At the beginning of the twentieth century, when there was a great interest in the Sanskrit texts in Europe, Bharati Krsna Swamiji tells us some scholars ridiculed certain texts which were headed 'Ganita Sutras'- which means mathematics. They could find no mathematics in the translation and dismissed the texts as rubbish. Bharati Krsna Swamiji, who was himself a scholar of Sanskrit, Mathematics, History and Philosophy, studied these texts and after lengthy and careful investigation was able to reconstruct the mathematics of the Vedas. According to his research all of mathematics is based on sixteen Sutras, or word-formulae.

Bharati Krsna wrote sixteen volumes expounding the Vedic system, which you can find in the book named Vedic Mathematics by Bharati Krsna Tirthaji. These sutras were unaccountably lost and when the loss was confirmed in his final years he wrote a single book: Vedic Mathematics, currently available. It was published in 1965, five years after his death. The term Vedic Mathematics now refers to a set of sixteen mathematical formulae or sutras and their corollaries derived from the Vedas.

You can find more details about Origin of Vedic Maths here at wikipedia - http://en.wikipedia.org/wiki/Bharati_Krishna_Tirtha's_Vedic_mathematics

Related posts:

1. Vedic Mathematics Techniques for Finding HCF
2. Vedic Mathematics Course
3. Benefits of Vedic Mathematics

This post in is in continuation of an earlier post named "Vedic Multiplication of two numbers close to hundred". Though you can understand this post stand alone, yet I'll recommend you to read the linked post before reading this one.

In this post I'll explain how to multiply two numbers lesser than the base (in this case 100). In the earlier post it was about both numbers more than 100.

Multiplication where both numbers are lesser than 100

Rule: Still remains the same as earlier. Here you go -

You will get the answer in two parts

First part, to get left hand side of the answer: Add the difference between 100 and either of the numbers to the other number

Second part, to get right hand side of the answer: multiply the difference from 100 of both the numbers

Example

93 x 94

First part: 93 - 100 = - 7; Add this to the other number, thus 94 + (- 7) = 87

Or you can start with the other number 94;

94 - 100 = - 6; Add this to the other number, thus 93 + (- 6) = 87

Result will be same in both the cases

Second part:

Multiply the difference from 100 of both the numbers.

Hence, (93 - 100) x (94 - 100) = -7 x -6 = 42

Combined effect:  87  | 42 = 8742

*| is just a separator.

Example

92 x 86

Step 1: 92 + (86 - 100) = 78

Step 2: (92 - 100) x (86 - 100) = -8 x -14 = 112

Combined effect will look like this: 78 | ₁12

Step 3: Add the 1 (digit at 100s place) of 112 to 78

Answer: 78 + 1 | 12 = 79 | 12 = 7912

When One number is lesser than 100 and the other is more than 100

Same Rule as Above

Example

96 x 103

First part: 96 - 100 = - 4; Add this to the other number, thus 103 + (- 4) = 99

Or you can start with the other number 103;

103 - 100 = 3; Add this to the other number, thus 96 + 3 = 99

Result will be same in both the cases

Second part:

Multiply the difference from 100 of both the numbers.

Hence, (96 - 100) x (103 - 100) = -4 x 3 = - 12

Combined effect:  99 | -12 = 8742

Now to remove negative sign from the right side, we have to take one from the left hand side. 1 when shifted from left to right becomes 100. Thus we’ll have:

Combined effect:  99 – 1 | 100 - 12 = 9888

*| is just a separator.

Example

89 x 113

= 89 + 13  | -11 x 13

= 102  |  -143

In this case, right side number is greater than 100, so we need to subtract it from next higher 100, i.e. 200. Hence, we’ve to take 2 from left hand side, so that we get 200 on the right hand side.

= 102 - 2 | 200 - 143

= 100 | 57

= 10057

Does this method makes such multiplications simpler for you? Leave a comment below to express your opinion.

