Faith

You did everything right but in the last but one step

(5/2-2)^2 = (5/2-3)^2

you need to sqare root on both sides

square root of (5/2-2)^2 is 5/2-2

but square root of (5/2-3)^2 is not 5/2-3 it is 3-5/2

as you know that if square root is positive sqare root

Hai friends we are doing some new project.

PROJECT: converting effluent into usable power......

1) filtering the drainage water to form sewage and purifie effluent

2)treating the sewage to get brikets or bio gas or electricity

3)generating electricity from treated effluent(water) by using turbines and generator.

Please ask any questions you have as we demonstrate our project, they may ask any questions.

Best questions will have more salutes.

IF YOU GIVE THIS ANSWER YOU WILL GET THREE SALUTES.

Can vaccum exerts upward pressure in atmosphere?

If so. By how much?

thank u

thank u

thank u

EK?DHIKENA P?RVE?A

The Sutra (formula) Ek?dhikena P?rvena means: ?By one more than the previous one?.

i) Squares of numbers ending in 5 :

Now we relate the sutra to the ?squaring of numbers ending in 5?. Consider the example 25².

Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure 'to multiply the previous digit 2 by one more than itself, that is, by 3'. It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand side) of the result is 5², that is, 25.

Thus 25² = 2 X 3 / 25 = 625.

In the same way,

35²= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;

65²= 6 X 7 / 25 = 4225;

105²= 10 X 11/25 = 11025;

135²= 13 X 14/25 = 18225;

Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

Algebraic proof:

a) Consider (ax + b)² ? a². x²+ 2abx + b².

This identity for x = 10 and b = 5 becomes

(10a + 5)²= a². 10² + 2. 10a . 5 + 5²

= a². 10² + a. 10² + 5²

= (a²+ a ) . 10²+ 5²

= a (a + 1) . 10²+ 25.

Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, -------,9 respectively. In such a case the number (10a + 5)²is of the form whose L.H.S is a (a + 1) and R.H.S is 25, that is, a (a + 1) / 25.

Thus any such two digit number gives the result in the same fashion.

     Example:    45 = (40 + 5)², It is of the form (ax+b)² for a = 4, x=10
                            and b = 5. giving the answer a (a+1) / 25
                                    that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.

b) Any three digit number is of the form ax²+bx+c for x = 10, a ? 0, a, b, c ? W.

Now (ax²+bx+ c)²= a²x⁴+ b²x²+ c²+ 2abx³ + 2bcx + 2cax²

= a²x⁴+2ab.x³+ (b² + 2ca)x²+2bc . x+ c².

This identity for x = 10, c = 5 becomes (a . 10²+ b .10 + 5)²

= a².10⁴+ 2.a.b.10³+ (b²+ 2.5.a)10²+ 2.b.5.10 + 5²

= a².10⁴+ 2.a.b.10³+ (b² + 10 a)10²+ b.10²+ 5²

= a².10⁴+ 2ab.10³ + b².10²+ a . 10³ + b 10²+ 5²

= a².10⁴+ (2ab + a).10³+ (b²+ b)10²+5²

= [ a².10²+ 2ab.10 + a.10 + b²+ b] 10²+ 5²

= (10a + b) ( 10a+b+1).10²+ 25

= P (P+1) 10²+ 25, where P = 10a+b.

Hence any three digit number whose last digit is 5 gives the same result as in (a) for P=10a + b, the ?previous? of 5.

Example : 165²= (1 . 10²+ 6 . 10 + 5)².

It is of the form (ax²+bx+c)²for a = 1, b = 6, c = 5 and x = 10. It gives the answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ?previous?. The answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.

Apply Ekadhikena purvena to find the squares of the numbers 95, 225, 375, 635, 745, 915, 1105, 2545.

ii) Vulgar fractions whose denominators are numbers ending in NINE :

We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion of such vulgar fractions into recurring decimals, Ekadhika process can be effectively used both in division and multiplication.

a) Division Method : Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator and denominator, i.e.,, 19 -1=18 places.

For the denominator 19, the purva (previous) is 1.

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.

