Sunday, January 13, 2008

Vedic Mathematics: A Boon Re-Discovered


Vedic Mathematics offers a different approach to the study of Mathematics based on pattern recognition. It is based on sixteen principles, which contain techniques for performing mathematical operations in easier and faster ways. This post emphasizes on the commonly used applications, made simpler by using Vedic Mathematics model.

















Square multiples of 5:


Divide the number into two parts. First part(x) is the digit formed by excluding the last digit 5 and the second part(y) is the last digit 5.



L.H.S. of square of xy = ( x *(x+1))
R.H.S. of square of xy = 25



Consider 125:
x = 12 and y = 5
L.H.S. of square of 125 = 12 * 13 = 156
R.H.S. of square of 125 = 25
=> square of 125 = 15625



Apply this technique to find the squares of the numbers 1025, 675, 95, 5, 12345 and verify the answers using any conventional method.









Multiplication of 2 numbers with digit count same or differing by 1:

1. Find the smaller of two numbers N1 and N2, say N1.
2. Let ‘a’ be the number of digits in N1.
3. Select the base ’y’ as 10(a-1) or 10a , depending N2 (the bigger number) has more proximity to which base.
4. If N2 is roughly equidistant to both, check N1 is nearer to which base.



Let D1 ( N1 - y) and D2 ( N2 - y) be the signed deviations of N1 and N2 from y.

The answer will have two parts as demarcated by slash below:


N1 D1
*
N2 D2


(N1 + D2) / (D1 * D2)

Product =(N1 + D2)* y + (D1 * D2)


NOTE:


  • (N1 + D2) is always equal to (N2 + D1).


  • Always maintain the sign of D1 and D2.


  • If either of N1 or N2 is negative, first apply the formula and then negate the product accordingly to achieve the final result.

Consider 94 * 998
Smaller number = 94, a= 2, y = 10 or 100
998 closer to 100 => y = 100
D1 = 94 - 100 = -6
D2 = 998 – 100 = 898




94 -6
*
998 898



(992) /(-5388)

Product = 992 * 100 + (-5388) = 93812



Consider 1034 * 1002
Smaller number = 1002, a= 4, y = 1000 or 10000
1034 closer to 1000 => y = 1000
D1 = 1034 - 1000 = 34
D2 = 1002 – 1000 = 2



1034 34
*

1002 2

(1036)/ (68)


Product = 1036 * 1000 + (68) = 1036018


Consider 75 * 85
Smaller number = 75, a= 2 y = 10 or 100
85 closer to 100 => y = 100
D1 = 75 - 100 = -25
D2 = 85 – 100 = -15




75 -25
*

85 -15


(60) /(375)


Product = 60 * 100 + (375) = 6375


Apply this technique to find the products of the following pairs and verify the answers using any conventional method:
1. 11112 * 9998
2. 18 * 14
3. -118 * -105
4. 875 * 994
5. -3 * 4


Division by 9:

1. Count the total number of digits in the dividend, say b.

2. Thumb rule: Quotient will have (b-1) or b digits and remainder will have 1 digit.

3. The first digit of the quotient is same as that of dividend.

4. Second digit of the quotient is the sum of the first and second digits of the dividend.

5. On similar lines, nth digit of the quotient is equal to the sum of all the digits in the dividend from most significant position till nth position. Vary n from 1 to (b-1) to find all the digits of the quotient.

6. If sum of digits at any position is more than 9, retain only the last digit of the sum, add the remaining number to the previously calculated digit of the quotient.

7. Remainder is the sum of all the digits of the dividend.

8. If remainder is greater than 9, re-apply steps 1 to 7 to get another quotient and remainder.

Add the new quotient to the previously calculated quotient to get the final quotient.

The final reminder is just the new remainder obtained in this step.


9. If remainder is equal to 9, increment quotient by 1 and set remainder to 0.


10. If the new remainder obtained in step 8 is still greater than 9, keep on applying steps 8 and 9 till new remainder obtained is less than 9.


