) = (−1)3 + 6. (−1)2 + 11( − 1) + 6
= – 1 + 6 – 11 + 6 = 0
Q(x) = x2+ 5x + 6
= x2 + 2x + 3 x + 6
= x (x + 2) + 3 ( x + 2)
= (x + 2) (x + 3).
α + β + γ = sum of roots = Σα = −b / a
Cubic Factorization 213
αβ + βγ + αγ = sum of product of two = Σαβ = c / a
αβγ = Product of three roots = −d / a
Vedic Method
Hence,
a = α + β + γ = sum of roots = coefficient of x2
b = αβ + βγ + αγ = sum of the product of two roots = coefficient of x
c = αβγ = Product of three roots
Verification:
Cubic Factorization 215
And α + β + γ = sum of roots = 2 + 4 + 6 = 12
Hence, the factors are–
x3 + 12x2 + 44x + 48 = (x + 2) (x +4) (x + 6)
Verification:
Example: Factorize x3 + 8 x2 + 19 x + 12
Verification:
Verification:
Example: Factorize x3 – 7x + 6
x3 – 7x + 6 = x3 + 0 x2 – 7 x + 6
Verification:
The above example illustrates that the Vedic Method to solve the
cubic factor is not only easy to understand but a time saving
technique because it involves no tedious or lengthy calculation,
but merely inspection, which helps you to reach the result in no
time. Always bear in mind, however, that this is valid as long
as the cubic polynomial is factorizable.
Cubic Factorization 217
16
Quadratic Equations
Introduction
foyksdue~
(Vilokanam)
vkuq:I;s 'kwU;eU;r~
( Anurupye Sunyamanyat)
Quadratic Equations 219
Vilokanam (foyksdue~) Sutra
Example: x + 2 + x + 1 = 37
x + 1 x + 2 6
Traditional Method:
Put x + 2 = a
x + 1
Hence,
a + 1/a = 37/6
or, a2 + 1 = 37
a 6
or, 6 a2 + 6 = 37 a
or, 6 a2 – 37 a + 6 = 0
or, 6 a2 – 36 a – a + 6 = 0
or, 6 a(a – 6) – 1 (a – 6) = 0
or, (6 a – 1) ( a – 6) = 0
or, a = 6 or 1/6
Now,
x + 2 = 6 or, x + 2 = 1
x + 1 x + 1 6
⇒ 6x + 6 = x + 2 ⇒ 6x + 12 = x + 1
⇒ 6x – x = 2 – 6 ⇒ 6x – x = 1 – 12
⇒ 5x = – 4 ⇒ 5x = – 11
⇒ x = – 4/5 ⇒ x = – 11/5
Vedic Method:
Look at the LHS; you observe that the LHS is the sum of two
reciprocals. The Vilokanam Vedic Sutra simply tells us to break the
Example: Solve, x + 2 + x + 1 = 37
x + 1 x + 2 6
x + 2 + x + 1 = 37 = 6 + 1
x + 1 x + 2 6 6
Now, equate either of the LHS parts to both the term of RHS
and solve them to find the value of x.
x + 2 = 6 or, x + 2 = 1
x + 1 x + 1 6
⇒ 6 x + 6 = x + 2 ⇒ 6x + 12 = x + 1
⇒ 6x – x = 2 – 6 ⇒ 6x – x = 1 – 12
⇒ 5x = – 4 ⇒ 5 x = – 11
⇒ x = −4/5 ⇒ x = −11/5
Example: Solve, x + 2 − x + 3 = 15
x + 3 x + 2 4
x + 2 − x + 3 = 15 = 4 − 1
x + 3 x + 2 4 4
Now, equate either of the LHS part to both the term of the RHS
and solve them to find the value of x.
Quadratic Equations 221
x + 2 = 4 or, x + 2 = 1
x + 3 x + 3 4
⇒ 4 x + 12 = x + 2 ⇒ 4x + 8 = x + 3
⇒ 4x – x = 2 – 12 ⇒ 4x – x = 3 – 8
⇒ 3x = − 10 ⇒ 3 x =− 5
⇒ x = −10/3 ⇒ x = − 5/3
60 = 2 x 30
= 3 x 20
= 4 x 15
= 5 x 12
= 6 x 10
Now, find the sum of the squares of these factors and check
when the sum is 169, the value of the numerator.
Now, equate either of the LHS parts to both the term of the RHS
and solve them to find the value of x.
Example: Solve, 3 x + 7 − 2x −9 = 56
2 x − 9 3x + 7 45
45 = 3 x 15
= 9 x 5
152 − 32 > 56
92 −52 = 56 (Numerator)
Hence, 56 = 9 − 5
45 = 5 9
3 x + 7 − 2x −9 = 56 = 9 – 5
2 x − 9 3x + 7 45 5 9
Now, equate either of the LHS parts to both the terms of the
RHS and solve them to find the value of x.