The sutra is applied in a different context. Now the method of division is as follows:

Step. 1 : Divide numerator 1 by 20.

i.e.,, 1 / 20 = 0.1 / 2 = .₁0 ( 0 times, 1 remainder)

Step. 2 : Divide 10 by 2

i.e.,, 0.0₀5( 5 times, 0 remainder )

Step. 3 : Divide 5 by 2

i.e.,, 0.05₁2 ( 2 times, 1 remainder )

Step. 4 : Divide ₁2 i.e.,, 12 by 2

i.e.,, 0.0526 ( 6 times, No remainder )

Step. 5 : Divide 6 by 2

i.e.,, 0.05263 ( 3 times, No remainder )

Step. 6 : Divide 3 by 2

i.e.,, 0.05263₁1(1 time, 1 remainder )

Step. 7 : Divide ₁1 i.e.,, 11 by 2

i.e.,, 0.052631₁5 (5 times, 1 remainder )

Step. 8 : Divide ₁5 i.e.,, 15 by 2

i.e.,, 0.0526315₁7 ( 7 times, 1 remainder )

Step. 9 : Divide ₁7 i.e.,, 17 by 2

i.e.,, 0.05263157 ₁8 (8 times, 1 remainder )

Step. 10 : Divide ₁8 i.e.,, 18 by 2

i.e.,, 0.0526315789 (9 times, No remainder )

Step. 11 : Divide 9 by 2

i.e.,, 0.0526315789 ₁4 (4 times, 1 remainder )

Step. 12 : Divide ₁4 i.e.,, 14 by 2

i.e.,, 0.052631578947 ( 7 times, No remainder )

Step. 13 : Divide 7 by 2

i.e.,, 0.052631578947₁3 ( 3 times, 1 remainder )

Step. 14 : Divide ₁3 i.e.,, 13 by 2

i.e.,, 0.0526315789473₁6 ( 6 times, 1 remainder )

Step. 15 : Divide ₁6 i.e.,, 16 by 2

i.e.,, 0.052631578947368 (8 times, No remainder )

Step. 16 : Divide 8 by 2

i.e.,, 0.0526315789473684 ( 4 times, No remainder )

Step. 17 : Divide 4 by 2

i.e.,, 0.05263157894736842 ( 2 times, No remainder )

Step. 18 : Divide 2 by 2

i.e.,, 0.052631578947368421 ( 1 time, No remainder )

Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving

0 __________________ . .
1 / 19 = 0.052631578947368421 or 0.052631578947368421

Note that we have completed the process of division only by using ?2?. Nowhere the division by 19 occurs.

b) Multiplication Method: Value of 1 / 19

First we recognize the last digit of the denominator of the type 1 / a9. Here the last digit is 9.

For a fraction of the form in whose denominator 9 is the last digit, we take the case of 1 / 19 as follows:

For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.

Therefore 2 is the multiplier for the conversion. We write the last digit in the numerator as 1 and follow the steps leftwards.

Step. 1 : 1

Step. 2 : 21(multiply 1 by 2, put to left)

Step. 3 : 421(multiply 2 by 2, put to left)

Step. 4 : 8421(multiply 4 by 2, put to left)

Step. 5 : ₁68421(multiply 8 by 2 =16, 1 carried over, 6 put to left)

Step. 6 : ₁368421( 6 X 2 =12,+1 [carry over]

= 13, 1 carried over, 3 put to left )

Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover]

= 7, put to left)

Step. 8 : ₁47368421 (as in the same process)

Step. 9 : 947368421 ( Do ? continue to step 18)

Step. 10 : ₁8947368421

Step. 11 : ₁78947368421

Step. 12 : ₁578947368421

Step. 13 : ₁1578947368421

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 : ₁2631578947368421

Step. 17 : 52631578947368421

Step. 18 : 1052631578947368421

Now from step 18 onwards the same numbers and order towards left continue.

Thus 1 / 19 = 0.052631578947368421

It is interesting to note that we have

i) not at all used division process

ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued to multiply the resultant successively by 2.

