MAGIC OF MATHEMATICS

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1. Square of numbers ending in 5

65 x 65 = (6 x (6+1) ) 25 = (6x7) 25 = 4225

45 x 45 = (4 x (4+1) ) 25 = (4x5) 25 = 2025

105 x 105 = (10 x (10+1) 25 = (10 x 11) 25 = 11025

2. When sum of the last digits is the base(10) and previous parts are the same

44 x 46 = (4 x (4+1)) (4 x 6) = (4 x 5) (4 x 6) = 2024

37 x 33 = (3 x (3+1)) (7 x 3) = (3 x 4) (7 x 3) = 1221

11 x 19 = (1 x (1+1)) (1 x 9) = (1 x 2) (1 x 9) = 209

3. 1 divided by 19, 29, 39,..............

Consider 1/19 since 19 is not divisible by 2 or 5 it is a purely a recurring decimal

take last digit 1

multiply this with 1+1 (one more) i.e 2 (this is the key digit) ==>21

multiply 2 by 2 ==> 421 multiplying 4 by 2 ==> 8421

multiply 8 by 2 ==> 68421 carry 1

multiply 6 by 2 =12 + carry 1= 13 ==> 368421 carry 1

continuing (till 18 digits =denominator-numerator)

the result is 0.052631578947368421

4. 1/19 using divisions

divide 1 by 2, answer is 0 with remainder 1 ==> .0

next 10 divided by 2 is 5 ==> .05

next 5 divided by 2 is 2 remainder 1 ==> 0.052

next 12 (remainder 2) divided by 2 is 6 ==> 0.0526

next 6 divided by 2 is 3 ==> 0.05263

next 3 divided by 2 is 1 remainder 1 ==> 0.052631

next 11 divided by 2 is 5 remainder 1 ==> 0.0526315

and so on...

5. 1/7 = 7/49 previous digit is 4 so multiply by 4+1 i.e. by 5

7-> 57 -> 857 -> 42857 -> 0.142857 (stop after 7-1= 6 digits)

Vedic Number Representation

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Vedic knowledge is in the form of slokas or poems in Sanskrit verse. A number was encoded using consonant groups of the Sanskrit alphabet, and vowels were provided as additional latitude to the author in poetic composition. The coding key is given as Kaadi nav, taadi nav, paadi panchak, yaadashtak ta ksha shunyam Translated as below ·

Varnmala

ka kha ga gha gna
cha chha ja jha inja
ta tha da dha na
pa pha ba bha ma
ya ra la va sha
sha sa ha chjha tra gna

letter "ka" and the following eight letters
letter "ta" and the following eight letters
letter "pa" and the following four letters
letter "ya" and the following seven letters, and
letter "ksha" for zero.

In other words,

ka, ta, pa, ya = 1
kha, tha, pha, ra = 2
ga, da, ba, la = 3
gha, dha, bha, va = 4
gna, na, ma, scha = 5
cha, ta, sha = 6
chha, tha, sa = 7
ja, da, ha = 8
jha, dha = 9
ksha = 0

Thus pa pa is 11, ma ra is 52. Words kapa, tapa , papa, and yapa all mean the same that is 11.

LOPANASTHAPANABHYAM

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Lopana sthapanabhyam means 'by alternate elimination and retention'.

Consider the case of factorization of quadratic equation of type ax2

+ by2 + cz2 + dxy + eyz + fzx This is a homogeneous equation of

second degree in three variables x, y, z. The sub-sutra removes the

difficulty and makes the factorization simple. The steps are as

follows:

i) Eliminate z by putting z = 0 and retain x and y and factorize

thus obtained a quadratic in x and y by means of ‘adyamadyena’

sutra.;

ii) Similarly eliminate y and retain x and z and factorize the

quadratic in x and z.

iii) With these two sets of factors, fill in the gaps caused by

the elimination process of z and y respectively. This gives actual

factors of the expression.

Example 1: 3x2 + 7xy + 2y2 + 11xz + 7yz + 6z2.

Step (i): Eliminate z and retain x, y; factorize
3x2 + 7xy + 2y2 = (3x + y) (x + 2y)

Step (ii): Eliminate y and retain x, z; factorize
3x2 + 11xz + 6z2 = (3x + 2z) (x + 3z)

Step (iii): Fill the gaps, the given expression
= (3x + y + 2z) (x + 2y + 3z)

Example 2: 12x2 + 11xy + 2y2 - 13xz - 7yz + 3z2.