Consider 988687/ 9
b = 6
vary n from 1 to ( 6-1)
1st digit , 2nd digit, 3rd digit, 4th digit, 5th digit of quotient / remainder
9, 17, 25, 31, 39 / 46



Applying step 6:



10, 7, 25, 31, 39/ 46

10, 9, 5, 31, 39/ 46
10, 9, 8, 1, 39/ 46
10, 9, 8, 4, 9/ 46



Applying step 8
46 / 9
b = 2
vary n from 1 to ( 2-1)
1st digit of quotient = 4
Remainder = 4 + 6 = 10 > 9
Final quotient =109849 + 4 = 109853, Final remainder = 10
Applying step 10
10 / 9
vary n from 1 to ( 2-1)
1st digit of quotient = 1
Remainder = 1 + 0 = 1

Final quotient =109853 + 1 = 109854 remainder = 1


Consider 126 / 9
b = 3 vary n from 1 to ( 3-1)
1st digit, 2nd digit of quotient / remainder
1 3 / 9
Remainder = 9
Applying step 9
Final quotient = 13 + 1 = 14, Final remainder = 0


Apply this technique to find the quotient and remainder for the following problems and verify the answers using any conventional method:
1. 312 / 9
2. 130038 / 9
3. 80 / 9
4. 7070 / 9
5. 132201 / 9




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Multiplication of 2 digit numbers:
Let ab and cd be the 2 two digit numbers.
ab * cd can be represented as:

a *c : ( a * d + c * b) : b * d

The product can be divided into 3 columns as described above.
if ( b * d) or (( a * d + c * b)) has more than 1 digit, retain the last digit and add the remaining digits to the immediate column.

Consider 24 * 36
Product :
2 * 3 : ( 2 * 6 + 4 * 3) : 4 * 6
6: (12 + 12) : 24
6: 24 : 24
6: 24 + 2 : 4
6: 26 : 4
6 + 2 : 6 : 4
8 : 6: 4 => 864

Apply this technique to find the products of the following pairs and verify the answers using any conventional method:
1. 99 * 99
2. 18 * 14
3. 48 * 47
4. 42 * 53
5. 84 * 64


To check your understanding of the above mentioned formulae, try to solve an easy tutorial from the academy of Vedic Mathematics:
http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp


To further appreciate how Vedic model helps in solving arithmetic problems faster, try the arithmetic game at the below mentioned link. This arithmetic game is a speed drill where you are given two minutes to solve as many arithmetic problems as you can.
http://zetamac.com/arithmetic/


Please do share your quiz scores together with your experiences with Vedic Mathematics.
Let’s hope these Vedic techniques enhance our logical reasoning and revive our interest in the enticing world of Mathematics!!


14 comments:

CresceNet said...

Hello. This post is likeable, and your blog is very interesting, congratulations :-). I will add in my blogroll =). If possible gives a last there on my site, it is about the CresceNet, I hope you enjoy. The address is http://www.provedorcrescenet.com . A hug.

Sandy said...

Quite informative. The irony of life is such that we rarely need to do such computations in our work life these days.

pilot-pooja said...

Thanks Crescenet for visiting my blog.
I too visited your website, nice one, though language is surely a barrier for me, could not decode which one it is.

Please share the vision of your website to help me understand better.

Thanks,
Pooja

pilot-pooja said...

Very true Sandy!
Even though we're engineers by background, still we hardly find any need to apply these in the daily applications!

I really wanted to revive the old memories where we tried our best to remember shortcuts and save our time!

I strongly feel the need to introduce this course in our primary school days!

Jas said...

boy! quite an effort! reminds me of PEC when infy was coming and then when I was preparing for CAT; i always used to find shortcuts and teacher dint quite like it. Not coz he was a stiff guy, but coz i was wrong half of the time :D
Lovely post, m bookmarking it.

pilot-pooja said...

Thanks Jasginder..

Appreciation from a marketing bond really means a lot.!

Aftab Ahmad said...

Hi Pooja, nice to know you are promoting vedic mathematics, I am also fond of the ancient system.

All the best for you effort.

Regards
Aftab

pilot-pooja said...

Many thanks Aftab!

I visited ur blog, very nformative n thoughful!
Good Day!

anji said...

This is cool, it would require some effort on my part :(

nice stuff though :)

Pilot-Pooja said...

Thanks Anji!

I must confess though that i m not giving due time and energy to this blog of mine!

Please share your blog link, i tried to access the same through your profile link, but was unable to do so because of some firewalls!

Ankur said...

Ohooo ,,,do we really need such huge calculations ..
not much articles on ur blog ...n u say blogging is ur hobby ..really ?

Pilot-Pooja said...

Hi Ankur,

Nice to hear from you!
Though i am not able to open you link currently because of some firewall issues at my end, will surely give a visit to ur blog to understand ur wave pattern!

I happen to be more active on my other blog!

But yes, you r right, i m not able to extract desired time for my hobbies since long!

Lavonne said...

Great work.

Pilot-Pooja said...

Thanks Lavonne.

I went through your informative blog link!


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