Observations :

a) For any fraction of the form 1 / a9i.e.,, in whose denominator 9 is the digit in the units place and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the repeating block?s right most digit is 1.

b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1) either in division or in multiplication.

c) Starting from right most digit and counting from the right, we see ( in the given example 1 / 19)

Sum of 1^st digit + 10^th digit = 1 + 8 = 9

Sum of 2^nd digit + 11^th digit = 2 + 7 = 9

- - - - - - - - -- - - - - - - - - - - - - - - - - - -
Sum of 9th digit + 18th digit = 9+ 0 = 9

From the above observations, we conclude that if we find first 9 digits, further digits can be derived as complements of 9.

i) Thus at the step 8 in division process we have 0.0526315₁7 and next step. 9 gives 0.052631578

        Now the complements of the numbers
                                0, 5, 2, 6, 3, 1, 5, 7, 8 from 9
                                9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order

                                        i.e.,, 0.052631578947368421

             Now taking the multiplication process we have

Step. 8 : ₁47368421

        Step. 9 :         947368421

    Now the complements of 1, 2, 4, 8, 6, 3, 7, 4, 9 from 9
                               i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.

0.052631578947368421.

d) When we get (Denominator ? Numerator) as the product in the multiplicative process, half the work is done. We stop the multiplication there and mechanically write the remaining half of the answer by merely taking down complements from 9.

e) Either division or multiplication process of giving the answer can be put in a single line form.

Algebraic proof :

Any vulgar fraction of the form 1 / a9 can be written as

1 / a9 =  1 / ( (a + 1 ) x - 1 ) where x = 10

                                     1
               =   ________________________
                  ( a + 1 ) x [1 - 1/(a+1)x ]

                     1^{=    ___________}[1 - 1/(a+1)x]^-1( a + 1 ) x

                       1
            = __________ [1 + 1/(a+1)x + 1/(a+1)x²+ ----------]
                ( a + 1 ) x

= 1/(a+1)x + 1/(a+1)²x² +1/(a+1)³x³+ ---- ad infinitum

= 10^-1(1/(a+1))+10^-2(1/(a+1)²)+10^-3(1/(a+1)³) + ---ad infinitum

This series explains the process of ekadhik.

Now consider the problem of 1 / 19. From above we get

1 / 19 = 10^-1 (1/(1+1)) + 10^-2 (1/(1+1)²) + 10^-3 (1/(1+1)³) + ----
( since a=1)

= 10^-1 (1/2) + 10^-2 (1/2)² + 10^-3 (1/3)³ + ----------

= 10^-1 (0.5) + 10^-2 (0.25) + 10^-3 (0.125)+ ----------

= 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - -

= 0.052631 - - - - - - -

Example1 :

1. Find 1 / 49 by ekadhikena process.

Now ?previous? is 4. ?One more than the previous? is 4 + 1 = 5.

Now by division right ward from the left by ?5?.

1 / 49 = .₁0 - - - - - - - - - - - -(divide 1 by 50)

= .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )

= .02₂0 - - - - - - --(divide 20 by 5, 4 times)

= .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )

= .0204₄0 -- - -- - ( divide 40 by 5, 8 times )

= .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )

= .020408₃1 - - - -(divide 31 by 5, 6 times, 1 remainder )

= .0204081₁ 6 - - - - - - - continue

= .02040816₁3₃2₂6₁5306₁1₁2₂2₂4₄4₄8 - -- - - - -

On completing 21 digits, we get 48

i.e.,,Denominator - Numerator = 49 ? 1 = 48 stands.

i.e, half of the process stops here. The remaining half can be obtained as complements from 9.
                                  .
         Thus 1 / 49 = 0.020408163265306122448
                                                                      .
                                  979591836734693877551

Now finding 1 / 49 by process of multiplication left ward from right by 5, we get

1 / 49 = ----------------------------------------------1

= ---------------------------------------------51

= -------------------------------------------₂551

= ------------------------------------------₂7551

= ---- ₄8₃9₄7₂9₄59₄1₁8₃3₃6₁7₂3₃4₄69₄3₃8₃7₂7551

i.e.,,Denominator ? Numerator = 49 ? 1 = 48 is obtained as 5X9+3

( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining half is automatically obtained as complements of 9.

    Thus 1 / 49 = ---------------979591836734693877551
                                .
                          = 0.020408163265306122448
                                                                    .
                                979591836734693877551

Example 2: Find 1 / 39 by Ekadhika process.

