What is Vedic Speed Mathematics?

Contents:

1. Introduction to Vedic Speed Mathematics

2. Features & Benefits of Learning “Vedic Speed Mathematics”

3. Some Solved Examples (Multiplication using Base Method)

4. Some Solved Examples (Multiplication using Criss Cross Method)

5. Some Solved Examples (Squares)

6. Some Solved Examples (Polynomials)

7. List of Formulas and their Meaning

8. List of Corollaries and their Meaning

1. Introduction to Vedic Speed Mathematics:

• Vedic Speed Mathematics is World’s Fastest Mental Calculation System.
• It is Amazingly Compact and Powerful System of Calculation.
• It is a collection of simple Formulas to solve mathematical problems in easy and faster way.
• It Consists of 16 Formulas (Aphorisms) and 16 Corollaries (Sub-formulas).
• It was rediscovered from the Vedas by Jagadguru Shankaryacharya Sri Bharati Krishna Tirthaji Maharaja of India.

2. Features & Benefits of Learning “Vedic Speed Mathematics”

• It is Very Simple, Efficient, Fast, Coherent, Flexible, Original & Straight Forward.
• It is Amazingly Compact and Powerful System of Calculation.
• It Leads to Improvement in Mental Ability, Sharpness, Creativity & Intelligence.
• Problems are reduced to One Line Answers.
• It can be Learnt and Mastered with Ease and in little Time.
• Covers from Basic Numeracy Skills to Advanced Math Topics.
• Faster Calculations and High Accuracy level when compared to the Conventional Methods.
• Increased Concentration and Confidence.
• Vedic Mathematics System also provides a set of Independent Cross Checking Methods.
• It Helps in Achieving Academic Success.

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3. Some Solved Examples (Multiplication using Base Method)

Case 1: When both numbers are lesser than the working base:

Working Procedure:

• Write multiplicand and multiplier one below the other.
• Write complements of multiplicand and multiplier to its right side with signs.
• Answer consists of two parts. Left and Right.
• Left Part: Evaluating any of the cross values.
• Right Part: Product of both complements (right side values).
• Caution: Total number of digits in the Right Part should be equal to total number of zeroes in the base. If lesser, add required number of zeroes before the right part. If greater then pass the carry (left most excess digits of right part) to left part.

Ex.1: 94×96

Base = 100

94     -6

96     -4

———

90 ¦ 24

9024

Ex.1: 94×96; Base is 100 as both the given numbers (94 and 96) are closer to 100. Complement of 94 is -6 (∵94-100=-6) and Complement of 96 is -4 (∵96-100=-4). Left Part is 90 (94-4 or 96-6). Right Part is 24 (-6*-4). So final answer is: 9024.

Ex.2: 90×89

Base = 100

90     -10

89     -11

———-

79  ¦ 110

79+1 ¦ 10

80 ¦ 10

8010

Ex.2: 90×89. Complement of 90 is -10 (∵90-100=-10) and Complement of 89 is -11 (∵89-100=-11). Left Part is 79 (90-11 or 89-10). Right Part is 110 (-10*-11). Here base is having two zeroes, so right part should be of two digits. But, Right Part is of three digits. So 1 is passed as carry to Left Part. So Left Part becomes 79+1=80 and Right Part becomes 10. So final answer is: 8010.

Ex.3: 997×993

Base = 1000

997     -3

993     -7

———-

990 ¦ 021

990021

Ex.3: 997×993; Base is 1000. Complement of 997 is -3 (∵997-1000=-3) and Complement of 993 is -7 (∵993-1000=-7). Left Part is 990 (997-7 or 993-3). Right Part is 021 (-3*-7). So final answer is: 99021. Note: Result of product of complements is 21. But we need to add one ZERO before 21. Because base is 1000 & having THREE zeroes.

Case 2: When both numbers are greater than the working base:

Working Procedure:

• Write multiplicand and multiplier one below the other.
• Write Surplus of multiplicand and multiplier to its right side with signs.
• Answer consists of two parts. Left and Right.
• Left Part: Evaluating any of the cross values.
• Right Part: Product of both surpluses (right side values).
• Caution: Total number of digits in the Right Part should be equal to total number of zeroes in the base. If lesser, add required number of zeroes before the right part. If greater then pass the carry (left most excess digits of right part) to left part.

Ex.1: 109×111

Base = 100

109  +9

111  +11

———-

120 ¦ 99

12099

Ex. 1: 109×111. Base is 100 as both the given numbers (109 and 111) are closer to 100. Surplus of 109 is +9 (∵109-100=+9) and Surplus of 111 is +11 (∵111-100=+11). Left Part is 120 (109+11 or 111+9). Right Part is 99 (9×11). Here base is 100 (Two Zeroes). Right part is having two digits. So no any further calculations are required. The final answer is 12099.

Ex.2: 117×110

Base = 100

117  +17

110  +10

———-

127¦170

127+1¦70

128 ¦70

12870

Ex. 2: 117×110. Base is 100 as both the given numbers (117 and 110) are closer to 100. Surplus of 117 is +17 (∵117-100=+17) and Surplus of 110 is +10 (∵110-100=+10). Left Part is 127 (117+10 or 110+17). Right Part is 170 (17×10). Here base is 100 (Two Zeroes). But Right part is having three digits. Leftmost digit of right part (here it is 1) is taken to Left part as carry. So Left part becomes 128 (127+1) and Right part becomes 70. So final answer is 12870.

