第一種情況：個位數互補 十位數相同的乘法

前部分 ：首位數加1再乘首位數

後部分：尾乘尾

例題1：62 × 68 = ?

前部分：(6 + 1) × 6 = 42

後部分：2 × 8 = 16

完整計算：62 × 68 = 4216

 a, b = {1, 2, 3, 4, 5, ...,9}

 a x 10 + b

 a x 10 + (10 - b)

 (a x 10 + b) x {a x 10 + (10 - b)}

               = (a x 10) x (a x 10) + (a x 10) x 10 - (a x 10) x b

                  + (a x 10) x b + b x (10 - b)

               = (a x 10) x {(a + 1) x 10} + b x (10 - b)

               = a x (a + 1) x 100 + b x (10 - b)

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第二種情況：十位數互補個位數相同的乘法

前部分：首位相乘加個位

後部分：尾乘尾

例題2：69 × 49 = ?

前部分：6 × 4 + 9 = 33

後部分：9 × 9 = 81

完整計算：69 × 49 = 3381

 a, b = {1, 2, 3, 4, 5, ...,9}

 a x 10 + b

 (10 - a) x 10 + b

 (a x 10 + b) x {(10 - a) x 10 + b}

               = (a x 10) x {(10 - a) x 10} + (a x 10) x b

                  + (b x 10 x 10) - (b x a x 10) + b x b

               = {a x (10 - a)} x 100 + (b x 10 x 10} + b x b

               = {a x (10 - a)} x 100 + (b x 100) + b x b

               = {a x (10 - a) + b} x 100 + b x b

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第三種情況： 互補數乘以疊數

前部分：互補數加1再乘以疊數的十位數

後部分：尾乘尾

例題3：82 × 33 = ?

前部分：(8 + 1) × 3 = 27

後部分：2 × 3 = 6 即06，在6前加一個0補足兩位數。

完整計算：82 × 33 = 2706

 a, b = {1, 2, 3, 4, 5, ...,9}

 a x 10 + (10 - a)

 b x 10 + b

 {a x 10 + (10 - a)} x (b x 10 + b)

               = (a x 10) x (b x 10 + b) + 10 x (b x 10 + b)

                  - a x (b x 10 + b)

               = (a x 10 + 10) x (b x 10 + b)

                  - a x (b x 10 + b)

               = {(a + 1) x 10} x (b x 10 + b)

                  - a x (b x 10 + b)

               = {(a + 1) x 10} x (b x 10) + {(a + 1) x 10} x b

                  - a x (b x 10) - a x b

               = (a + 1) x b x 100 + a x 10 x b + 1 x 10 x b

                  - a x b x 10 - a x b

               = (a + 1) x b x 100 + 10 x b - a x b

               = (a + 1) x b x 100 + (10 - a) x b

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第四種情況：兩位數乘法，二個數字離100不遠，補數計算

前部分：原本數字

後部分：100減去原本數字

例題1：96 × 88 = ?

前部分：96

前部分：88

後部分：96 - 100 = -4

後部分：88 - 100 = -12

完整計算：

百位計算：前部分後部分交錯相減 (96-12)或(88-4)=84

個位計算：兩後部分相乘 -4 x -12 = 48

得到 84 48

 (100 - a)      a

 (100 - b)      b

 (100 - a)(100 - b) = 10000 - 100 * (a + b) + ab

               = 100 (100 - a - b) + ab

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Indian Multiplication Table 19 x 19

14 x 12 = ?

step 1：  14 +  2 =  16

step 2：  16 x 10 = 160

step 3：    4 x   2 = 8

step 4： 160 +  8 = 168

 10 + a

 10 + b

 (10 + a) x (10 + b) = (10 + a) x 10 + 10 x b + a x b

               = (10 + a + b) x 10 + a x b

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Vedic Mathematics – Multiplication – Nikhilam Method

Steps to be followed for multiplication

1. The two numbers to be multiplied are written one below the other. Their deviations are written in front of the respective numbers.
2. Answer space is divided into left hand side (LHS) and right hand side (RHS) by a slash.
3. RHS of the answer is the product of the deviations of the numbers and will contain the digits equal to the number of zeroes of the base.
4. LHS of the answer is the sum of one number with the deviation of the other.
5. If RHS contains less number of digits than the number of zeroes in the base , the remaining digits are filled by giving zero or zeroes on the left side of RHS. If the number of digits are more than the number of zeroes in the base , the excess digit or digits are to be added to LHS of answer.
6. Remove the slash to get the answer.