Now by multiplication method, Ekadhikena purva is 3 + 1 = 4

1 / 39 = -------------------------------------1

= -------------------------------------41

= ----------------------------------₁641

= ---------------------------------₂5641

= --------------------------------₂25641

= -------------------------------₁025641

Here the repeating block happens to be block of 6 digits. Now the rule predicting the completion of half of the computation does not hold. The complete block has to be computed by ekadhika process.

Now continue and obtain the result. Find reasons for the non?applicability of the said ?rule?.

Find the recurring decimal form of the fractions 1 / 29, 1 / 59,
1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether the rule of completion of half the computation holds good in such cases.

Note : The Ekadhikena Purvena sutra can also be used for conversion of vulgar fractions ending in 1, 3, 7 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - - - -, 1 / 7, 1 / 17, - - - - - by writing them in the following way and solving them.

NIKHILAM NAVATAS?CARAMAM DASATAH

The formula simply means : ?all from 9 and the last from 10?

The formula can be very effectively applied in multiplication of numbers, which are nearer to bases like 10, 100, 1000 i.e., to the powers of 10 . The procedure of multiplication using the Nikhilam involves minimum number of steps, space, time saving and only mental calculation. The numbers taken can be either less or more than the base considered.

The difference between the number and the base is termed as deviation. Deviation may be positive or negative. Positive deviation is written without the positive sign and the negative deviation, is written using Rekhank (a bar on the number). Now observe the following table.

Number	Base	Number ? Base	Deviation

14	10	14 - 10	4
			_
8	10	8 - 10	-2 or 2
			__
97	100	97 - 100	-03 or 03

112	100	112 - 100	12
			___
993	1000	993 - 1000	-007 or 007

1011	1000	1011 - 1000	011

Some rules of the method (near to the base) in Multiplication

a) Since deviation is obtained by Nikhilam sutra we call the method as Nikhilam multiplication.

Eg : 94. Now deviation can be obtained by ?all from 9 and the last from 10? sutra i.e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06.

b) The two numbers under consideration are written one below the other. The deviations are written on the right hand side.

Eg : Multiply 7 by 8.

Now the base is 10. Since it is near to both the numbers,
7
we write the numbers one below the other.                          8
-----
Take the deviations of both the numbers from
the base and represent                                                        _
7   3
_
Rekhank or the minus sign before the deviations               8    2
------
------

or 7 -3
8 -2
-------
-------

or remainders 3 and 2 implies that the numbers to be multiplied are both less than 10

c) The product or answer will have two parts, one on the left side and the other on the right. A vertical or a slant line i.e., a slash may be drawn for the demarcation of the two parts i.e.,

(or)

d) The R.H.S. of the answer is the product of the deviations of the numbers. It shall contain the number of digits equal to number of zeroes in the base.

                                  _
i.e., 7    3
_
8    2
_____________
/ (3x2) = 6

Since base is 10, 6 can be taken as it is.

e) L.H.S of the answer is the sum of one number with the deviation of the other. It can be arrived at in any one of the four ways.

i) Cross-subtract deviation 2 on the second row from the original number 7 in the first row i.e., 7-2 = 5.

ii) Cross?subtract deviation 3 on the first row from the original number8 in the second row (converse way of (i))
i.e., 8 - 3 = 5

iii) Subtract the base 10 from the sum of the given numbers.
i.e., (7 + 8) ? 10 = 5

iv) Subtract the sum of the two deviations from the base.
i.e., 10 ? ( 3 + 2) = 5

Hence 5 is left hand side of the answer.

                                              _
Thus     7    3
_
8    2
¯¯¯¯¯¯¯¯¯¯¯¯
5 /

Now (d) and (e) together give the solution
_
7    3                7
_
8    2    i.e., X 8
¯¯¯¯¯¯¯         ¯¯¯¯¯¯
5 / 6               56

f) If R.H.S. contains less number of digits than the number of zeros in the base, the remaining digits are filled up by giving zero or zeroes on the left side of the R.H.S. If the number of digits are more than the number of zeroes in the base, the excess digit or digits are to be added to L.H.S of the answer.

