# [BEST] Vedic Maths Multiplication Makes You Crazy 2021

In mathematics multiplication is one of the four elementary operations of arithmetic, the other ones being addition, subtraction and division. The result of a multiplication operation is called a product.

In Vedic Maths, there are various methods of multiplication which can make your calculation easy and quick. Regular practice of these multiplication methods enable you to choose the one that best suits a particular problem.

These methods are not only helpful to all kinds of students, but also helpful for people of all walks of life, e.g. office people, homemakers, teachers etc. Because in some way other people need to multiply in their day to day life.

Below are some simple multiplication methods that can really improve your calculation speed.

Here we are going to cover different multiplication methods:

## Multiplication Methods

### Multiplication With 99, 999, 9999, 99999

We have already learnt about Base and Compliment in our previous post. These two systems we are going to use many times. So you must be well versed.

Every number has its base like any single digit number or double digit number base is 10, 20 or 100, etc. The Bases of 1, 2, 3………6, 7, 8, 9 are 10

For example:

6 base is 10

8 base is 10

17 base is 20

87 base is 100

We can take the nearest 10s, 100s or 1000s. E.g. The base of 75 is 100, but we will take the nearest 10s. So 75 nearest 10s is 80. So we will take 80 as a base for 75 for calculation.

Now,

If we take 20 as a base of say 17, then what is the difference between these two, it is 3. 20-17=3. This difference is called complement.

For example, if we take 30 as the base of 26. Then the complement is 4. Therefore it is very easy to find a complement for small numbers, but what if there are big numbers?

It will take time to solve and there are chances for mistakes. e.g., complement (difference) of 6834 and 10000 is 3166. Which technique do you use? Traditional methods take time but if you apply Vedic maths technique it will take not more then 2 sec.

10000

-6834

_______

3166    (Takes more time)

Vedic Maths method:

9 9 9 10   (solve it in your mind)

-6 8 3 4

_________

3 1 6 6    (Applying ‘All from 9 and the last from 10’ method) takes 2 sec.

Now find complement of: (just for practice)

76, 576, 457, 687, 7865, 2345, 5873, 77654

Now, let’s use this complement method in real problems.

Suppose you multiply 543 with 999, then do you think it’s easy. Yes it is, but it will take time and in between you may make mistakes.

So how to solve it?

543 × 999

9 9 9

×5 4 3

______    ( first line)

_____×    ( second line)

___× ×     ( third line)

________

542457     ( final answer) Takes a lot of time and gives confusions. ☺☺

Vedic maths method: ??

Example 1:

543 × 999

First write down one number lesser than 543, that is 542, then apply the complement method. Deduct 543 from 9, 9, 10 (All from 9 and the last from 10′ method)

First step: write one number lesser than 543 that is 542

Second step: deduct 543 from 9 9 10

9 | 9 | 10  (solve in your mind)

-5   4     3

4    5     7

Final step: Write 542 first and then 457

(Remember in vedic maths always start from left side and write the actual value)

Isn’t it so exciting?

Same way you can solve any number with 99, 999, 9999, 99999 etc

67 × 99

9 | 10

Ist step:    6    7

3    3

Final step:  6 6 3 3 – Answer

⬇️

(One less than 67)

Example 2:

367 × 9999

In this case, there is one extra 9. So what you do is put one 0 before 367, then it becomes 0367, then apply the complement method.

9   9   9   10

0   3   6     7

_____________

9   6   3     3

Final step –  write one less than 367 and attach 9633.

Example 3:

367 × 99

In this case the number of 9 is less. Here the system of solving is a bit different. Two 9’s means we have to apply a complement for two numbers only. That means you don’t have to take a full 367, but only 67. 3 is remaining. So add 1 to 3, which become

3 + 1 = 4, then deduct 4 from 367. It will become 367 – 4 = 363.

First step:

3    6  7

↘️

-4  (3 + 1)

_______

3   6  3 ……(i)

Second step:

Now apply the complement system.

9 10 – all from 9 last from 10

6  7

_______

3  3 …….(ii)

Final step: join (i) and (ii)

3 6 3 3 3- Answer

(i)    (ii)

Similarly,

45678 × 99999 = 4567754322 – Answer

67 × 99 = …..

785 × 999 = ….

87432 × 99999 = ….

468 × 9999 = …..

53 × 999 = ….

642 × 99 = …..

Balancing Rule

Example 1:

2 3 | 45 If we balance this it will become 275.

How?

