Interesting Maths

CHAPTER - 1
Diamond Of Vedic Mathematics- Number “9”
Friends, I am sure that the ‘Magical Methods’
explained in this book are very easy to work with and you
will be thrilled after learning & understanding these
methods. Try to teach these methods to as many people as
you can.
To develop the interest in ‘Vedic Mathematics’, I
take an example of ‘Table - 99’.
Have a view at following example :-
99 - Multiplicand
x 20 - Multiplier
19/80 - Product
The clue to the answer is found by the multiplier 20 only.
The product is divided into two parts by a slash ( / ) sign.
L.H.S =19 & R.H.S. = 80
By more observation you can see that L.H.S = Multiplier-1
(i.e.20-1) & R.H.S. = 100-20 (Multiplicand 99 is close to
100, so we take 100 as base).
If you have understood the above principle hats off, you
can now Mentally Calculate 99 into any 2 digit number.
Let us see some more example.
99 99 99
X 30 x 85 x 63
29/70 84/15 62/37
9 (30-1) (100-30) 9 (85-1) (100-85) 9 (63-1)
(100-63)
With two example you will master the table 99.
99 99
X 08 similarly x 95
07/92 94/05
L.h.s =8-1=07 where o is called zero deficiency r.h.s. =
95-1=94
R.h.s. =100-8=92
r.h.s. =100-5=05
Thus to conclude i may, vedic mathematics is magic until.
You understood & it is mathematics there after. The sutra
on which the above problem works is explained in the
chapter multiplication. Now we go to individual
applications of each sutras.
99999 999
9999
X12345 x678
x 2566
12344/87655 677/322
2566/7435
Nikhilam sutra :- all from 9 & last from10
Chapter
Cube roots of exact cubes
The technique used in this chapter for extraction of cube
roots is very simple & interesting. You just need to
memories the following cubes & rest all is easy for you.
cube last digit
13 = 1 1
23 = 8 8
33 = 27 7
43 = 64 4
53 = 125 5
63 = 216 6
73 = 343 3
83 = 512 2
93 = 729 9
03 = 0 0
Silent features for finiding cube root:-
1. The cube of numbers ending in 0,1,4,5,6 &9
have their cube roots also ending in the same digits
respectively.(b)
2. The cubes ending in2,3,7& 8 have their cube
roots also ending in 8,7,3,&2 respectively.(b)
3. The numbers og digits in a cube root of a
numbers in the same as like number of 3 digit groups in the
given numbers including a single or a two digit group if
there in any.
Thus we start from the right hand side of the cube &put a
comma when the 3 digits are over.
Eg:- thus 117649 will be written as 117,649 & thus will
have 2 digits in its cube-root
12167 will be written as12,167 & thus will have 2 digits in
its cube-root
12977875 will be written as12,977,875 & thus will have 3
digits in its cube-root
125 has 3 digits only, 20 there will be 1 digit in its cube
root i.e.5 as you see above.
Thus to summaries we say that the grouping of digits is
done from right to left for non decimal & from left of the
decimal towards right in the case of decimals.
Let us take a example to find cube root.
59319
Starting from the right hand side of the numbers we put
comma after every three digits are over. The numbers of
groups formed specify the number digits in the cube – root.
Thus we writ 59,319 & so there are 2 digits in the cube –
root.
Now first see the last digits of the numbers i.e 9 going
though the first two points of salient features & the table
provided above, you can now easily guess. The last digit of
the cube root. Thus we get 9 as last digit of cube root
Now take the first group i.e. 59
From the above table just find out cube of number which
is less than 59. Since 27 (cube of 3) in ten than 59 we get 3
as the first digit of cube root.
Thus the equation looks as follows
59 319
3 9
Cube root is 39.
Case ii:-
474, 552
We write the numbers as 474, 552.
Since there are 2 groups so there are two digits in cube
root.
The last digit of the numbers is 2, reading the first two
points of salient features & the table provided. We can now
easily judge that 8 is the last digit of cube root.