The general form of the multiplication under Nikhilam can be shown as follows :

Let N₁ and N₂ be two numbers near to a given base in powers of 10, and D₁ and D₂ are their respective deviations from the base. Then N₁ X N₂ can be represented as

Case (i) : Both the numbers are lower than the base. We have already considered the example 7 x 8 , with base 10.

Now let us solve some more examples by taking bases 100 and 1000 respectively.

Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is as follows:

Ex. 2: 98 X 97 Base is 100.

Ex. 3: 75X95. Base is 100.

Ex. 4: 986 X 989. Base is 1000.

Ex. 5: 994X988. Base is 1000.

Ex. 6: 750X995.

Case ( ii) : Both the numbers are higher than the base.

The method and rules follow as they are. The only difference is the positive deviation. Instead of cross ? subtract, we follow cross ? add.

Ex. 7: 13X12. Base is 10

Ex. 8: 18X14. Base is 10.

Ex. 9: 104X102. Base is 100.

                               104    04
102    02
¯¯¯¯¯¯¯¯¯¯¯¯
106 / 4x2    =    10608      ( rule - f )
¯¯¯¯¯¯¯¯¯¯¯¯

Ex. 10: 1275X1004. Base is 1000.

                              1275   275
1004   004
¯¯¯¯¯¯¯¯¯¯¯¯
1279 / 275x4   =    1279 / ₁100     ( rule - f )
____________    =    1280100

Case ( iii ): One number is more and the other is less than the base.

In this situation one deviation is positive and the other is negative. So the product of deviations becomes negative. So the right hand side of the answer obtained will therefore have to be subtracted. To have a clear representation and understanding a vinculum is used. It proceeds into normalization.

Ex.11: 13X7. Base is 10

Note : Conversion of common number into vinculum number and vice versa.

Eg :
__
9 = 10 ? 1 = 11
_
98 = 100 ? 2 = 102
_
196 = 200 ? 4 = 204
_
32 = 30 ? 2 = 28
_
145 = 140 ? 5 = 135
_
322 = 300 ? 22 = 278. etc

The procedure can be explained in detail using Nikhilam Navatascaram Dasatah, Ekadhikena purvena, Ekanyunena purvena in the foregoing pages of this book.]

Ex. 12: 108 X 94. Base is 100.

Ex. 13: 998 X 1025. Base is 1000.

Algebraic Proof:

Case ( i ):

Let the two numbers N1 and N2 be less than the selected base say x.

N₁ = (x-a), N₂ = (x-b). Here a and b are the corresponding deviations of the numbers N₁ and N₂ from the base x. Observe that x is a multiple of 10.

Now N₁ X N₂ = (x-a) (x-b) = x.x ? x.b ? a.x + ab

= x (x ? a ? b ) + ab. [rule ? e(iv), d ]

= x [(x ? a) ? b] + ab = x (N₁?b) + ab[rule?e(i),d]

or = x [(x ? b) ? a] = x (N₂ ? a) + ab. [rule ?e (ii),d]

x (x ? a ? b) + ab can also be written as

x[(x ? a) + (x ? b) ? x] + ab = x[N₁+N₂ ? x] + ab [rule ? e(iii),d].

A difficult can be faced, if the vertical multiplication of the deficit digits or deviations i.e., a.b yields a product consisting of more than the required digits. Then rule-f will enable us to surmount the difficulty.

Case ( ii ) :

When both the numbers exceed the selected base, we have N₁ = x + a, N₂ = x + b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds good, of course with relevant details mentioned in case (i).

Case ( iii ) :

When one number is less and another is more than the base, we can use (x-a)(x+b) = x(x?a+ b)?ab. and the procedure is evident from the examples given.

          Find the following products by Nikhilam formula.

1) 7 X 4                 2) 93 X 85             3) 875 X 994

4) 1234 X 1002     5) 1003 X 997        6) 11112 X 9998

7) 1234 X 1002     8) 118 X 105

Nikhilam in Division

Consider some two digit numbers (dividends) and same divisor 9. Observe the following example.

i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4.

since 9 ) 13 ( 1

9
____
4

ii) 34 ÷ 9, Q is 3, R is 7.

iii) 60 ÷ 9, Q is 6, R is 6.

iv) 80 ÷ 9, Q is 8, R is 8.