Step (i): Eliminate z i.e., z = 0; factorize
12x2 + 11xy + 2y2 = (3x + 2y) (4x + y)

Step (ii): Eliminate y i.e., y = 0; factorize
12x2 - 13xz + 3z2 = (4x -3z) (3x – z)

Step (iii): Fill in the gaps; the given expression
= (4x + y – 3z) (3x + 2y – z)

Example 3: 3x2+6y2+2z2+11xy+7yz+6xz+19x+22y+13z+20

Step (i): Eliminate y and z, retain x and independent term
i.e., y = 0, z = 0 in the expression (E).
Then E = 3x2 + 19x + 20 = (x + 5) (3x + 4)

Step (ii): Eliminate z and x, retain y and independent term
i.e., z = 0, x = 0 in the expression.
Then E = 6y2 + 22y + 20 = (2y + 4) (3y + 5)

Step (iii): Eliminate x and y, retain z and independent term
i.e., x = 0, y = 0 in the expression.
Then E = 2z2 + 13z + 20 = (z + 4) (2z + 5)

Step (iv): The expression has the factors (think of

independent terms)
= (3x + 2y + z + 4) (x + 3y + 2z + 5).

In this way either homogeneous equations of second degree or general

equations of second degree in three variables can be very easily

solved by applying ‘adyamadyena’ and ‘lopanasthapanabhyam’ sutras.

ADDITION AND SUBTRACTION

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ADDITION:

In the convention process we perform the process as follows.

234 + 403 + 564 + 721

write as 234
403
564
721

Step (i): 4 + 3 + 4 + 1 = 12 2 retained and 1 is carried over to left.

Step (ii): 3 + 0 + 6 + 2 = 11 the carried ‘1’ is added

i.e., Now 2 retained as digit in the second place (from right to left) of the answer and 1 is carried over to left.

step (iii): 2 + 4 + 5 + 7 = 18 carried over ‘1’ is added

i.e., 18 + 1 = 19. As the addition process ends, the same 19 is retained in the left had most part of the answer.

thus 234
403
564
+721
_____
1922 is the answer

we follow sudhikaran process Recall ‘sudha’ i.e., dot (.) is taken as an upa-sutra (No: 15)

consider the same example

i) Carry out the addition column by column in the usual fashion, moving from bottom to top.

(a) 1 + 4 = 5, 5 + 3 = 8, 8 + 4 = 12 The final result is more than 9. The tenth place ‘1’ is dropped once number in the unit place i.e., 2 retained. We say at this stage sudha and a dot is above the top 4. Thus column (1) of addition (right to left)
.
4
3
4
1
__
2

b) Before coming to column (2) addition, the number of dots are to be counted, This shall be added to the bottom number of column (2) and we proceed as above.

Thus second column becomes
.
3 dot=1, 1 + 2 = 3
0 3 + 6 = 9
6 9 + 0 = 9
2 9 + 3 = 12
__
2

2 retained and ‘.’ is placed on top number 3

c) proceed as above for column (3)

2 i) dot = 1 ii) 1 + 7 = 8
4 iii) 8 + 5 = 13 iv) Sudha is said.
.
5 A dot is placed on 5 and proceed
7 with retained unit place 3.
__
9 v) 3+4=7,7+2=9 Retain 9 in 3rd digit i.e.,in 100th place.

d) Now the number of dots is counted. Here it is 1 only and the number is carried out left side ie. 1000th place
..
Thus 234
403
.
564
+721
_____
1922 is the answer.

Though it appears to follow the conventional procedure, a careful observation and practice gives its special use.

eg (1):
.
437
. .
624
.
586
+162
______
1809

Steps 1:

i) 2 + 6 = 8, 8 + 4 = 12 so a dot on 4 and 2 + 7 = 9 the answer retained under column (i)

ii) One dot from column (i) treated as 1, is carried over to column (ii),

thus 1 + 6 = 7, 7 + 8 = 15 A' dot’; is placed on 8 for the 1 in 15 and the 5 in 15 is added to 2 above.

5 + 2 = 7, 7 + 3 = 10 i.e. 0 is written under column (ii) and a dot for the carried over 1 of 10 is placed on the top of 3.

(iii) The number of dots counted in column (iii) are 2.

Hence the number 2 is carried over to column (ii) Now in column (iii)

2 + 1 = 3, 3 + 5 = 8, 8 + 6 = 14 A dot for 1 on the number 6 and 4 is retained to be added 4 above to give 8. Thus 8 is placed under column (iii).

iv) Finally the number of dots in column (iii) are counted. It is ‘1’ only. So it carried over to 1000th place. As there is no fourth column 1 is the answer for 4th column. Thus the answer is 1809.

Example 3:

Check the result verify these steps with the procedure mentioned above.

The process of addition can also be done in the down-ward direction i.e., addition of numbers column wise from top to bottom

Example 1:

Step 1: 6 + 4 = 10, 1 dot ; 0 + 8 = 8; 8 + 4 = 12;

1 dot and 2 answer under first column - total 2 dots.

Step 2: 2+2 ( 2 dots) = 4; 4+9 = 13: 1 dot and 3+0 = 3; 3+8 = 11;

1 dot and 1 answer under second column - total 2 dots.

Step 3: 3+2 ( 2 dots ) = 5; 5+6 = 11:1 dot and 1+7 = 8; 8+7 = 15;

1 dot and 5 under third column as answer - total 2 dots.

Step 4: 4 + 2 ( 2 dots ) = 6; 6 + 5 =11:

1 dot and 1+3 = 4; 4+2 = 6. - total 1 dot in the fourth 6 column as answer.