The first groups is 474. From the table provided, we see
that 373 (cube of 7) lets below 474. Thus 7 is the first digit
of cube root.
474 , 552
7 8
Cube root is 78.
Now practice the following examples on your own.
Case iii:-
614 , 125
8 5
Cube root is 85.
Case 4:-
2 , 197
1 3
Cube root of 2197 is 13.
Case v:-
19 , 683
2 7
Cube root of 19683 is 27
Case vi:-
42, 875
3 5
Cube root of42875 is 35.
Case vii:-
884, 736
9 6
Cube root of 884736 is 96.
Friends , i believe that you are now in a condition to easily
calculate mentally the cube root of a 6 digit number in just
6 records!!! To make a note above technique is valid only
for exact cubes. But do not worry, the technique for
calculating cube roots of any numbers is get to some & will
be expounded at a later stage.
When you might be wondering , whether there is a one line
method to find cube –root of any general numbers (may or
may not be a exact cube) bt the vedic sutras, for your
information the answer ia yes! And will be dealt with, at an
appropriate place, in a later stage.
Simultaneous equations
Chapter-
Students & professionals, both come across simultaneous
equations frequently. The current method of simultaneous
equations method by which we frame new equations
involving only x or y coefficients is tiresome & can load
to manual errors. The vedic system use the cross
multiplication method, which gives one line mental answer
for the coefficient thus saving time & errors .
Let us start with an example.
3x – 4y = 4
2x – 3y = 6
Here we apply the ‘paravartya’ rule which means transpose
& divide. It enable us to calculate value of ‘x’ by mere
mental dritrmetic. X= numerator for numerator.
Denominator
We adopt the following procedures .
3x – 4y = 4
2x – 3y = 6
Numerator=(coefficients of y in 1st row x constant in 2nd
row)-
(coefficients of y in 2nd row x constant in 1st row)
Note: the coefficients are taken along with this signs (+or-)
intact
\numerator =(-4x6) – (-3x4) = -24 +12 =-12
For denominator we adopt like following procedure.
3x – 4y = 4
2x – 3y = 6
Denominator = (coefficients of y in 1st row x coefficient
of x 2nd row) –( coefficients of y in 2nd row x coefficients
of x 1st row)
Note: the coefficients are taken along with this signs (+ or -
) intact.
Denominator = (-4x2) – (-3x3) =
= -8 + 9
=1
\x=numerator =-12 = -12
denominator 1
Ituting the value of x in 1st eqn we can now find the value
of y.
3(-12) -4y=4
-36-4y=4
-4=4+36=40
Y=-10
\x=-12 & y =-10
Ii) 6x-3y=3
4x+2y=14
X = (-3x14) – (2x3) = -42-6 = -48 =2
(-3x4) – (2x6) = -12-12 = -24
x in 1st eqn we get value for y.
6 x 2-3 x y =3
\y= -9 =3
-3
\x=2 & y=3
Iii) 2x-8y=20
3x+3y=-3
X = (-8x-3) – (-3x-20) = +24-60 = -36 = 2
(-8x-3 ) – (-3 x 2) = -24+6 = -18
. In 1st eqn
2 x 2 -8y = -20
-8y - -24
Y = 3
\x=2 & y=3
Sunyam anyat: if one is in ratio, the other one is zero. Lets
us clear the above sutra by an example.
3x + 4y = 6
6x + 3y = 12
In the above two equators, we see that the x-coefficients
are the same ratio to each other as the independent terms
are to each other. Thus by the above sutra, if one
coefficient is in ratio, the coefficient in ‘ 0.’
Thus y=0
value of y in 1st eqn we getx
3x + 4 x 0 =6
\x =2
Ii) 43x + 86y =43
86x + 72y = 86
Since x is in same ratio to each other as the independent
terms are to each other, y=0.