Related posts:

1. Vedic Multiplication of two numbers close to Hundred
2. Vedic Multiplication by 9, 99, 999 and so on
3. Multiply 2 numbers, sum of whose unit places is 10

1. To get the first part of the answer, add the digit at the units place to 25 and write the sum
2. To get the second part, calculate the square of units place digit and write it

It’ll be easier to understand this with an example.

To find the square of 57 –

First part: 25 + digit at units unit in 57 = 25 + 7 = 32

Second part: square of 7 = 49

Combining both the parts – 3249 is the answer.

Another way to look at it can explain you the logic behind this technique – 57 can be written as

(50+7)^2

= 50^2 +2*50*7 + 7^2

= 2500 + 100*7 + 7^2

=100*(25+7) + 7^2

You can replace 7 by any other number in unit’s place and get answer for it.

Related posts:

1. Shortcut to Find Square of a Number
2. Vedic Multiplication of two numbers close to Hundred
3. Multiply 2 numbers, sum of whose unit places is 10

This vedic maths trick will help you in multiplying two numbers when these numbers start with the same digit. For example 34 x 37; see their ten’s digit (starting digit) is same. Another example can be 234 x 232, see their hundred’s and ten’s digits (starting two digits) are same.

In one of the earlier post a similar method was described. In that like this trick the starting digit(s) should be same but at the same time the sum of digit at unit’s place should be 10 – please check that out - Vedic multiplication

Learn Multiplication

34 x 37

To multiply 34x37, we know they are in the base 30. Hence the reference point (base) will be 30.

Step 1.

Determine how much more is 34 from 30. The answer is 4

Determine how much more is 37 from 30. The answer is 7

Step 2.

Either add 4 to 37 = 41 or 7 to 34 = 41.

The result will be same always.

Step 3.

Multiply the resultant number from step 2 by the base, which is in this case 30

41x30 = 41x3x10 = 123x10 = 1230

Step 4.

Add to the resultant of step 3 the product of the numbers obtained from step 1. This will give you the answer.

1230+ (4x7) = 1230 + 28 = 1258

Another example,

23 x 29

From step 1 and step 2 above, 23 + 9 = 32 or 29 + 3 = 32

From step 3 : 32 x 20 = 640

From step 4 : 640 + (3x9) = 667

One more example,

234 x 232

From step 1 and step 2 above, 234 +2 = 236 or 232 + 4 = 236

From step 3 : 236 x 230 = 54280

From step 4 : 54280 + (4x2) = 54288

using the above method can also be used for multiplying two numbers close to humdred

Related posts:

1. Vedic Multiplication of two numbers close to Hundred
2. Speed Multiplication by 111 : Vedic Maths
3. Vedic Multiplication by 11

This post is one of the many areas where Vedic Mathematics really surpasses traditional methods as you shall soon see. This post is about dividing any number by 9.

We will start by taking an example

Divide 200103002 by 9

We can initiate solving this problem by writing it as below -

9) 2 0 0 1 0 3 0 0 | 2

The symbol in front of the final 2 is not the number “1”, but a vertical bar “|”. This last position will hold the remainder, if any.

There are only two steps in this procedure.

Step 1:  Bring down the 2. It will look like this:

9) 2 0 0 1 0 3 0 0 | 2

2

Step 2:   Add the answer so far, the “2”, to the number on the upper right. For this example, we would add the “2” from the answer to “0”, the number on the above right. So we have:

9) 2 0 0 1 0 3 0 0 | 2

2 2

Now just repeat this process. Add the 2 to the 0, add the 2 to the 1, etc. We will end up with:

9) 2 0 0 1 0 3 0 0 | 2

2 2 2 3 3 6 6 6 | 8

Try a little advanced problem

9) 3 2 3 6 0 5 2 | 2

Step 1: Bring down the first number in the divisor.

9) 3 2 3 6 0 5 2 | 2

3

Step 2: Add the number on the above right and repeat. If we get carrys, put them in and don’t worry about them for now. So we will have:

9) 3   2   3   6   0   5   2 |   2

3   5   8 14 14 19  21 | 23

So, what do we do with all this stuff? Any number that has a carry, needs to be added to the left. Let’s do this starting with the “21” just after the bar.