Let see

Write the first digit as it is and carry tens digit to the next and add. Here first digit is 5

2 3 | 4 5  =  2 7(4+3)5= 275

|     |

Example 2:

23 | 38 | 14

It’s very simple: drop 4 carry 1 to next number 38, it will become 38+1=39, again drop 9, carry remaining 3 to next number 23, it will become 23+3=26. So the answer will be 2694

2 3  |  3 8  |  1 4

|      |   |      |⬇

________________

2(3+3)6 (8+1)9 4

### Multiplication With Single Digit

i)    512 × 5

Solution:

5   1    2

× 5

_________

25 | 05 | 10 (now apply Balancing Rule)

|    |  |    |

ii)  3245 × 6

Solution:

3    2    4     6

× 6

______________

18 | 12 | 24 | 36

|    | |    | |    |

1  9      4     7  6  (Answer)

iii)  5678 × 9

Solution:

5     6     7    8

× 9

_____________

45 | 54 | 63 | 72

|    | |    | |   |

5    1     1    0   2  (Answer)

⬇️

(7+3=10- drop 0 and carry 1 to next that is 6+1=7 then add 7 with next 4 it will become 11, drop 1 carry another 1 to next)

### Multiplication With Double Digit

#### Multiplication within and between 12, 13,14 …… 19

i) 12 × 13

Step 1- 12 + 3 = 15

Step 2-  3 × 2 = 6

ii)  12 × 14

12 + 4= 16, 4 × 2= 8

iii) 14 × 16

14 + 6= 20, 6 × 4= 24

20 | 24 (use Balancing rule)

So now practice :

14 × 18 = ……

15 × 19 = ……

16 × 18 = ……

13 × 19 = ……

19 × 19 = …..

#### Multiplication With 11

Rule- If there is 11 or two times one then put one 0 before the actual  number and one 0 after the number. If there is three times one 111, then put two 0 before the number and two 0’s after the number, if there is four times one 1111, then put three 0’s before the number and three 0’s after the number, and so on.

i)  26 × 11

In this case there are two times one then we will write 0260.

0 2 6 0    now add two numbers from right

First step:       Add 0 + 6 = 6

Second step:  Add 6 + 2 = 8

Third step:      Add 2 + 0 = 2

ii)  26 × 111

In this case there are three times one then as per the rule we will write 002600. Now add three numbers from the right.

002600

Ist-     0 + 0 + 6 = 6

IInd-   0 + 6 + 2 = 8

IIIrd-   6 + 2 + 0 = 8

IV-      2 + 0 + 0 = 2

Now do it yourself

iii) 26 ×1111= …..

00026000 (add four numbers from the right)

iv) 57 × 111 = ……

v)  643 × 11 = …..

vi) 6789 × 1111 = …..

iv) 57 × 111 = ……

v)  643 × 11 = …..

vi) 6789 × 1111 = …..

#### Multiplication With 22, 33, 44, …. 99

i) 26 × 22

First method: Apply same formula like 11

26 × 2 (11) (take 2 common)

52 × 11

0520

Second method:

|2        6|

↗️↖️

×|2        2|×

____________

4 | 16 | 12 (apply Balancing rule)

Ist step:   Multiply 6 ×2 =12

IInd step: Multiply 2 × 2 =4

IIIrd step: Cross multiply 2 × 2 = 4 and 2 × 6 = 12, than add 4 + 12 = 16

iii) 56 × 44

• 56 × 4 (11)-  (take 4 common)
• 224 × 11   (56 × 4 = 224)
• 02240

iv) 46 × 222

46 × 2(111)

–   009200

Now practice:

iv) 76 × 66 = …..

v)  87 × 99 = …..

vi) 73 × 88 = ….

#### Multiplication of two digits where unit place numbers are the same and sum of tens place numbers are 10

Example 1:

43 × 63, in this case unit place numbers are same that is 3 and sum of tens place number is 10 – (4 + 6 = 10)

First step: Multiply unit place numbers and write the answer as it is.

Second step: Multiply ten digit numbers and add one unit place number 3 (In case unit place number is 4 then add 4, or in case it is 5 , 6, 7 etc then add accordingly)

4  3

× 6  3

_______

27 09

(Here 27 = 6 × 4 + 3, and 09 = 3 × 3, 0 is there because we need two digits)

Example 2:

7 8

× 3 8

______

29 64

#### Multiplication of two digits if the sum of unit place is 10 and tens place numbers are the same.

Example 1:

6 7 × 6 3

First step: Multiply unit place numbers 7 × 3 = 21, write  down as it is.

Second step: Add 1 to upper tens place number that is 6 + 1 =7, then multiply 6 × 7 = 42

6 7

× 6 3

______

42 21

Example 2:

8 6

× 8 4

______

72 24

{72 = 8 × 9(8+1), 24 = 6 × 4}

38 × 32 = …..

56 × 54 = …..

73 × 77 = …..

67 × 47 = …..

84 × 26 = …..

### Multiplication by factors

Example 1:

44 x 14

Since we know 14 = 2 x 7, d. In order to multiply any number by 14, we will first multiply it by 2 and then multiply the answer by 7.