Now we have another type of representation for the above examples as given hereunder:

i) Split each dividend into a left hand part for the Quotient and right - hand part for the remainder by a slant line or slash.

Eg. 13 as 1 / 3, 34 as 3 / 4 , 80 as 8 / 0.

ii) Leave some space below such representation, draw a horizontal line.

Eg. 1 / 3 3 / 4 8 / 0

______ , ______ , ______

iii) Put the first digit of the dividend as it is under the horizontal line. Put the same digit under the right hand part for the remainder, add the two and place the sum i.e., sum of the digits of the numbers as the remainder.

Eg.
1 / 3         3 / 4         8 / 0
1              3              8
______ ,   ______ ,  ______
1 / 4         3 / 7         8 / 8

Now the problem is over. i.e.,

13 ÷ 9 gives Q = 1, R = 4

34 ÷ 9 gives Q = 3, R = 7

80 ÷ 9 gives Q = 8, R = 8

Proceeding for some more of the two digit number division by 9, we get

             a) 21 ÷ 9  as
9) 2 / 1     i.e     Q=2, R=3
2
¯¯¯¯¯¯
2 / 3

b) 43 ÷ 9  as
9) 4 / 3     i.e     Q = 4, R = 7.
4
¯¯¯¯¯¯
4 / 7

The examples given so far convey that in the division of two digit numbers by 9, we can mechanically take the first digit down for the quotient ? column and that, by adding the quotient to the second digit, we can get the remainder.

Now in the case of 3 digit numbers, let us proceed as follows.

i)
9 ) 104 ( 11                 9 ) 10   / 4
99                                1 / 1
¯¯¯¯¯¯            as            ¯¯¯¯¯¯¯
5                                 11 / 5

ii)
9 ) 212 ( 23                 9 ) 21 / 2
207                               2 / 3
¯¯¯¯¯              as            ¯¯¯¯¯¯¯
5                                23 / 5

iii)
9 ) 401 ( 44                 9 ) 40 / 1
396                                 4 / 4
¯¯¯¯¯             as            ¯¯¯¯¯¯¯¯
5                                 44 / 5

Note that the remainder is the sum of the digits of the dividend. The first digit of the dividend from left is added mechanically to the second digit of the dividend to obtain the second digit of the quotient. This digit added to the third digit sets the remainder. The first digit of the dividend remains as the first digit of the quotient.

Consider 511 ÷ 9

Add the first digit 5 to second digit 1 getting 5 + 1 = 6. Hence Quotient is 56. Now second digit of 56 i.e., 6 is added to third digit 1 of dividend to get the remainder i.e., 1 + 6 = 7

Thus
9 ) 51 / 1
5 / 6
¯¯¯¯¯¯¯
56 / 7

Q is 56, R is 7.

Extending the same principle even to bigger numbers of still more digits, we can get the results.

Eg : 1204 ÷ 9

i) Add first digit 1 to the second digit 2. 1 + 2 = 3

ii) Add the second digit of quotient 13. i.e., 3 to third digit ?0? and obtain the Quotient. 3 + 0 = 3, 133

iii) Add the third digit of Quotient 133 i.e.,3 to last digit ?4? of the dividend and write the final Quotient and Remainder. R = 3 + 4 = 7, Q = 133

In symbolic form 9 ) 120 / 4
13 / 3
¯¯¯¯¯¯¯¯
133 / 7

Another example.
9 ) 13210   / 1          132101 ÷ 9
gives
1467   / 7          Q = 14677, R = 8
¯¯¯¯¯¯¯¯¯¯
14677 / 8

In all the cases mentioned above, the remainder is less than the divisor. What about the case when the remainder is equal or greater than the divisor?

Eg.
                 9 ) 3 / 6              9)   24 / 6
3                      2 /   6
¯¯¯¯¯¯        or        ¯¯¯¯¯¯¯¯
3 / 9 (equal)           26 / 12 (greater).

We proceed by re-dividing the remainder by 9, carrying over this Quotient to the quotient side and retaining the final remainder in the remainder side.