Quadratic Equations 223
3 x + 7 = 9 or, 3 x + 7 = 5
2 x −9 5 2 x −9 9
⇒ 15 x + 35 = 18 x – 81 ⇒ 27 x + 63 = 10 x −45
⇒ 15 x – 18 x = – 81– 35 ⇒ 27 x – 10x = −45 −63
⇒ −3x = − 116 ⇒ 17 x = −108
⇒ x = 116/3 ⇒ x = −108/17
2 + 3 = 4 + 1
x+2 x+3 x+4 x+ 1
N1 + N2 in LHS = 2 + 3 = 5
N1 + N2 in RHS = 4 + 1 = 5
Since they are equal, we have to equate the sum of the
denominator equal to zero.
D1 + D2 = 0
⇒x + 2 + x + 3 = 0
⇒ 2 x + 5 = 0
⇒ x = −5/2
Therefore, two roots are x = 0 and x = −5/2
Example 2: a + b + b + c = 2b + a + c
x+ a+ b x + b + c x + 2b x + a + c
Solution: Ratio of constant term in LHS =
a + b + b + c = 1 + 1 = 2
a + b b + c
Ratio of constant term in RHS =
2b + a + c = 1 + 1 = 2
2b a + c
Since the ratio of LHS and RHS are the same, x = 0 (By Sunyam
Anyat Vedic Sutra). The second root will be extracted using another
Vedic Sutra – Sunyam Sam samuccaye.
N1 + N2 in LHS = a + b + b + c = a + 2b + c
N1 + N2 in RHS = 2b + a + c = a + 2b + c
Since they are equal, we have to equate the sum of the
denominator equal to zero.
D1 + D2 = 0
⇒x + a + b + x + b + c = 0
⇒ 2 x + a + 2b + c = 0
⇒ x = − (a + 2b + c)/2
Hence, two roots are– x = 0 and x = − (a + 2b + c)/2
Quadratic Equations 225
Sunyam Sam Samuchchaye ('kwU;e lkE; leqPp;s)
N1 + N2 = D1 + D2 = 0
N1 + N2 = 3x+4 + 5x + 6 = 8x + 10
D1 + D2 = 6x + 7 + 2x + 3 = 8x + 10
⇒ 8x + 10 = 0
or, x = –5/4
N1 – D1 = N2 – D2 = 0
⇒ N1 – D1 = 3x + 4 – 6x – 7 = 0
⇒ –3x – 3 = 0
⇒x = 1
Hence, the two roots of the above equation are x = –5/4 and -1.
N1 + N2 = D1 + D2 = 0
N1 + N2 = 3x+ 6 + 5x + 4 = 8x + 10
D1 + D2 = 6x + 3 + 2x + 7 = 8x + 10
⇒ 8x + 10 = 0
⇒ or, x = –5/4
N1 – D1 = N2 –D2 = 0
⇒ N1 – D1 = 3x + 6 – 6x – 3= 0
⇒ – 3x + 3 = 0
⇒x = 1
or, N2 – D2 = 5x + 4 – 2x – 7 =0
⇒ 3x – 3 = 0
⇒ x = 1
Hence the two roots of the above equation are x = –5/4 and 1.
Quadratic Equations 227
17
Introduction
Casting out Nines literally means to throw nines. Now let us focus
on its working.
• Add the digits of a number across, dropping out 9, to
get a single figure. If it is not a single figure, add the
digits obtained so as to get a single figure less than 9.
• 9 is not taken into account in this process, as a digit
sum of 9 is the same as a digit sum of zero.
• If the number is made up of all 9’s and/or all sub-additions
of 9’s, then its number digit is zero.
• In subtraction, while applying this rule for verification,
you may encounter a negative number. Add 9 to the
negative number to make it positive. Example – if your
answer is – 5 , convert it to positive integer by adding
9, i.e. – 5 = – 5 + 9 = 4
Solution:
= 4 + 3 + 8 = 15
Digit sum of 15 = 1 + 5 = 6
2 3 4 5 6 8 9
Rule:
Rule:
Rule:
Example: Verify, 8 5 9 4 2 x 3 0 5 4 = 2 6 2 4 6 0 8 6 8
The casting out nines method described in the very beginning will
suffice to check the division operation effectively.
Rule:
Verification:
Verification:
Verification:
Rule:
Number 1 2 3 4 5 6 7 8 9
Square 1 4 9 16 25 36 49 64 81
DS 1 4 0 7 7 0 4 1 0
Verification:
Verification:
Verification:
Verification:
Rule:
Find the DS of the cube of the number and check whether the result
obtained after cubing the number is the same as the result you
have extracted. Always remember that the final Digit Sum of the
LHS and RHS should be of one digit. Moreover, the combination
of 9 and multiples of 9 should be discarded at very first sight,
so that the journey to reach the final digit sum becomes easier.