\ 43xx + 86 x 0 =43
\ x=1
\ x=1 & y=0
Iii) 142 x + 72y = 216
799x + 216 = 648
Since y is in same ratio to each other as the independent
terms are to each other we conclude that x=0
valve of x in 1st eqn
142 x 0 + 72y = 216
\y =3
\ x=0 & y=3
Sankalana – vyavakalanalhyam :- by addition &
subtraction
The above up sutra is helpful wherever the x& y
coefficients are found interchanged by simply adding or
subtracting the two equations give the values of (x+y) &
(x-y)
Repeating the above process one more time gives us values
of x & y.
23x – 33y = -53 (1)
33x – 23y = -3 (2)
By adding the above two equation 1 &2 we get
56x – 56y = -56
\56 (x-y) = -56
X- = -1 (3)
& by subtracting the above two equation 1&2 we get
-10x – 10y = -50
-10 (x+y) = -50
\x+y = 5 (4)
Adding equation 3 & 4 we get
2x = 4
\ x = 2 & y = 3
Introduction about vedic mathematics:-
The main research on vedic mathematics is done by his
holies jagadguru sankaracarya , sri bharti krishna , tirthaji
maharaja of govardhana matha puri (1884-1960). He had
used 16 sutras and sub sutras (corollaries) which are listed
in this chapter.
Sutra literally means ‘ thread’ but tirthaji maharaja
employs the word ‘ aphorism’ because it does not show
how a calculation is to be made but only throws up a
pointer or direction in which the calculation can proceed.
Thus the same sutra can be employed for a variety of
applications. The sutras are original in sanskrit and so the
english version of one which are used in this book are
given below
Sutra (word-formula) used in
Sub-sutra (corollary)
1. Ekadhikena pu rvena multiplication,
divisibility, recurring
“by one more than the decimals.
privious one”
anurupyrna multiplication,
division, cubing
(corollary)
“proportionately”
2.nikiilam multiplication,
division
navataxaranam
dasatah
“all from nine & last ten”
Sir multiplication, division
(corollary) multiplication
‘remainder remains
Constant
3. Urdhava-tiryagbhyam multiplication
‘vertically & cross- division
wise’
4. Paravartya yojayet division
‘transpose & divide
5.yavadunam multiplication
tavadunam(corollary)
“whatever the extent
of its deficiency lessen
It stil further to that
Very extent.
6.antyayordasake api multiplication
(corollary)
A.squaring of a number
whose last digits add
to 10 and whose
previous part is
exactly the same
7. Yavadunam squaring
‘deficiency’ cubing
8. Desanyankena recurring decimals
caramena
“the remainder
by the last digit”
vilokanam (corollary)
‘by inspection or division
observation’
9.ekany-------- multiplication
‘multiplication
whenever the
multiplied- digit
entirely of 9’5”
10.dhvajanka division
‘on top of the flag’
Ekadhikena purvena – “by one more than the previous
one”
Valgar fractions.
By more observation of the sutra we see that it has used the
preparation “by” at the start at indicate that the arithmetical
operation prescribed is either multiplication or division.
For in the case of addition & multiplication, ‘to’ and aaa
respectively would have been the appropriate preposition
to use thus as the matter of selection we use ‘division’
technique for
take example of 1/19
=1 dividend & 19 divisor.
The last digit of the denominator in this case being 9 & the
previous one being 1 “one more than the previous one”
evidently means 2.
By the vedic one line method we write the answer as
follows-
1 = 0.0526315789473 68421 in vedic form we
write as follows
19
1 = 0.1 = 10 05 12 06 03 11 15 17 18 quotient
reminder
quotient
09 14 7 13 16 8 4 2 1 reminder
Can you believe it !!! Even the calculation through which
you are checking has ten digit answer & we re upto 18
digit answer.
Explanation:-
I. Since the denominator has a single 9, we shift the
decimal in the numerator by one place to the left making
number = 0.1
II. Drop 9 from the denominator and increase the
penultimate digit (i,e.1) of denominator by one so that the
vulgar fraction now reads 0.1.