9) 3   2   3   6   0   5   2 |   2

3   5   9   5   6   1    1 | 23

Now, notice that the remainder of 23 needs to be reduced until it is below a 9. There are two multiples of 9 in 23 with 5 left over. Therefore, we carry over 2 to the other side of the bar and add it to the 1. We will then have the answer:

9) 3   2   3   6   0   5   2 |   2

3   5   9   5   6   1    3 |   5

There is a little more work involved, but, still a lot less than the conventional way.

Hope to hear some feedback from you in comments or facebook

Author - Vineet Patawari

Related posts:

1. Division in Vedic Mathematics
2. Find the remainder – Vedic Algebra
3. Speed Multiplication by 111 : Vedic Maths

Long Division reduced to one-line shortcut

Example 1:  716769 ÷ 54.

Reduce the divisor 54 to 5 pushing the remaining digit 4 “on top of the flag” (Dhvajanka so to say).

Corresponding to the number of digits flagged on top (in this case, one), the rightmost part of the number to be divided is split to mark the placeholder of the decimal point or the remainder portion.

Let us walk through the steps of this example:
716769 ÷ 54 = 13273.5

1. 7 ÷ 5 = 1 remainder 2. Put the quotient 1, the first digit of the solution, in the first box of the bottom row and carry over the remainder 2
2. The product of the flagged number (4) and the previous quotient (1) must be subtracted from the next number (21) before the division can proceed. 21 - 4 x 1 = 1717 ÷ 5 = 3 remainder 2. Put down the 3 and carry over the 2
3. Again subtract the product of the flagged number (4) and the previous quotient (3), 26 - 4 x 3 = 1414 ÷ 5 = 2 remainder 4. Put down the 2 and carry over the 4
4. 47 - 4 x 2 = 3939 ÷ 5 = 7 remainder 4. Put down the 7 and carry over the 4
5. 46 - 4 x 7 = 1818 ÷ 5 = 3 remainder 3. Put down the 3 and carry over the 3
6. 39 - 4 x 3 = 27. Since the decimal point is reached here, 27 is the raw remainder. If decimal places are required, the division can proceed as before, filling the original number with zeros after the decimal point27 ÷ 5 = 5 remainder 2. Put down the 5 (after the decimal point) and carry over the 2
7. 20 - 4 x 5 = 0. There is nothing left to divide, so this cleanly completes the division

Example 2:  45026 ÷ 47

Reduce the divisor 47 to 4 pushing the remaining digit 7 “on top of the flag” (Dhvajanka so to say).

Corresponding to the number of digits flagged on top (in this case, one), the rightmost part of the number to be divided is split to mark the placeholder of the decimal point or the remainder portion.

Let us walk through the steps of this example:

45026 ÷ 47 = 958.0

1. 4 ÷ 4 = 0 remainder 4. Put the quotient 0, the first digit of the solution, in the first box of the bottom row and carry over the remainder 4
2. The product of the flagged number (7) and the previous quotient (0) must be subtracted from the next number (45) before the division can proceed. 45 - 7 x 0 = 4545 ÷ 4 = 9 remainder 9. Put down the quotient 9 and carry over the remainder 9.
3. Again subtract the product of the flagged number (7) and the previous quotient (9), 90 - 7 x 9 = 2727 ÷ 4 = 5 remainder 7. Put down the quotient 5 and carry over the remainder 7.
4. 72 - 7 x 5 = 3737 ÷ 4 = 8 remainder 5. Put down the quotient 8 and carry over the remainder 5.
5. 56 - 7 x 8 = 0there is nothing left to divide, so this cleanly completes the division.

In reply to the query of Pratiush

To divide 716769 by 156:  Split divisor as 15 and 6

---------------------     11        17       12             12      15        9

15 6               71         6          7         6               9        0        0

----------------- 4         5          9         4               6        7        3

Answer:  4594.673

Remarks:  In the first step we have written that 4 ÷ 4 = 0 remainder 4 instead of 4 ÷ 4 = 1 remainder 0. Otherwise, in the following step, we would have to subtract 7*1=7 from 05 which is not possible.