For the convenience of readers, the digit sum of cubes from 1
to 9 is given here.
Number 1 2 3 4 5 6 7 8 9
Cube 1 8 27 64 125 216 343 512 729
Digit Sum 1 8 0 1 8 0 1 8 0
Verification:
LHS = Digital sum of (13)3 = (1 + 3) 3 = 1
RHS = Digital sum of 2197 = 2 + 1 + 9 + 7 = 1
LHS = RHS
Result verified
Verification:
Verification:
Trigonometry
Introduction
a c
C b A
TRIGONOMETRIC RATIO VALUE VEDIC TRIPLET CASE
Sin A a / c 1st value / 3rd value
Cos A b/c 2nd value / 3rd value
Tan A a/ b 1st value / 2nd value
Cot A b /a 2nd value / 1st value
Sec A c / b 3rd value / 2nd value
Cosec A c / a 3rd value / 1st value
In the above table, the side a, b and c are taken as triplet. You
must be remembering the famous Pythagoras theorem which
states – ‘In a right angle triangle the square of hypotenuse is
equal to the sum of square of other two sides.’
2 2
Hypotenuse = Perpendicular + Base2
c2 = a2 + b2
a b c
3 4 5
5 12 13
6 8 10
7 24 25
8 15 17
9 40 41
9 12 15
10 24 26
11 60 61
12 15 18
3n 4n 5n
Trigonometry 241
angle triangle, two of its legs are given as 3 and 4 and you need
to find its largest leg i.e. Hypotenuse. Traditionally, what will
you do? You will use Pythagoras’ theorem to find the third side.
On the other hand, if you are familiar with the above triplets,
then in a fraction of a second you can say that the length of
the third side is 5.
Now let us see how the triplet helps us to compute the values
of other trigonometric ratios. From the above table on triplets,
you can find t if–
a = 8 b = 15 then c = 17.
Now move to the trigonometric table and find the value of other
trigonometric ratios in no time.
Trigonometry 243
Sin A = a / c = 8 / 17
Cos A = b / c = 15 / 17
The above example are enough to prove that the triplet method
of Vedic mathematics is interesting, easy to understand and time
saving. Now, let us extend the value to twice the angle.
Take an example.
Traditional method–
720 1519 1681
So,
Cos2A = 2nd value / 3rd value = 1519 /1681
We have so far seen the triplets for angle A and 2A – let’s extend
it for 3A to find the value of Sin 3A, cos 3A and tan 3A.
Trigonometry 245
Example: If tan A = 7/ 24 find sin 3A and cos 3A ?
Solution: The triplet for angle A –
Tan A = 7/24 = 1st value / 2nd value
If the triplet for the angle A is a b and c then the triplets for
A/2 is–
a b + c and (b + c)2 + a2
a b + c and (b + c)2 + a2
12 5 + 13 182 + 122
12 18 468
12 18 6√13
Angle a b c
A a b c
2A 2 ab b – a2
2
c2
3A 3ac2 – 4 a3 4b3 – 3 bc2 c3
A/2. a b+c (b + c)2 + a2
The above table will help you immensely to find the different
trigonometric ratios with ease and thus save your precious time.
Once you are well equipped with the method of finding square
and cube you can find the calculation involved in sin 3A, cos
2A etc quite easily. Now let us extend the concept of triplets in
finding the value of compound angles.
But before that, you have to find cos A and cos B and put these
values in the desired formula. Now let us see how the Vedic
method helps you to find the value of compound angle.
If the triplets of angle A is x, y, z and triplet for B is X, Y
and Z, then the triplet for the angle A+B is given by:
Trigonometry 247
A x y z
B X Y Z
A+ B yX + x Y yY – xX z Z
Example: If sin A = 3/5 and sin B = 8/17, then find the value
of Sin (A+B) and cos (A+B)
Solution: Let us first draw the triplet table for the angle A and B
A 3 4 5
B 8 15 17
A + B 3 x 15 + 4 x 8 4 x 15 – 3 x 85 x 17
= 77 = 36 = 85
Example: If sin A = 7/25 and sin B = 8/17 then find the value
of Sin (A+B) and cos (A+B)
Solution: Let us first draw the triplet table for the angle A and B
A 7 24 25
B 8 15 17
A + B 7 x 15 + 24 x 8 24 x 15 – 7 x 8 25 x 17
= 297 = 304 = 425
A x y z
B X Y Z
A − B xY −Xy xX + yY z Z
Example: If sin A = 7/25 and sin B = 8/17 then find the value
of Sin (A −B) and cos (A−B)
Solution: Let us first draw the triplet table for the angle A and B
A 7 24 25
B 8 15 17
A − B 7 x 15 −24 x 8 24 x 15 + 7 x 8 25 x 17
= −77 = 416 = 425
Example: If sin A = 3/5 and sin B = 8/17 then find the value
of Sin (A−B) and cos (A−B)?