              9 ) 3 / 6                  9 ) 24 / 6
/ 3                         2 / 6
¯¯¯¯¯¯¯                   ¯¯¯¯¯¯¯¯
3 / 9                       26 / 12
¯¯¯¯¯¯¯                   ¯¯¯¯¯¯¯¯
4 / 0                       27 /   3

Q = 4, R = 0 Q = 27, R = 3.

When the remainder is greater than divisor, it can also be represented as

                            9 ) 24 / 6
2 / 6
¯¯¯¯¯¯¯¯
26 /1 / 2
/ 1
¯¯¯¯¯¯¯¯
1    / 3
¯¯¯¯¯¯¯¯
27    / 3

Now consider the divisors of two or more digits whose last digit is 9,when divisor is 89.

We Know 113 = 1 X 89 + 24, Q =1, R = 24

10015 = 112 X 89 + 47, Q = 112, R = 47.

Representing in the previous form of procedure, we have

                89 ) 1 / 13                 89 ) 100 / 15
/   11                        12 / 32
¯¯¯¯¯¯¯                    ¯¯¯¯¯¯¯¯¯¯
1 / 24                        112 / 47

But how to get these? What is the procedure?

Now Nikhilam rule comes to rescue us. The nikhilam states ?all from 9 and the last from 10?. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply nikhilam formula on 89 and get the complement 11.Further while carrying the added numbers to the place below the next digit, we have to multiply by this 11.

           89 ) 1 / 13         89 ) 100 / 15
¯¯
/ 11         11      11 /            first digit 1 x 11
¯¯¯¯¯¯¯¯
1 / 24                    1 / 1      total second is 1x11
22    total of 3^rd digit is 2 x 11
¯¯¯¯¯¯¯¯¯¯
112 / 47

What is 10015 ÷ 98 ? Apply Nikhilam and get 100 ? 98 = 02. Set off the 2 digits from the right as the remainder consists of 2 digits. While carrying the added numbers to the place below the next digit, multiply by 02.

Thus
98 )   100   / 15
¯¯
02        02 /                 i.e., 10015 ÷ 98 gives
0 /   0              Q = 102,     R = 19
/    04
¯¯¯¯¯¯¯¯¯¯
102   / 19

In the same way
897 ) 11 / 422
¯¯¯
103 1 / 03
/ 206
¯¯¯¯¯¯¯¯¯
12 / 658

gives 11,422 ÷ 897, Q = 12, R=658.

In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the case of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the divisors complement from 10. In case of more digited numbers we apply Nikhilam and proceed. Any how, this method is highly useful and effective for division when the numbers are near to bases of 10.

* Guess the logic in the process of division by 9.

* Obtain the Quotient and Remainder for the following problems.

1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97

4) 2342 ÷ 98 5) 113401 ÷ 997

6) 11199171 ÷ 99979

Observe that by nikhilam process of division, even lengthier divisions involve no division or no subtraction but only a few multiplications of single digits with small numbers and a simple addition. But we know fairly well that only a special type of cases are being dealt and hence many questions about various other types of problems arise. The answer lies in Vedic Methods.

What is king kong?

THE ROAD TO SUCCESS.
The Road to success is not straight...
There is a curve called failure...
A loop called confusion...
Speed bumps called friends...
Red lights called enemies...
Caution lights called family...
You will have flats called jobs,

BUT...
If you have a spare called determination ...
An engine called perseverance ...
Insurance called faith...
A driver called God...
You will make it to a place called Success!

A well-known speaker started off his seminar by holding up a $20 bill in
the room of 200, he asked, "Who would like this $20 bill?"
Hands started going up.
He said, "I am going to give this $20 to one of you but first let me do
this."
He proceeded to crumple the dollar bill up. He then asked, "Who still
wants it?"
Still the hands were up in the air.
"Well," he replied, "What if I do this?" And he dropped it on the ground and started to grind it into the floor with his shoe. He picked it up, now all crumpled and dirty. "Now who still wants it?"
Still the hands went into the air.
"My friends, you have all learned a very valuable lesson. No matter
what I did to the money, you still wanted it because it did not decrease in value. It was still worth $20.
Many times in our lives, we are dropped, crumpled, and ground into the dirt by the decisions we make and the circumstances that come our way.
We feel as though we are worthless. But no matter what has happened or what will happen, you will never lose your value.
You are special - Don't ever forget it!
"Never let yesterday's disappointments overshadow tomorrow's dreams"

P.S. Please pass this to everyone in you know. You will never know the lives it touches, the hurting hearts it speaks to, or the Hope that it can bring!