Example 1
Step I : Here numbers (105 , 107) are near base 100. So RHS will have two digits. D₁ x D₂ = 05 x 07 = 35 (RHS)
Step II : (N₁+D₂) = 105 + 07 = 112 OR (N₂+D₁) = 107 + 05 = 112 (LHS)
Step III : Product 11235

Example 2
Step I : Here numbers (98 , 93) are near base 100. So RHS will have two digits. D₁ x D₂ = (-02) x(-07) = 14 (RHS)
Step II : (N₁+D₂) = 98 + (-07) = 91 OR (N₂+D₁) = 93 + (-02) = 91 (LHS)
Step III : Product 9114

Example 3
Step I : Here numbers (992 , 996) are near base 1000. So RHS will have three digits. D₁ x D₂ = (-008) x(-004) = 032 (RHS)
Step II : (N₁+D₂) = 992 + (-004) = 988 OR (N₂+D₁) = 996 + (-008) = 988 (LHS)
Step III : Product 988032

Example 4
Below is the example of 2 digit number multiplication which has base other than 100 and this base need to be modified. Consider example of multiplying 46 x 45. Here base is 50.
Step I : 46 x 45
Step II : A = 46 - 50 = -4
Step III : B = 45 - 50 = -5
Step IV : C = -4 x -5 = 20
Step V : D = 46 - 5 = 45 - 4 = 41
Step VI : Result = 50 x 41 + 20 = 2070

Example 5
Consider example of multiplying 39 x 44. Here base is 40.
Step I : 39 x 44
Step II : A = 39 - 40 = -1
Step III : B = 44 - 40 = 4
Step IV : C = -1 x 4 = -4
Step V : D = 44 - 1 = 39 + 4 = 43
Step VI : Result = 40 x 43 - 4 = 1716

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Vedic Mathematics – Division – Nikhilam Method

Division by Vedic methods - PDF

1. Nikhilam Sutra (Specific Technique)
2. Paravartya Sutra (Specific Technique)
3. Anurupyena Sutra (Specific Technique)
4. Direct Flag Method (General Technique)
5. Ekadhikena Purvena (Specific Technique)

Nikhilam Sutra (Specific Technique)

Nihilam Sutra - slideshow
Fastest Vedic methods for Division - Nikhilam Sutra - youtube
Fastest division tricks for Division by 9 - youtube

Nihilam Sutra is a Specific Method to Divide Numbers using Vedic Mathematics. This Vedic Maths Division Method can be applied when Divisor is closer to power of 10 BUT less than that of it. Using Nikhilam Sutra, you can easily divide when divisor is like 98, 92, 995, 89997, etc.

Paravartya Sutra (Specific Technique)

Fastest Vedic methods for Division - Paravartya Sutra - youtube

Paravartya Sutra is a Specific Method for division in Vedic Maths. This Vedic Maths Division Method can be applied when Divisor is closer to power of 10 BUT greater than that of it. Using Paravartya Sutra, you can easily divide when divisor is like 123, 104, 1112, etc.

Anurupyena Sutra (Specific Technique)

Fastest Vedic methods for Division - Anurupyena Sutra - youtube

Anurupyena Sutra is another Specific Vedic Maths Division Tricks which shows how to divide numbers when Nikhilam and Paravartya are not applicable. Using Anurupyena Sutra, we multiply Divisor by a factor so that either Nikhilam or Paravartya Sutra can be applied.

Vinculum Process of Division

Vinculum is another division in vedic maths tricks which can be applied when Divisor has digits greater than 5. Using Vinculum Process, convert those bigger digits to smaller digit and then apply Nikhilam Sutra or Paravartya Sutras of Division.

Direct Flag Method (General Technique)

Direct Flag Method - youtube

Direct Division (Flag Method) is a general method of Division in Vedic Mathematics shows shortcut to divide any types of numbers. It is a shortcut method for division of large numbers.

Ekadhikena Purvena (Specific Technique)

Vedic Division using "Ekadhikena Purvena" - youtube

It is another shortcut method of division in Vedic Maths when Divisor is ending with 9.

Quick Square method

Quick Square around 50
Quick Square around 100