Solution: Let us first draw the triplet table for the angle A and B
A 3 4 5
B 8 15 17
A − B 3 x 15 − 4 x 8 4 x 15 + 3 x 8 5 x 17
= 13 = 84 = 85
Hence, sin (A− B) = 13/85
Cos (A − B) = 84 /85
Hope you have enjoyed the journey of Trigonometry using the
triplet method and the traditional method. A little practice and
memorization of the triplet will ease the process of calculation
almost one tenth that of the traditional method.
Trigonometry 249
Questions for Practice
1. Addition
1) 6 6 6 2) 6 9 8 0
– 4 8 2 – 5 7 9 8
3) 9 8 6 7 9 8 5 4 6 4) 8 7 0 5 6 9 7 6 9
– 6 8 9 7 9 7 9 7 9 – 8 2 4 8 9 3 9 8 7
5) 4 7 9 8 6 7 6) 7 5 6 2 7 9
– 3 6 4 7 8 0 – 2 9 8 7 0 5
7) 9 2 6 7 9 9 8 2 5 8) 4 6 8 7 9 3
– 5 4 9 8 9 9 9 9 9 – 3 5 9 7 0 2
9) 4 0 0 0 0 0 0 0 10) 7 0 0 0 0 0 0 0 0
– 7 8 8 9 6 3 2 – 8 8 8 8 8 8
3. Multiplication
a) 36 x 34 b) 87 x 83 c) 128 x 122
d) 112 x 998 e) 688 x 988 f) 107 x 95
g) 9997 x 9998 h) 252 x 248 i) 148 x 149
j) 506 x 494 k) 2487 x 9999 l) 87904 x 99999
m) 8284 x 99 n) 43427 x 9999 o) 144x 9999
p) 279 x 331 q) 7628 x 4287 r) 144 x 66
s) 248 x 128 t) 82765 x 42897 u) 5628 x 3047
v) 983 x 994 x 1005 w) 1003 x 1007 x 1009
x) 876 x 602 y) 56 x 66 x 65 x 64
z) 992 x 994 x 996 x 998
5. Multiplication in Algebra
6. Division
a) 15 b) 25 c) 35 d) 65 e) 85 f) 95
g) 105 h) 115 i) 125 j) 44 k)76 l) 54
m) 29 n) 168 o) 185 p) 97 q) 99 r) 109
s) 168 t) 2356 u) 6254 v) 2591
8. Square Root
10. Cube
Find the cube of the following by using the appropriate Vedic Sutra.
a) 13 b) 19 c) 25 d) 36 e) 46 f) 54 g) 69
h) 87 i) 104 j) 113 k) 208 l) 315 m) 86 n) 97
o) 98 p) 16 q) 94 r) 96 s) 97 t) 102
a) 23 b) 17 c) 14 d) 98 e) 42
f) 21 g) 24 h) 73 i) 76 j) 55
a) x3 + 13x2 + 31x – 45
b) x3 – 2x2 – x + 2
c) x3 – 3x2 – 9x – 5
d) y3 – 2y2 –29y – 42
e) x3 – 10x2 – 53x – 42
f) x3 – 23x2 + 142x – 120
g) y3 – 7y + 6
Solve for x
18. Trigonometry
Answer
Chapter 1: Addition
Chapter 3: Multiplication
Quotient Remainder
a) 5 262
b) 2390 722
c) 1 3457
d) 1 2204
e) 233 3711
f) 22 22068
g) 124 111
h) 10 9
i) 21 4216
j) 22 135
k) 24 24
l) 13219 1866
m) 5 8207
n) 124 0
Chapter 7: Square
a) 15 b) 62 c) 32 d) 12 e) 47 f) 23 g) 17 h) 78
a) x = 42/13, y = 0
b) x = 1; y = 1
c) u = 1/5; v = ¼
d) x = 2; y = 1
e) x = 6; y = 0
f) x = 1; y = 8
g) x = 3; y = 2
h) x = –19/4;y = 9/2
i) x = 1/3; y = ½
j) x = 36; y = 27
a) (x – 1) (x + 5 ) (x + 9)
b) (x – 2) (x – 1) ( x +1)
a) x = –6, –4
b) x = ¼, 1/13
c) x = 4, 13/2
d) x =3, –1/2
e) x = 2; y = –3
f) x = 4±√3
g) x = ½, 4/3
h) x = 3, −4
i) x = 3, –9/2
j) x = 0, 1
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