## Vedic Maths Tricks for Multiplication

Both the videos given below cover Vedic Maths Multiplication Trick i.e. The Criss-Cross Method or Urdhva Tiryak Sutra.

Each video is made using different tools and aids. I would request you to share your opinion on which format of the recording did you like more. Please share your views by posting a comment below. I intend to make more such videos after getting your feedback.

## Vedic Maths Multiplication Tutorial: Video 2

The videos are posted without any sort of editing. Kindly ignore all kind of disturbances and aberrations. Your feedback will help me in improving the quality of future video tutorials, which will be posted for free on Quickermaths.com.

## Quick Multiplication up to 20 x 20

“I’m having trouble above 10×10.”

This was a statement I heard many times while interacting with students preparing for competitive examinations including CAT. This was in response to my appeal to them to memorize tables up to 20×20.

Today I am posting here on QuickerMaths.com, the method which I recommend to my students too.

How to multiply up to 20×20 in your head?

Assumption: You know your multiplication table reasonably well up to 10×10.

I am trying to explain this with an example, Read More

## Checking of Calculations: Casting Out Nines

While doing arithmetic calculations, we should normally check our calculation. But the checking should not be as tedious as the original problem. To solve this problem I am explaining below a very frequently used method which is discussed in Vedic Mathematics as well as by many other mathematicians.

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.

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.

Concept: CHECKING OF CALCULATIONS
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.
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)
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.
Concept: CHECKING OF CALCULATIONS
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.
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)
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.

## Vedic Multiplication of two numbers close to Hundred

Vedic Method of Multiplication: Base System of multiplication

Application: Multiplication of two numbers close to Hundred

Case 1: Both numbers greater than 100.

Rule: 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

103 x 104 = 10712

The answer is in two parts: 107 and 12,

107 is just 103 + 4 (or 104 + 3), and 12 is just 3 x 4.

Similarly 107 x 106 = 11342

107 + 6 = 113 and 7 x 6 = 42

123 x 103 = 12669

(123 + 3) | (23 x 3) = 126 | 69 =12669 .

If the multiplication of the offsets is more than 100 then this method won’t work. For example 123 x 105. Here offsets are 23 and 5.

Multiplication of 23 and 5 is 115 which are more than 100. So this method won’t work.

But it can still work with a little modification. Consider the following examples:

Example 1

122 x 123 = 15006

Step 1: 22 x 23 = 506 (as done earlier)

Step 2: 122 + 23 (as done earlier)

Step 3: Add the 5 (digit at 100s place) of 506 to step 2

Answer: (122 + 23 + 5) | (22 x 23) = 150 | 06 = 10506

Example 2

123 x 105 (Different representation but same method)

123 + 5 = 128

23 x 5 = 115

128 | 115

= 12915

In the next post I’ll tell you about vedic multiplication, i.e.,  how to multiply two numbers lesser than the base (in this case 100).

Here’s the promised post for you – http://www.quickermaths.com/base-method-of-multiplication/

If you liked this method of vedic multiplication included in ancient Vedic Maths, Please leave a comment to let us know.

## Vedic Multiplication by 9, 99, 999 and so on

When any number has to be multiplied by a series of 9s, like 9, 99, 999, 9999 and so on than we can apply this very simple vedic maths technique to increase your speed of calculation.

Multiplication with 9/ 99 / 999 and so on.

we know, 789 × 999 = 788,211

You will get the answers in two parts,

• The left hand side of the answer: subtract 1 from 789, which is 788
• The right hand side of the answer subtract 789 from 1000 = 1000-789= 211

Thus, 999 x 789 = 789-1   |  1000-789 = 788, 211 (answer)

{for the right hand side of the answer, 789 should be subtracted from (999+1)}

or,  99999 x 78 = 78-1   | 100000 – 78

= 7799922

{78 should be subtracted from (99999+1)}

Another example:

1203579 × 9999999 = 1203579-1   | 10000000- 1203579

=120357887964 21

Number in red is 1 less than 1203579. Number in blue is (10000000-1203579). Hence the answer.

This method has to be altered a little bit when number of 9s are lessers than the number of digit in the divisor.

1432  x 9 = 1432 (10 – 1) = 14320 – 1432 = 12888

So for multiplication with 9, put a zero after that number and subtract the number itself from that.

Likewise for 99 put two zeroes after that number .

3256 x 99 = 325600 – 3256 =  322344

## How to Find the Average Speed?

Lot of students gets confused while finding out the average speed, when various distances are travelled with different speed. Say for example, trip to Agra from Delhi is made at an average speed of 40 km/hr and the trip back at an average speed of 60 km/hr. Find their average speed for the entire trip. (Hint: It’s Not 50 Km/hr)

Rule: If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

Or, If a person travels half the distance at a speed of x km/hr and the other half at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

So answer to the above question:

(2*60*40)/(60+40)  = 48 Km/hr

If a person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz/xy + yz + xz km/hr.

This problem can also be dealt with assuming some hypothetical distance.

You can find thousands of such amazing faster calculation tricks in the MAGICAL BOOK on QUICKER MATHS – by M. Tyra

If you’ve a question related to time and speed or any other topic in maths, you can post it on QuickerMaths.com’s question answer platform

## Vedic Multiplication by 11

Speed Vedic Multiplication Trick

Vedic Multiplication by 11

Step 1.

Assume that there are two invisible 0 (zeroes), one in front and one behind the number to be multiplied with 11

say if the number is 234, assume it to be  0 2 3 4 0

Step 2.