2
III. We now divided 0.1 by 2 which is a very simple and
working divisor.
IV. On dividing 0.1 by 2 we get quotient q=0 &
remainder r=1. We therefore, set 0 down as the first digit of
the quotient and prefix the remainder 1 to that very digit of
the quotient and thus obtain 10 as our next dividend. (10)
V. Dividing this 10 by 2, we get 5 as the record digit of
the quotient and there is no remainder to the prefixed there
to, we take up that digit 5 itself as our next dividend (1.05)
VI. Do the next quotient digit is 2 and the remainder is
1. We therefore put 2 down as the third digit of quotient
and prefix the remainder 1 to that quotient digit 2 and thus
have 12 as our next dividend (1 .05 1 2)
VII. This gives us 6 as quotient digit and remainder is 0.
So we set 6 down as the fourth digit of the quotient and as
there is no remainder to the prefixed thereto, we take 6
itself as our next digit for division which gives the next
quotient digit as 3.(105 12 6 311)
VIII. Dividing by 2 & prefixed to quotient we get new
dividing by 2 we get 5 as quotient and 1 as remainder. So
the new dividend is 15. 1 0 5 1 2 6 3 1 1 1 5
IX. Carrying this process of straight, continuous
division by 2, we get 2 as the 17th quotient-digit and 0 as
remainder.
X. Dividing 2 by 2, we get 1 as 18th quotient digit and 0
as remainder but it is repletion of what we started with.
Thus the decimal begins to repeat itself from here. So we
stop the mental division process and put down the usual
recurring symbols (dots) on the 1st & 18th digit to show that
the whole of it is a circulating decimal.
A further short-cut :- let us put down the first 9 digit of
the answer in one horizontal row above and the other 9
digits exactly below the first 9 digits.
0.052631578
947368421
999999999
By more observation we see that each set of digits in the
upper row and lower row total 9. Thus it means that when
just half the work has been completed by the great vedic
one line method the other half is mechanically available to
us by subtracting from 9 each digits already obtained and
this means a reduction of work still further by 50%
But now you should know when the t is exactly half
finished. Do here it is as soon as we reach the difference
between the numerator & denominator (i,e. 19-1=18), we
shall have completed exactly half the work!!!. So in above
example when we reached 18 as dividend we stop the
work, thus if you see now, the vulgar fractions such as 1 ,
1
19 29
1 are solved in one simple line & oven young boys man
do it.
19
Case 2:- 1 = 0.1 = 0.10131424 8 22 17 25 18 6 22 20 628
9 6 5 5 1 7 2 4 1 3 8 9 3 1
1. Drop 9 from the denominator and increase the
penultimate digit (i.e. 2) of denominator by 1 so that the
vulgar fraction now reads 0.1/3.
2. Dividing 0.1 by 3 we get q=0 & r=1. Thus the
new dividing is 10 which is again divided by 3.
3. Continue the process up to we get the dividend
as 28 because numerator – denominator = 29-1=28
and now we know that after we get 28, the further digit are
obtained by merely subtracting the 14 digits each by 9.
Case 3:-
1 = 0.1 = 0.1011213253854945055055167187696747
89 9
25720880887 9 8 8 7 6 4 0 4 4 9 4 3 8 2 0 2 2
47191
Note:- if you notice that in the above examples the last
answer is found to be 1. Product of the last digit of the
denominator and the last digit of the decimal equivalent of
the fraction in question must invariably and in 9.
Therefore, as the last digit of the denominator in this case
is 9, it automatically follows that the last digit of decimal
equivalent is found to be 1 ( so that the product of the
multiplicand and the multiplies concerned may end in 9)
Let us consider some more auxiliary fractions
63 = 6.3 = 6.3 = 0.74 45 33 52 103
139 13.9 14
Friends, if you have carefully studied the earlier case
studies, then the above example is self explanatory. Do
practice the following examples.