One day a very wealthy family took their son on a trip to the country with
the firm purpose of showing him how very poor people can be. They spent a day
and a night on the farm of a very poor family. When they got back from their
trip the father asked his son, "How was the trip?"

"Very good Dad!"

"Did you see how poor people can be?" the father asked.

"Yeah!"

"And what did you learn?"

The son answered, "I saw that we have a dog at home, and they have four.
We have a pool that reaches to the middle of the garden, they have a creek
that has no end. We have imported lamps in the garden, they have the stars.
Our patio reaches to the front yard, they have a whole horizon."

When the little boy was finished, his father was speechless. His son added,
"Thanks Dad for showing me how poor we are!"

Isn't it true that it all depends on the way you look at things?

Faith

A guy's in his house when horrendous rains come up. The water starts rising, and before you know it, we're talking major flood. Roads are covered.
Nothing's moving. Pretty soon, a boat comes along.
Guy in the boat yells, 'Come on - we're here to save you. Get in the boat.'
Guy in the house says, 'No...I've got faith that God will save me.'
The boat leaves. The water keeps rising. The guy is forced up the second floor of his house by the flood waters. Another boat comes along. The guy in the boat yells, 'Come on! It's getting worse. If you don't get in the boat, you're going to drown.'
From the second floor window the guy says, 'No...I'll be ok. I've got faith in God that he'll save me.' The boat leaves. Water's rising. The guy's on the roof. A helicopter hovers overhead and the pilot shouts out, 'This is your last chance. Climb up the ladder. If you don't come
now you're going to drown.'
The guy says from the roof, 'No, Thanks. God will save me.'
The pilot shrugs his shoulders and splits. The water rises. The guy drowns. Ascends to the pearly gates. He asks St. Peter, 'What happened? I've been devoted to God and had absolute faith that he would save me. Why did he let me down?'

And St. Peter tells him, 'What more do you want? God sent you two boats and a helicopter!?'

thank u
dhanyavaadamulu

WHY ARE YOU MY FRIEND

A story tells that two friends were walking through the desert. In a specific point of the journey, they had an argument, and one friend slapped the other one in the face. The one, who got slapped, was hurt, but without anything to say, he wrote in the sand: "TODAY, MY BEST FRIEND SLAPPED ME IN THE FACE". They kept on walking, until they found an oasis, where they decided to take a bath. The one who got slapped and hurt started drowning, and the other friend saved him. When he recovered from the fright, he wrote on a stone: "TODAY MY BEST FRIEND SAVED MY LIFE" The friend who saved and slapped his best friend, asked him, "Why, after I hurt you, you wrote in the sand, and now you write on a stone?" The other friend, smilingly replied:

"When a friend hurts us, we should write it down in the sand, where the winds of forgiveness erase it away. And when something great happens, we should engrave it in the stone of the memory of the heart, where no wind can erase it"

Learn to write in the sand, when you have differences and hurt feelings with your friend. Learn to write in stone when your friend had done some thing really good to you.

WHY ARE YOU MY FRIEND

A story tells that two friends were walking through the desert. In a specific point of the journey, they had an argument, and one friend slapped the other one in the face. The one, who got slapped, was hurt, but without anything to say, he wrote in the sand: "TODAY, MY BEST FRIEND SLAPPED ME IN THE FACE". They kept on walking, until they found an oasis, where they decided to take a bath. The one who got slapped and hurt started drowning, and the other friend saved him. When he recovered from the fright, he wrote on a stone: "TODAY MY BEST FRIEND SAVED MY LIFE" The friend who saved and slapped his best friend, asked him, "Why, after I hurt you, you wrote in the sand, and now you write on a stone?" The other friend, smilingly replied:

"When a friend hurts us, we should write it down in the sand, where the winds of forgiveness erase it away. And when something great happens, we should engrave it in the stone of the memory of the heart, where no wind can erase it"

Learn to write in the sand, when you have differences and hurt feelings with your friend. Learn to write in stone when your friend had done some thing really good to you.