This post is contributed by Nandeesh Nagarajaia. He is a Chemical Engineer who did his B.Tech from NIT Suratkal.  He is now in IT field as Assistant General Manager(Systems) in Hindustan Copper Limited.

Other guest posts by Nandeesh -

1. Quick Method to evaluate polynomials - Horner's Method
2. Quick calculations for extremely large numbers
3. Shortcut multiplication for approximate numbers
4. Heron's method of finding roots

Related posts:

1. Vedic Division by Nine
2. Divisibility Rules for 7 , 11 and 13
3. Decimal Fraction Rules

Before going further on the method to find the cube root, please make a note of the following points –

1) Cube of a 2-digit number will have at max 6 digits (99^3 = 970,299). That implies if you are given with a 6 digit number, its cube root will have 2 digits.

2) This trick works only for perfect cubes, it will not work for any arbitrary 6-digit

3) It works only for integers

Now let us start with the trick to find cube root of a 5 or 6 digit number in vedic mathematic way.

Say you have to find the cube root of 54872. It is known that it’s a perfect cube.

Now divide this number into two parts. The right hand side should always have 3 digits. Remaining digits will come in left hand side. Do it as shown below.

54            |             872

You know the answer will have 2 digits. Digit at tens place and digit at units place. We will get the digit at tens place using the left hand side of the original number (54) and digit at units place using right hand side of the number (872)

Step 1.

Memorize these tables (very soon you will know why) –

Table 1: Cube of 1 to 10

Table 2: Unit’s digit of Cube Roots

Step 2.

For left hand side we need to use table 1. We have to see between which 2 numbers in the 2^nd column do 54 lies. In this case it lies between 27 and 64. So we will take the cube root of the smaller number i.e. 27 which is 3.

So 3 is the tens digit of the answer.

Step 3.

For right hand side we need to use table 2. Since our original number (the perfect cube) ends in 2 (see 54872), its cube root will ends in 8.

Thus the units digit will be 8.

Combining the results we get the answer as 38.

Thus (54872)^1/3  =  38

Try for perfect cubes like 185193, 42875, 1728.

You might also be interested in the trick of finding square root of any number

I hope you liked the simple trick to find the cube root. Leave your comments below -

Related posts:

1. Shortcut to Find Square of a Number
2. Shortcut to find the Cube of a number
3. Speed Multiplication by 111 : Vedic Maths

I have got a mail from some QuickerMaths follower, to illustrate usage of Vedic Mathematics in branches of mathematics other than arithmetic. This post is for that purpose only. Here I am highlighting the usage of Vedic Mathematics in finding out the remainder when an algebraic expression is divided by another.

Finding out the remainder becomes extremely easy using Vedic Maths.

So lets begin with a simple example -

Find the remainder when

x³ + 4x² + 6x - 7 is divided by (x + 5)

Solution:

Step I: Put divisor equal to 0 .i.e.

x + 5 = 0

x = -5

Step II: The remainder will be f(x).

f (-5) = (-5)³ + 4(-5)² + 6(-5) - 7

= -125 + 100 - 30 – 7

= -62

Example 2: (Mx³ + 3x² -3) and (2x³ – 5x +M) leaves the same remainder when divided by (x -4) Find the value of M.

Solution:

Let R1 and R2 be remainder for 1^st and 2^nd equation simultaneously

Rl = f (4) = M (4)³ + 3(4)² - 3 = 64M+ 45

R2 = f (4) = 2(4)³ - 5(4) + M = M + 108

They leave the same remainder. So,

Since, Rl = R2. We have

64 M + 45= M +108

Or, 63 M = 63

M = 1

Example 4: (Mx³ + x² - 2x – N) is exactly divisible by (x - 1) and (x + 1). Find the value of M and N.

Soln: When the expression is exactly divisible by any divisor, the remainder will be zero.

Now, the remainder, when the divisor is x-1, is

f (l) = M + 1 - 2 - N = 0          .