65 = 6.5 = 6.5 0.94 106 87 36 82
139 13.9 14
83 = 8.3 = 8.3 0.85 105 07 70 104 146
149 14.9 15
All the above cases have denominator ending in 9 this does
not mean that the above rule is applicable only for above
cases. For your information & pleasure the name rule
applies for digits in denominator ending with 8,7,6 etc, but
with a plight change.
Denominator ending with 8:-
Please observe carefully the following example & you will
understand the rule on your own.
+4 +2 +5 +6
63 = 6.3 = 6.3 = 0.34 82 95 106 quotient
148 1 4.8 15
remainders
In case of denominator digits ending with 8 (1less than9),
the steps are as follows:
1. Placing of the remainder in front of the quotient
remains the same as explained in the earlier cases.
2. In the quotient digit, 1 time (9-8=1) of the
quotient digit is added to every step and divided by the
divisor.as in the above case. We found our first quotient
q1=4 and remainder r1=3. Our gross dividend comes out to
be 34 in which we add the quotient digit as per rule 2, to
make it 38 and then divide it by 15.
In the next step q2=2 & r2= 8, thus our 2nd gross dividend
will be 82 + q2 = 84, divide this by 15. Continue the above
steps to find the solution to the required number of decimal
places you desire.
Some more examples will make you more comfortable
with rule.
+5 +1 + 6 + 9
61 = 6.1 = 6.1 = 0. 15 81 106 49
118 11.8 12
89 = 8.9 = 8.9 = +4 +4 +9 +4
198 19.8 20 0.94 184 89 184
Denominator ending with 7 :-
Let me clear the rule by a simple example
+8 +10 +2 +9
53 = 5.3 = 5.3 = 0. 54 25 112 69 quotient
117 11.7 12 remainder
By more observation you can guess that for the above case
the quotient digit is multiplied by 2 (9-7=2) and added to
the quotient
Q1 = 4
R1 = 5
New divided = 54 + 201 = 54 + 8 = 62
And now divided 62 by 12 & proceed as earlier method.
Denominator ending with 6 :-
+15 +27 +15 + 6
75 = 7.5 = 7.5 = 0. 105 39 15 42
126 12.6 13
For the above case, the quotient digit is multiplied by 3 (:9-
6=3)
Q1 = 5, r1 = 10
:new dividend = 105 + 3q = 105 + 15 = 120
Rest all the steps are same as before.
Denominators ending in 1:-
Let us take an example
3 4 0 4 2
93 = 93-1 = 92 = 9.2 0. 86 135 89 105 67
141 141-1 140 14
We find reduce 1 from both numerator & denominator. The
gross dividend will be prefixing the remainder to the
compliment from 9 of each quotient digit.
Thus in the above example we get 8 as remainder & 6 as f
quotient. Compliment of 6 from 9 is 3 and then the gross
dividend is 83 and then to be divided by 14.
Now we get r2 = 13 & q2 = 5 . Compliment of q2 from 9 is
4 thus giving second gross dividend as 134
2 4 3 2 4 3
84 = 84-1 = 83 = 8.3 = 0. 67 75 86 67 75 86
111 111-1 110 11
Friends, we have discussed in detail the auxiliary & vulgar
fraction & i belive that you might be thrilled by doing the
above examples, to add to your pleasure & information, the
next chapter de with division technique of large numbers
and it is my word that if you do there both chapter
thoroughly then you people will never use calculators for
division in yours day to day practice.
Division “gowning gem of vedic mathematics”
The magical method explained in this chapter in one of the
crowning beauties of vedic mathematics. It is a ap
beautiful & very easy method to understood and thus will
make you put your fingers in mouth seeing the “vedic one
line mental answer”
The conventional format for division you people are using
is as mentioned below.
division ) dividend ( quotient
remainder
The vedic format for division is as follows.