64 x 14 = 44 x (2 x 7)

Step 1:

First multiply 64 by 2, i.e, 64 x 2 = 128

Step 2:

Multiply the number obtained in step 1 with 7 to get the answer, i.e., 128 x 7 = 896

Answer: 64 x 14 = 896

Example 2:

Step 1:

69 x 49 = 69 x (7 x 7)

Step 2:

Multiply 69 by 7, i.e, 69 x 7 = 483

Step 3:

Multiply the number obtained in step 2 with 7 to get the answer, i.e., 483 x 7 = 3381

### Multiplication involves 25 or 50

Example:

26 x 25

We know that 25 is one-quarter of 100, that is, 100 /4 = 25. So instead of multiplying by 25, you multiply the number by 100 and then divide the answer by 4.

Step 1:

26 x 25 = 26 x 100/4

Step 2:

Multiply the number by 100

26 x 100 = 2600

Step 3:

Divide the answer obtained in Step 2 with 4

2600/4 = 650

Answer- 26 x 25 = 650

Similarly, when multiplying with 50, the fraction will become 100/2. The rest of the method will remain the same.

### Multiplication Using Base Method

Base is a number that is close to multiples of 10.

For example, 12 x 14 – In this, the base will be 10.

#### Multiplication With Same Base

Case 1:

In case of surplus.

Example 1:

Multiply 12 by 14

Step 1:

Identify the base. In this case 10 is close, this will be the base.

Step 2:

The difference between number 12 and base is (+2). Since the difference is positive, we call it surplus (the negative difference is called deficit).

Similarly, the difference between number 14 and base is (+4).

Step 3:

Add either (+2) to 14 or (+4) to 12. This process is called cross addition. Either way, we get 16 as an answer.

Step 4:

Now multiply the surpluses, i.e, 2 and 4

2 x 4 = 8

(Note-  in case base is 100 then you need to write 08 instead of just 8, in case of 1000, you need to write 008 etc.)

Step 5:

Join the answers obtained in step 3 and 4, i.e., 16 + 8 = 168

It’s very easy and interesting. Isn’t it?

Example 2:

Multiply 43 x 44

Step 1: You take 40 base because 43 and 44 are closer to 40. So, 43 is 3 more than 40 and 44 is 4 more than 40.

Step 2:

Surplus

⬇️

4 3     —  3

4 4↗️ —  4

Step 3: Cross addition gives us 47.

By multiplying 47 with 40, we get 1880

(Note- In case base is 10, 100 or 1000 etc. Then you need not to do this multiplication like 47 × 40. So anything more then 10 or 100 like 20, 30, 200, 300 etc we need to multiply)

Step 4: Multiply surpluses 3 x 4 = 12

Step 5: 1880 + 12 = 1892. This is our answer.

Case 2:

In case of deficits.

Example:

Multiply 38 x 37

Step 1: Since 30 will not be close to the numbers, we will choose 40 as our base.

Step 2: 38 – 40 = (-2), 37 – 40 = (-3)

Step 3: Cross subtraction gives us 35 (38 – 3) or (37 – 2), take anyone out of this.

35 x 40 = 1400

Step 4: (-2) x (-3) = (+6)

Step 5: 1400 + (+6) = 1406

Therefore, Answer is 38 x 37 = 1406

Case 3:

If one number is in surplus from base and another number is deficit from base.

106 × 98

Here you take base 100

Step 1:  106 —- (+6)

98 —- (-2)

Step 2:  Cross addition of 98 and 6 or cross subtraction of 106 and 2 gives you 104

But in this case we take one less then 104 that become 103 ( it is because product of +6 and -2 is -12, so if the product is positive then there is no need to deduct 1)

1 0 6 — (+6)

9 8 — (-2)

____________

1 0 4      -12

Step 3: Here you don’t write 12 directly to answer, instead you use complement method that is

1 0 6 — (+6)

9 8 — (-2)

_____________

1 0 4 —-  9  10 (All from 9 last from 10)

1    2

______

8    8

Step 4: Combine results obtained from step 2 and step 3, i.e, 103 (one less than 104) and 88

#### Multiplication With Different Base

Example:

998 × 97

Step 1: In this case there are two different bases. For 998 you take 1000 base and for 97 you take 100 base.

