# VEDIC MATHEMATICS

"India was the motherland of our race and Sanskrit the mother of Europe's languages. India was the mother of our philosophy, of much of our mathematics, of the ideals embodied in Christianity...of self-government and democracy. In many ways, Mother India is the mother of us all." - Will Durant, American Historian 1885-1981

What is an ancient system of mathematics that is being taught in some of the most prestigious institutions in England and Europe and not in India? It is Vedic mathematics-a long forgotten technique for mathematical calculations! It is amazing how with the help of 16 sutras and 16 upa-sutras you will be able to solve/calculate complex mathematical problems-mentally! The basic roots of Vedic mathematics lie in Vedas jus as basic roots of Hinduism. Vedic Maths form part of Jyothisha which is one of the six Vedangas. To many Indians Vedic and Sanskrit slokas/manthras are relevant only for religious purposes/occasions. But Vedas (written around 1500-900 BCE) in fact are a treasure house of knowledge and human experience-both secular and spiritual. Here you will will get an idea about the power of Vedic Mathematics.

Vedic Number Representation

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

As the name suggests "Vedic Mathematics" has originated from "Vedas".It has originated from " Atherva Veda" the foourth Veda.It helps to solve problems at a much faster speed which one can realize only when one practises.
The ancient system of "Vedic Mathematics" was rediscovered from the vedas between 1911 and 1918 by Sri Bharti Krsna Tirathaji.According to his research all of Mathematics is based on 16 sutras(word formulas)and 13 subsutras.The simplicity of "Vedic Mathematics" means that answer can be obtained in one line. Being a mental system students progress faster in terms of mental ability.

We will now start learning some simple techniques of multiplication:

## Finding square of 2 digit numbers

### Numbers ending with 0

Its quiet simple known to everybody.Need not explain!
Example:302 = 900

### Numbers ending with 1

Let the number be $x$.Since its a number ending with $1$ ,hence its $1$ more than the number ending with $0$.Square of $x-1$ may be obtained easily.Now $x^2$ may be obtained as $x^2=(x-1)^2+(x-1)+x\,$
Or we can follow stepwise procedure as given below:

1. Subtract $1$ from the number and square it.
2. Add the original number and the number minus $1$.
3. Add the results of step 1 and step 2.
 Example: $41^2 = 40^2+40+41\,$
$= 1600+81\,$
$= 1681\,$
Example: $71^2 = 70^2+70+71\,$
$= 4900+141\,$
$= 5041\,$


### Numbers ending with 2

Let the number be $x$.Since its a number ending with $2$ ,hence its $2$2 more than the number ending with 0.Square of x-2 may be obtained easily.Now $x^2$ may be obtained as: $x^2=(x-2)^2+2((x-2)+x)\,$

1. Subtract $2$ from the number and square it.
2. Add the original number and the number minus $2$ and multiply by $2$.
3. Add the results of step 1 and step 2.
  Example: $42^2=40^2+2(40+42)\,$
$=1600+2.82\,$
$=1764\,$
Example: $72^2=70^2+2(70+72)\,$
$=4900+284\,$
$=5184\,$


### Numbers ending with 5

Consider any two digit number x = a|b. This means a number "x" whose digit at unit's place is "b" and digit at ten's place is "a"i.e. 27=2|7,or we can say that "|" is digit separator.Its square may be given as a.(a+1)|25.That is the last two digits are always 25 and the digit at hundred's place is the product of the number at ten's place "a" multiplied by "a+1".

  Example:$35^2=3.(3+1)|25=1225\,$
$75^2=7.(7+1)|25=5625\,$


### Numbers ending with 9

1. Add 1 to the number and square it.
2. Add original number and (number+1).
3. Subtract number obtained in step 2 from number obtained in step 1.