Flag
Divisor
Dividend
Remainder
Quotient
Let me take a example to explain the above vedic format
which will reduce considerably the lab of operation as
well as time of the conventional format.
Eg. Divide 2,35,796 by 46
As per vedic division format we get
6
4
23
5
5
3
11
7
1
27
9
3
:
:
0
6
3-remainder
5 1 2 6 . 0- quotient
Please study the following steps carefully.
1. The divisor 46 is split into two parts 4 & 6.
We put down only the first digit i.e.4 in the divisor column
and put the other digit i.e. 6 “on top of the flag” by the
“dhvajanka sutra” as shown above.
2. Thus doing the above step, we are left by a
single digit divisor i.e. 4 and the entire division is to be
done by 4!
3. Number of digit on the remainders side is
always equal to the number of digits in the flag. This
decides the location of decimal point! As one digit has
been put on the top, we allot one place at the right end of
the dividend to the remainder portion of the answer and
mark it off from the digits by a vertical dot.
4. Since 2 cannot be divided by 4, we take 23
of the dividend to be divided by 4. Thus we get the
quotient q1=5 & remainder r1=3. We put 5 down as first
quotient digit and just prefix the remainder 3 up before the
5 of dividend.
5. Thus we get the new dividend as 35 from
this, we however , deduct the product of the indexed 6 (
flag digit) and the first quotient digit (q1) 5 i.e. 35- (6x5) =
5. The remainder 5 as circled in our actual net dividend.
6. Now 5 is divided by 4 which gives us q2 = 1
& r2=1, to be placed in their respective places as done in
above steps. Now we get 17 as gross dividend through
which we subtract product of flag digit & record quotient
digit i.e.1. I.e. 17 – (6 x 1 ) = 11 thus like 2nd net dividend
in 11 to be divided by 4
7. On dividing 11 by 4 we get
q3 = 2 & r3 = 3 dividend = 39
: 3rd net dividend = 39 – ( 6x 2) =27
8. Divide 27 by 4, we get
q4=6, r4= 3 & dividend = 36
: 4th net dividend =36 – (6 x 6) =0
9. Divide 0 by 4, we get
q5=0, r5 = 0.
as remainder is 0 we say that 2,25,796 is exactly divisible
by 46 & the answer is 5126.
Divide
25,116 by 42
41 33 0
2 25 1 1 : 6
4 5 5 : 1
5 9 8 . 0
1) Divisor 42 split into two parts. First digit
4 is taken as divisor & second digit 2 in kept as flag.
2) Since 2 cannot be divided by 4 so 25
taken as first dividend.
3) Divide 25 by 4. We get 01=6 & r1 = 1.
Now dividend is 11.
\ net 1st dividend = 11-(6x2) = -1.
Rule:- if the net divided obtained is a negative numbers
then reduce 1 from the previous quotient digit & work
again.
4) Thus the first quotient 6 is reduced by 1 for the
above example to give
Q1 = 5 & r1 = 5 \ new dividend = 51
Thus net 1st dividend = 51 – (5x2) = 41.
5) divide 41 by 4
Rule:- the quotient should always be a single digit even it it
comes a two digit number. Thus the largest quotient
number will be 9.
Now 4 goes 10 times for 41, but as per the above said rule
take the quotient as 9.
\q2= 9, r2 = 5 – dividend = 51
Thus net 2nd dividend = 51- (9x2) = 33.
6) divide 33 by 4. Q3 = 8 & r3 = 1 dividend = 16
\3rd net dividend = 16 –(8x2) = 0.
\25,116 = 598.0
Note: the best part of vedic divisor, is that it gives it gives
one line mental answer in which actual division is done by
only 1 digit divisor.
Divide:- 7,92,456 by 93
48 19 9 3 30 21 24 54
3 79 2 4 5 : 6 0 0 0 0
9 7 3 1 0 3 3 3 6
8 5 2 1 . 0 3 2 2
1) 93 split into two parts. The first digit 9 is divisor &
second digit 2 in flag digit.