\M - N = 1                 ………….(1)

And the remainder, when the divisor is x + 1, is

f( -1) = M( -1)³ + (-1)² - 2(-1) - N = 0

-M + 1 + 2 - N = 0

M + N = 3                     …..(2)

Solving (1) & (2), we have,

M = 2, N = 1

Thanks to Nehul from Nagpur for asking this question. If you have any similar question, go to Contact Page and post your queries/suggestions.

Take Care. God Bless!!

Vineet Patawari

Related posts:

1. Vedic Division by Nine
2. Divisibility Rules for 7 , 11 and 13
3. Division in Vedic Mathematics

To appreciate the Vedic Maths process of finding the HCF you first need to know the other methods taught in school. I am giving you two other methods to compare with.

Example 1: Find the H.C.F. of x^2 + 5x + 4 and x^2 + 7x + 6.

Example 1: Find the H.C.F. of x^2 + 5x + 4 and x^2 + 7x + 6.

1. Factorization method:x^2 + 5x + 4 = (x + 4) (x + 1)

x^2 + 7x + 6 = (x + 6) (x + 1)

H.C.F. is ( x + 1 ).

2. Continuous division process.

x^2 + 5x + 4 ) x^2 + 7x + 6 ( 1

x^2 + 5x + 4___________2x + 2 ) x^2 + 5x + 4 ( ½x

x^2 + x__________4x + 4 ) 2x + 2 ( ½2x + 2______0
Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.

Now see Vedic Maths way of finding HCF of 2 algebraic expressions.

i.e. x+1 is the HCF

Isn't it much simpler than the above 2 methods.

Now see some more examples -

Related posts:

1. Division in Vedic Mathematics
2. Finding Cube Root – Vedic Maths Way
3. Vedic Mathematics Course

Vedic Sutra: Vedic Mathematics Technique

Beejank: The Sum of the digits of a number is called Beejank. If the addition is a two digit number, then these two digits are also to be added up to get a single digit.

Find the Beejank of 632174.

As above we have to follow

632174  --> 6 + 3 + 2 + 1 + 7 + 4 --> 23 --> 2 + 3 --> 5

But a quick look gives 6 & 3 ; 2 & 7 are to be ignored because 6+3=9,2+7=9.

Hence remaining 1 + 4 --> 5 is the beejank of 632174.

Checking of Addition

Thumb Rule: Whatever we do to the number, we also do to their digit sum: then the result                 we get from the digit sum of the number must be equal to the digit sum of the answer.

For example: The number: 12+45+96+75+25 =253

The digit sum = 3+9+6+3+7 =28=10=1

Answer’s digit sum: 2+5+3 =10=1 (verified)

Another example:  3.5+23.4+17.5 = 44.4

The digit sum: 8+9+13=8+9+4=21=3

Answer’s digit sum: 12=3 (verified)

Casting Out Nines

This method is also known as "casting-out-nines". The method involves converting each number into its "casting-out-nines" equivalent, and then redoing the arithmetic. The casting-out-nines answer should equal the casting-out-nines version of the original answer. Below are examples for using casting out nines to check addition.

We get the casting-out-nines equivalent of a number by adding up its digits, and then adding up those digits, until you get a one digit number. If our answer is 9, then that becomes 0. As a short cut, we don't have to add in any of the 9's in our work, as these are the equivalent of 0. We can just "cast out" those 9's. For example, 19 becomes 1, without even adding 1 and 9 and getting 10, and then adding 1 and 0 and getting 1. As a further short cut, we can group numbers together which add up to 9, and replace them with 0. 2974 becomes 4, because we can cast out the 9 and the 2+7 (which is also 9 or 0). Well, let's try an arithmetic problem:

137892     3

+ 92743   + 7

------    --

230635     1

3+7=10, casting out 9 we get 1.

This rule is also applicable to subtraction, multiplication and up to some extent to division also

In the next post I will explain the use of this method for all of them.

Related posts:

1. Quick calculations for extremely large numbers
2. Speed Multiplication by 111 : Vedic Maths
3. Vedic Multiplication by 11