998 — (-2) (1000 base)

97 — (-3) (100 base)

Step 2: Take ratio of 1000 and 100

R = 1000/100

R = 10

Step 3: Multiply either ratio 10 with 97, it will become 970 and then do cross subtraction with -2, it will become 968 or multiply ratio 10 with -3 which will become -30 and then do cross subtraction with 998, it will become 968

9 9 8 — (-2)

9 7 — (-3)

____________

9 6 8    06 (consider lower base for putting 0’s. Here two bases are 1000 and 100. So take the lower base which is 100. In 100 there are two 0’s, that means two digits should be there, so you write 06, because 06 are two digit number)

#### Multiplication of different bases with one surplus and one deficit.

Example:

9998 × 104

First step: Take bases

9998—- (-2)  (10000 base)

104 —- (+4) (100 base)

Second step: Find out base ratio

Base Ratio R = 10000/100

R = 100

Third step:

Multiply 100 with 104 which  will become 10400 and cross subtraction with -2, it will become 10398, or multiply 100 with 4 which will become 400 and cross add with 998 which  will become 10398.

9 9 9 8 — (-2)

1 0 4 — (+4)

_______________

1 0 3 9 8   -08

1 0 3 9 8     9 10 (complement system)

-1     0   8

______________

1 0 3 9 7     9   2

#### Multiplication With Sub-Base

204 × 203

Step 1: Find out the nearest base. In this case the nearest base is 200 not 100 because it is far.

Step 2: Find out the differences.

Cross addition either 203 with 4 or 204 with 3 it will give 207

204 — 4 (200 base)

203 — 3

___________

207     12 (203+4=207)

×2 (because of 200 base)

____________

414     12

### Multiplication With Criss-Cross Method

Multiply unit digit numbers, and ten digit  numbers. Cross multiply and add. Apply Balancing rule.

Case 1:

Example:

4 6 × 7 8

First step:

Multiply 6 and 8 becomes 48, —(i)

Multiply 4 and 7 gives 28,       —(ii)

Cross multiply 4 and 8 gives 32

Cross multiply 6 and 7 gives 42

Add 32 + 42 gives 74            —(iii)

Second step: Write  down the results

4    6

×   7    8

_________

(ii) | (iii) | (i)

28 | 74 | 48  (apply Balancing  rule)

|   |  |    |

___________

3   5    8   8

Case 2:

Example 1:

234 × 356

First step:

Multiply 4 and 6 gives 24 —— (i)

Cross multiply 6 and 3 gives 18 and cross  multiply 5 and 4 gives 20, add 20 and 18 gives 38 ………. (ii)

Cross multiply 6 and 2 gives 12, cross multiply 3 and 4 gives 12, multiply 5 and 3 gives 15, add 12, 12 and 15 gives 39 ….. (iii)

Cross multiply 5 and 2 gives 10, cross multiply 3 and 3 gives 9, add 10 and 9 gives 19 …… (iv)

Multiply 2 and 3 gives 6 …….. (v)

2           3          4

3           5          6

_______________

(v) | (iv) | (iii) | (ii) | (i)

6 | 19  | 39  | 38 | 24

|    | |    |  |    |  |    |

8     3      3     0   4

Example 2:

513 × 371

5        1        3

3        7        1

________________

15 | 38 | 21 | 22 | 3

### Multiplication With 21, 31, 41….91, 211, 311,….911, 2111, 3111…9111

It is similar to multiplication with 11. Put one 0 before and after the number (in case of 211 put two 0’s before and after the number and so on).

Example 1:

2345 × 21

First step: put one 0 before and after the number

0 2 3 4 5 0 × 2

Second step: Multiply with 2 and add with the next number. Like 2 × 0 + 5, 2 × 5 + 4….

0 2 3 4 5 0

× 2

____________

4   9   2  4  5 – Answer

Example 2

3456 × 31

Step 1: put one 0 before and after the number and multiply with 3

034560 × 3

Step 2: Multiply with 3 and add with the next number. Like 3 × 0 + 6, 3 × 6 + 5, …..

0 3 4 5 6 0

× 3

___________

1 0 7 1 3  6 – Answer

Your assignment: do it for 31, 41, 51, 61..91

Example 3:

1234 × 211

Step 1: Apply the same procedure like 21. Only difference is put two 0’s before and after the number and multiply by 2

00123400 × 2

Step 2: Multiply with 2 and add with next two numbers. Like 2 × 0 + 0 + 4, 2 × 0 + 4 + 3 …

0 0 1 2 3 4 0 0

× 2

________________

2  6  0 3  7  4

》 Answer of 1234 × 211 is 260374

Same way you can do for anyone of these. Practice for 311, 411…..3111, 4111…9111 etc.

### Question 1: How to do multiplication by 11 in Vedic Maths

Answer: If there is 11 or  two times onethen put one 0 before the actual  number and one 0 after the number. If there is three times one 111, then put two 0 before the number and two 0’s after the number, if there is four times one 1111, then put three 0’s before the number and three 0’s after the number, and so on.
Then from right add in twos, or threes depending upon number of 0’s.

i)  26 × 11
In this case there are two times one then we will write 0260.

0 2 6 0    now add two numbers from right

First step:       Add 0 + 6 = 6
Second step:  Add 6 + 2 = 8
Third step:      Add 2 + 0 = 2