Ex. 3: 1020×1033

Base = 1000

1020  +20

1033  +33

———-

1053 ¦ 660

1053660

Ex. 3: 1020×1033. Base is 1000 as both the given numbers (1020 and 1033) are closer to 100. Surplus of 1020 is +20 (∵1020-1000=+20) and Surplus of 1033 is +33 (∵1033-1000=+33). Left Part is 1053 (1020+33 or 1033+20). Right Part is 660 (20×33). Here base is 1000 (Three Zeroes). Right part is having three digits. So no any further calculations are required. The final answer is 1053660.

4. Some Solved Examples (Multiplication using Criss Cross Method)

Formula Used: 3. Ūrdhva – tiryagbhyām (ऊर्ध्वतिर्यग्भ्याम्)

Meaning: Vertically & Crosswise

How to Remember? Here you no need to remember any formulas, just you need to understand pattern. Go through graphical representation of various cases and understand pattern. The first part will be product of respective first digits of both multiplier and multiplicand. Last Part will be product of respective last digits of both multiplier and multiplicand. Second Part will be applying criss cross on first two digits of both multiplier and multiplicand. Second last Part will be applying criss cross on second last digits of both multiplier and multiplicand AND SO ON…

Case 1: Two Digit Numbers (2D×2D and 2D×1D) {D: Digit}

 First Part a    b  c    d (a×c)

 Second Part a    b   c    d (a×d) + (b×c)

 Third Part a    b   c    d (b×d)

Ex.1: 42 × 57

(4×5) ¦ (4×7 + 2×5) ¦ (2×7)

20 ¦ 28+10 ¦ 14

20 ¦ 38 ¦ 14

20 ¦ 38+1 ¦ 4

20 ¦ 39 ¦ 4

20+3 ¦ 9 ¦ 4

23 ¦ 9 ¦ 4

2394

Ex.1: Put values as per formula. Evaluate all parts. All Parts except first should contain only one digit. Start observation from Right to Left. If you find more than one digit, then pass leftmost excess digits to its immediate left part.

Ex.2: 84 × 36

(8×3) ¦ (8×6 + 4×3) ¦ (4×6)

24 ¦ 48+12 ¦ 24

24 ¦ 60 ¦ 24

24 ¦ 60+2 ¦ 4

24 ¦ 62 ¦ 4

24+6 ¦ 2 ¦ 4

30 ¦ 2 ¦ 4

3024

Ex.3: 67 × 89

(6×8) ¦ (6×9 + 7×8) ¦ (7×9)

48 ¦ 54+56 ¦ 63

48 ¦ 110 ¦ 63

48 ¦ 110+6 ¦ 3

48 ¦ 116 ¦ 3

48+11 ¦ 6 ¦ 3

59 ¦ 6 ¦ 3

5963

Ex.4: 76 × 59

(7×5) ¦ (7×9 + 6×5) ¦ (6×9)

35 ¦ 63+30 ¦ 54

35 ¦ 93 ¦ 54

35 ¦ 93+5 ¦ 4

35 ¦ 98 ¦ 4

35+9 ¦ 8 ¦ 4

44 ¦ 8 ¦ 4

4484

5. Some Solved Examples (Squares)

• What is Square: a square is the result of multiplying a number by itself.
• For ex. Square of 3 is 9 (3×3); Square of -45 is 2025 (-45×-45)
• Square of 12 is 144 (12×12); Square of -12 is 144 (-12×-12)

Squares Using ‘One More than the Previous One’ Formula:

Formula used is: 1 Ekādhikena Pūrvena  (एकाधिकेन पूर्वेण)

Meaning: One More than the Previous One

Note: This formula is used to obtain square of number ending with digit 5 (Ex. 15, 125, 345, 4585, 6485, 9745 etc.).

Working Procedure:

• Answer consists  of two Parts (Left and Right Part)
• Right Part is 25 (Square of 5).
• Left Part is Product of number (obtained by discarding last digit 5 of given number) with its next number in the number line.

Ex.1: What is Square of 15?

1  ¦  5

1×2  ¦ 25

2 ¦ 25

225

Ex.2: What is Square of 25?

2  ¦  5

2×3  ¦ 25

6 ¦ 25

625

Ex.2: Right Part is 25 (Square of 5). Left Part is 6 {2×3; multiplication of number with its next number in the number line}.

Ex.3: What is Square of 75?

7  ¦  5

7×8  ¦ 25

56 ¦ 25

5625

Ex.3: Left part is 7 and right part is 5. Multiply 7 with its next number in the nuber line (8). It gives 56. Right part is 25 (52). After removing vertical line we get 5625, which is the square of 75.

Ex.4: What is Square of 95?

9   ¦  5

9×10 ¦ 25

90   ¦ 25

9025

Ex.5: What is Square of 115?