Start from the right, add the two adjacent digits and keep on moving left

02340

Add the last zero to the digit in the ones column (4), and write the answer below the ones column. Then add 4 with digit on the left i.e. 3. Next add 3 with 2. Next 2 with 0.

0+4 = 4

4+3 = 7

3+2 = 5

2+0 = 2

So answer is 2574

Similarly,

36 x 11 = 0+3   |   3+6   | 6+0  = 396

74 x 11 =0+ 7 |  7+4 |  4+0 =  7  | 11 |  4 = 814   (1 of 11 is carried over and added to next digit, so 7+1 = 8 )
6349 x 11 = (0+6)  |  (6+3)   |   (3+4)   |   (4+9)  |   9+0 =  69839

This method works for all the number, no matter how long or short, times 11. Just try it yourself and get amazed at the simplicity of the concept.

In the next post will learn Vedic Multilplication by 111, 1111, 11111, and so on.

## Vedic Maths Subtraction

#### Learn Amazingly Fast Vedic Mathematics Subtraction

Very often we have to deduct a number from numbers like 1000, 10000, 100000 and so on.

This Vedic Maths Subtraction method found as sutra in ancient vedas, is given below is very useful for such subtractions.

Memory Trick: ALL FROM 9 AND THE LAST FROM 10

Use the formula all from 9 and the last from 10, to perform instant subtractions.

For example 1000 – 357 = ?      (subtraction from 1000)

We simply take each figure in 357 from 9 and the last figure from 10.
Step 1. 9-3 = 6
Step 2. 9-5 = 4
Step 3. 10-7 = 3

So the answer is 1000 – 357 = 643
And that’s all there is to it!

This always works for subtractions from numbers consisting of a 1 followed by noughts: 100; 1000; 10,000 etc.
Similarly 10,000 – 1049 = 8951      (subtraction from 10000)

9-1 = 8
9-0 = 9
9-4 = 5
10-9 = 1

So answer is 8951,

For 1000 – 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.
So 1000 – 83 becomes 1000 – 083 = 917

Corollary: If last term is 0, keep that last term as 0 and subtract the last non Zero term from 10 .

Illustration: 10000 – 920 = 10000 – 0920 = (9-0) (9-9) (10-2) 0 =9080

Illustration: 100000 – 78010 = (9-7) (9 – 8 ) (9- 0) (10 – 1) 0 = 21990

## Multiply 2 numbers, sum of whose unit places is 10

Vedic Multiplication: This method of multiplication which is from Vedic Maths will make it very easy to multiply two numbers when sum of the last digits is 10 and previous parts are the same

You will get the answer in two parts.

First part, to get left hand side of the answer: multiply the left most digit(s) by its successor

Second part, to get right hand side of the answer: multiply the right most digits of both the numbers.

Example

First part: 4 x (4+1)

Second part: (4 x 6)

Combined effect:  (4 x 5)  | (4 x 6) = 2024

*| is just a separator. Left hand side denotes tens place, right hand side denotes units place

More Examples

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

As you can see this method is corollary of  “Squaring number ending in 5”

It can also be extended to three digit numbers like :

E.g. 1: 292 x 208.

Here 92 + 08 = 100, L.H.S portion is same i.e. 2

292 x 208 = (2 x 3) x 10 | 92 x 8  (Note: if 3 digit numbers are multiplied, L.H.S has to be multiplied by 10)

60 | 736 (for 100 raise the L.H.S. product by 0) = 60736.

E.g. 2: 848 X 852

Here 48 + 52 = 100,

L.H.S portion is 8 and its next number is 9.

848 x 852 = 8 x 9 x 10 | 48 x 52 (Note: For 48 x 52, use methods shown above)

720 | 2496

= 722496.

[L.H.S product is to be multiplied by 10 and 2 to be carried over because the base is 100].

Eg. 3: 693 x 607

693 x 607 = 6 x 7 x 10 | 93 x 7 = 420 / 651 = 420651.

Note: This Vedic Maths method can also be used to multiply any two different numbers, but it requires several more steps and is sometimes no faster than any other method. Thus try to use it where it is most effective

How do you like this Vedic Maths technique, please let us know. You can also share this with your friends.

## How to Quickly Find Square of Any Number Ending in 5

Finding square of any number with unit’s digit being 5 is the most common, yet very interesting trick of Vedic Maths.  Using this technique you can find the square of any number ending in 5 very easily.  Also explore a quick method of squaring numbers ending in 9. Given below is the step by step explanation of this Vedic Maths Method.

Let us take a 2 digit number in generic form, say the number is a5 (=10a+5), where a is the digit in ten’s place

Square of a5= a x (a+1) | 25

That means a is multiplied by the next higher number, i.e. (a+1). Now let’s take example of a real number ending in 5, say 45.

452 = Left hand side of the answer will be 4 multiplied by its successor i.e. 5 and the right hand side part will always be 25 for squares of numbers of which the unit’s digit is 5.

Giving the answer a x (a+1) | 25 ( |     stands for concatenation}

i.e. 4  x  (4+1) | 25 = 4 x 5 | 25 = 2025

Similarly we can proceed for 3 digit numbers ending in 5

Few more examples:

952=9 x 10 | 25 =9025

1252 = 12 x 13 | 25 = 15625

5052 = 50 x 51 | 25 = 255025

Test yourself

Find out the square of 85, 245, 145, 35, 15, and 95?

Answer: 7225, 60025, 21025, 1225, 225, 9025

Please let us know if you like this Vedic Maths trick