2)since 7 cannot be divided by 9 so take first dividend as
79. On dividing 79 by 9 we get first quotient * remainder.
Q1 = 8 & r1 = 7 giving new dividend as 72
\net 2nd dividend = 72-8 x 3 = 48
3) on dividing 48 by 9 we get
Q2 = 5 & r2 = 3 \dividend = 35
\net 3rd dividend = 34-5x3 = 19
4) on dividing 19 by 9, we get
Q3 = 2 & r3 = 1 giving dividend = 15
\net 4th dividend = 15-2x3=9
5) on dividing 9 by 9 we get
Q4 = 1 & r4 = 0 - dividend = 6
\net 5th dividend = 6-1x3 = 3
6) on dividing 3 by 9 we get
Q5 = 0 & r5 = 3 - dividend = 30
\net 6th dividend = 30-3x0 = 30
7) on dividing 30 by 9 ,we get
Q6 = 3 & r6 = 3 \dividend = 30
\net 7th dividend = 30-3x3 =21
\ 7,92,456 = 8521.032258
93
Divide :- 1,50,381by 651
20 6 0 0
51 15 0 3 : 8 1
6 3 2 : 0 0
2 3 1 . 0 0
1) the divisor 651 is divided into two parts. The first digit 6
is taken as divisor & other two digit i.e.51 taken as flag
digits.
2) as two parts digit are taken as flag digits, we allot two
places at the right end of the dividend and mark it off by a
semi colon. The decimal is immediately marked in quotient
row below the semi-colon.
3) since 1 cannot be divided by 6, so we take 15 as our first
dividend on dividing 15 by 6 we get
Q1=2 & r1=3 giving next dividend 30.
Rule:- from 30, we subtract 10 the product of the first flag
digit (5) & first quotient digit (2) and get the net 2nd
dividend as 20.
4) on dividing 20 by 6, we get
Q2=3 & r2=2 giving next dividend as 23,
Rule:- from 23 we deduct 17 ( the cross – products of two
flag digits 51 and two quotient digits 2 & 3 i.e.)
5 1 = 5 x 3 + 1 x 2 = 17
2 3
Thus 3rd net dividend = 23-17 =6
5) on dividing the 6 by divisor 6, we get
Q3=1 & r3 = 0 giving dividend as 08
Rule:- same above step, from 08 we deduct 8 (crossproduct
of two flag digits 3 & 1 )
5 1 = 5 + 3 = 8
3 1
Thus 4th net dividend = 0
6) on dividing 0 by 6, we get
R4=0 & q4=0 giving next dividend as 01
Rule:- same as 4 th step, from 01 we deduct 01 ( the cross
– products of flag digits& quotient digits 1&0)
5 1 = 0x5 + 1 = 1
1 0
5th net dividend = 0
Thus we conclude 1,50,381 = 231.00
651
Now i take a very hard problem which you people will face
rarely & so i am sure that if you understand it properly
then you will become master in vedic division technique.
Divide:- 202272 by 258
27 16 3 0
58 2 0 2 2 : 7 2
2 6 11 : 8 3
7 8 4 . 0 0
(1) (9) (7) (1)
1) The divisor 258 is divided into two parts. The first
digit (2) is taken as divisor & other two digit i.e. 58 kept as
flag digits.
2) As two digits are taken as flag digits, we allot two
places at the right end of dividend and mark it off by a
semi – colon. The decimal is immediately marked below
the semi- colon in the quotient row.
3) Since 2 can be divided by 2, so we take our first
dividend as 2. On dividing 2 by, we get
Q1 = 1 & r = 0 giving next dividend as 0.
Now we deduct 5 x 1 = 5 from 0, & we get 2nd net
dividend as -5.
sanket
Digitally signed
by sanket
DN: cn=sanket,
o=Reliance,
c=IN
Date:
2005.05.21
12:17:15 +
05'30'
the end