Step 5: 1 dot in the fourth column carried over to 5th column (No digits in it) as 1

Thus answer is from Step5 to Step1; 16512

Example 2:

Steps

(i): 8 + 9 = 17; 7 + 4 = 11; 1 + 1 = (2) (2dots)

(ii): 7 + 2 = 9; 9 + 1 = 10; 0 + 8 = 8, 8 + 9 = 17, (7) (2dots)

(iii): 2 + 2 = 4; 4 + 6 = 10; 0 + 0 = 0; 0 + 7 = (7) (1 dot)

(iv): 3 + 1 = 4; 4 + 4 = 8; 8 + 3 = 11; 1 + 1 = (2) (1 dot)

(v): 1

Thus answer is 12772.

Add the following numbers use ‘Sudhikaran’ whereever applicable.

1. 2. 3.
486 5432 968763
395 3691 476509
721 4808 +584376
+609 +6787 ¯¯¯¯¯¯¯¯
¯¯¯¯¯ ¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯
¯¯¯¯¯ ¯¯¯¯¯¯

Check up whether ‘Sudhkaran’ is done correctly. If not write the correct process. In either case find the sums.

SUBTRACTION:

The ‘Sudha’ Sutra is applicable where the larger digit is to be subtracted from the smaller digit. Let us go to the process through the examples.

Procedure:

i) If the digit to be subtracted is larger, a dot ( sudha ) is given to its left.

ii) The purak of this lower digit is added to the upper digit or purak-rekhank of this lower digit is subtracted.

Example (i): 34 - 18

34
.
-18
_____
.
Steps: (i): Since 8>4, a dot is put on its left i.e. 1

(ii) Purak of 8 i.e. 2 is added to the upper digit i.e. 4
_
2 + 4 = 6. or Purak-rekhank of 8 i.e. 2 is
_
Subtracted from i.e. 4 - 2 =6.

Now at the tens place a dot (means1) makes the ‘1’ in the number into 1+1=2.This has to be subtracted from above digit. i.e. 3 - 2 = 1. Thus

34
.
-18
_____
16

Example 2:

63
.
-37
_____
.
Steps: (i) 7>3. Hence a dot on left of 7 i.e., 3

(ii) Purak of 7 i.e. 3 is added to upper digit 3 i.e. 3+3 = 6.

This is unit place of the answer.

Thus answer is 26.

Example (3) :

3274
..
-1892
_______

Steps:

(i) 2 < 4. No sudha . 4-2 = 2 first digit (form right to left)
.
(ii) 9 > 7 sudha required. Hence a dot on left of 9 i.e. 8

(iii) purak of 9 i.e. 1, added to upper 7 gives 1 + 7 = 8 second digit
.
(iv) Now means 8 + 1 = 9.
.
(v) As 9 > 2, once again the same process: dot on left of i.e., 1

(vi) purak of 9 i.e. 1, added to upper 2 gives 1 + 2 = 3, the third digit.
.
(vii) Now 1 means 1+1 = 2

(viii) As 2 < 3, we have 3-2 = 1, the fourth digit

Thus answer is 1382

Vedic Check :

Eg (i) in addition : 437 + 624 + 586 + 162 = 1809.

By beejank method, the Beejanks are

437 4 + 3 + 7 14 1 + 4 5

624 6 + 2 + 4 12 1 + 2 3

586 5 + 8 + 6 19 1 + 9 10 1 + 0 1

162 1 + 6 + 2 9

Now

437 + 624 + 586 + 162 5 + 3 + 1 + 9 18 1 + 8 9

Beejank of 1809 1 + 8 + 0 + 9 18 1 + 8 9 verified

Eg.(3) in subtraction :

3274 – 1892 = 1382

now beejanks

3274 3 + 2 + 7 + 4 3 + 4 7

1892 1 + 8 + 9 + 2 2

3292-1892 7-2 5

1382 1 + 3 + 8 + 2 5 Hence verified.

Mixed addition and subtraction using Rekhanks:

Example 1 : 423 - 654 + 847 - 126 + 204.

In the conventional method we first add all the +ve terms

423 + 847 + 204 = 1474

Next we add all negative terms

- 654 - 126 = -780

At the end their difference is taken

1474 - 780 = 694

Thus in 3 steps we complete the problem

But in Vedic method using Rekhank we write and directly find the answer.

4 2 3
_ _ _
6 5 4

8 4 7
_ _ _
1 2 6

2 0 4
_____
_
7 1 4 This gives (7 -1) / (10 - 1) / 4 = 694.

Example (2):

6371 – 2647 + 8096 – 7381 + 1234
____ ____
= 6371 + 2647 + 8096 + 7381 + 1234
_ _ _ _ _ _ _ _
= (6+2+8+7+1)/(3+6+0+3+2)/(7+4+9+8+3)/(1+7+6+1+4)
_
= 6 / 4 / 7 / 3

= (6 – 1) / (10 – 4) / 73

= 5673