In other words $x^2=(x+1)^2-(x+x+1)\,$

  Example:$39^2=40^2-(39+40)\,$
$=1600-79=1521\,$
$69^2=70^2-(69+70)\,$
$=4900-139=4761\,$


### Numbers Starting with 5

These are numbers where digit at ten's place is 5 that is numbers of the type 5|a ,where a is the digit at unit's place(or we can say that "|" is a digit separator). Its square is given by $25+a|a^2$.

1. Note that this will be a 4 digit number.
2. First two digits from left are given by $a^2$
3. Next two digits from left are $25+a$
  Example:$52^2=25+2|2^2\,$
$=27|04\,$
$=2704\,$
$56^2=25+6|6^2\,$
$=31|36\,$
$=3136\,$


### The General Rule

Consider any two digit number $x=a|b$. This means a number $x$ whose digit at unit's place is $b$ and number at ten's place is $a$.Its square may be given as $a\cdot(x+b)|b^2$

 Example:$13^2=1\cdot(13+3)|3^2$
$=16|9\,$
$=169\,$
$25^2=2\cdot(25+5)|5^2$
$=60|25\,$
$=625\,$  (there can not be two digits at unit's place ,so 2 is carried over)
$34^2=3\cdot(34+4)|4^2$
$=3\cdot38|16$
$=114|16\,$
$=1156\,$


### Another Alternative

Consider any two digit number $x= a|b$. This means a number "x" whose digit at unit's place is "b" and number at ten's place is "a".Its square may be obtained as follows:

1. Digit at unit's place is $b^2$
2. Digit at ten's place is $2.a.b$(Keep the right-most digit and the left digit will be carried over)
3. Digit at hundred's place is $a^2$ + carry over from step 2
  Example:$53^2=5^2|2.5.3|3^2\,$
$25|30|9 = 2809\,$
$74^2=7^2|2.7.4|4^2\,$
$=49|56|16\,$
$= 49|57|6\,$
$= 5476\,$


## Multiplication of Two Digit numbers

### Multiplication of Any Two Digit Numbers by 99

Let $x=a|b$ be any number where $a$ is the digit at ten's place and $b$ is the digit at unit's place e.g. the number 76=7|6 or we can say that "|" is a digit separator. Then
$x\cdot99=a|(b-1)|(9-a)|(10-b)$

 Example: $76\cdot99=7|(6-1)|(9-7)|(10-6)=7524$
$54\cdot99=5|(4-1)|(9-5)|(10-4)=5346$
$87\cdot99=8|(7-1)|(9-8)|(10-7)=8613$


### Same digit at Ten's place & sum of digits at Unit's place is 10

Let the two numbers be x=a|b and y=a|d Which means that digits at ten's place of these numbers is "a" and the digits at unit's place are "b" and "d" resp. with the condition that b+d=10.The product x.y is simply calculated as $x.y=a.(a+1)|b.d\,$

Example:$46.44 = (4.5)|6.4 = 2024\,$
$73.77 = (7.8)|7.3 = 5621\,$
$82.88 = (8.9)|2.8 = 7216\,$


### Any Two 2 Digit Numbers

Let the two numbers be x=a|b and y=c|d where as explained earlier a and c are digits at ten's place and band d are digits at unit's place. Now the product $x\cdot y=a\cdot c|(a\cdot d)+(b\cdot c)|b\cdot d$. Let me explain this in the following steps:

1. The product will be a 4 digit number
2. Digit at unit's place $b\cdot d$(if these are 2 digits one will be carried over)
3. Digit at ten's place is $(a\cdot d)+(b\cdot c)$+carried over from step 1((if these are 2 digits one will be carried over)
4. Digit at remaining two places are given by $a\cdot c$+carry over from step 3.
 Example: $23\cdot46 =2\cdot4|2\cdot6+3\cdot4|3\cdot6$
$=8|24|18\,$
$=(8+2)|(4+1)|8\,$
$=1058\,$
$58\cdot73 =5\cdot7|5\cdot3+8\cdot7|8\cdot3$
$=35|71|24\,$
$=(35+7)|(1+2)|4\,$
$=4234\,$