11  ¦  5

11×12 ¦ 25

132  ¦ 25

13225

Squares Using ‘Complements/Surpluses’:

(यावदूनं तावदूनीकृत्य वर्गं च योजयेत्)    Meaning: Lessen by the Deficiency and set up the square of that deficiency.

Note: This corollary is used to obtain square of numbers which are closer to bases.

Case 1: When Number is below the Working Base:

Working Procedure:

• Note given number, its Base and Complement.
• Answer consists of Two Parts (Left Part and Right Part)
• Right Part is square of Complement.
• Left Part = (Given Number + Complement).
• Note: Total number of digits in the Right Part should be same as total number of zeroes in the base. If lesser add required number of zeroes, if greater pass the carry (leftmost excess digits) to left part.

Ex.1: What is Square of 94?

Base: 100

Complement: -06

94-6 ¦ -62

88 ¦ 36

8836

Ex.2: What is Square of 97?

Base: 100

Complement: -03

97-3 ¦ -32

94 ¦ 09

9409

Ex.3: What is Square of 87?

B:100; C: -13

87-13 ¦ -132

74 ¦ 169

74+1 ¦ 69

75 ¦ 69

7569

Ex.4: What is Square of 893?

B:1000; C:-107

893-107 ¦ -1072

786 ¦ 11449

786+11 ¦ 449

797 ¦ 449

797449

Ex.5: What is Square of 9790?

B:10000; C:-210

9790-210 ¦ -2102

9580 ¦ 44100

9580+4 ¦ 4100

9584 ¦ 4100

95844100

Case 2: When Number is above the Working Base:

Working Procedure:

• Note given number, its Base and Surplus.
• Answer consists of Two Parts (Left Part and Right Part)
• Right Part is Square of Surplus.
• Left Part = (Given Number + Surplus).
• Note: Total number of digits in the Right Part should be same as total number of zeroes in the base. If lesser add required number of zeroes, if greater pass the carry (leftmost excess digits) to left part.

Ex.1: What is Square of 108?

Base: 100

Surplus:+08

108+8 ¦ 82

116  ¦ 64

11664

Ex.2: What is Square of 103?

Base: 100

Surplus: +03

103+3 ¦ 32

106  ¦ 09

10609

Ex.3: What is Square of 1104?

B:1000; S:+104

1104+104 ¦ 1042

1208  ¦ 10816

1208+10 ¦ 816

1218 ¦ 816

1218816

Ex.4: What is Square of 1250?

B:1000; S:+250

1250+250¦2502

1500 ¦ 62500

1500+62 ¦ 500

1562 ¦ 500

1562500

Squares using Criss Cross Method:

• Square is the result of multiplying a number by itself.
• Square of A is (A × A); Square of 98456 is (98456 × 98456); Square of 64578965 is (64578965 × 64578965)
• For finding Squares of any complex (bigger) numbers, apply Criss Cross Method. We already Studied Multiplication using Criss Cross Method and refer the same for finding squares of any number.

Ex.1:What is Square of 83?

8  3

8  3

—————-

64   24+24   9

64   48   9

64+4  8   9

68  8   9

6889

Ex.2:What is Square of 678?

6   7   8

6   7   8

——————–

36;  42+42; 48+49+48; 56+56; 64

36   84   145  112   64

45   99   156   118   64

459684

459684

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• Contact: Mr. Chaitanya Patil (Author); +91-97640-58-654 OR Direct WhatsApp OR [email protected]
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6. Some Solved Examples (Polynomials)

Multiplication using Criss Cross Method

Formula used is: 3)  Ūrdhva – tiryagbhyām; (ऊर्ध्वतिर्यग्भ्याम्)

Meaning: Vertically & Crosswise

Note: Read Multiplication using Criss Cross Method from Multiplication Unit.

Working Procedure:

• Write coefficients of both polynomials one below the other separated by spaces. Write coefficient as zero if any term is absent.
• Apply Criss Cross method on coefficients.
• No need to alter any parts (it means don’t transfer leftmost excess digits to its immediate left part).
• Last part is constant. Go on incrementing powers of variable by 1 from right side. Second last is x, then x2, x3, x4, x5 and so on.

CASE 1: (2×2; 2×1)

 First Part a    b  c    d (a×c)

 Second Part a    b c    d (a×d) + (b×c)

 Third Part a    b  c    d (b×d)

Ex.1: (x+3) (x+5)

1   3

1   5

—————–

(1×1) | (1×5 + 1×3) | (3×5)

1 | 8 | 15

x2    x1   x0

x2 + 8x + 15

Ex.2: (2x+7) (-3x-4)

2      7

-3    -4

——————

(2×-3) | (2×-4 + 7×-3) | (7×-4)

-6 | -29 | -28

x2    x1    x0

-6x2 – 29x – 28

Ex.3: (7x-6) (2x+8)

7     -6

2     8

—————-

(7×2) | (7×8 + -6×2) | (-6×8)

14 | 44 | -48

x2    x1   x0

14x2 + 44x – 48

Ex.4: (3x-7) (9x)

3     -7

9     0

(3×9) | (3×0 + -7×9) | (-7×0)

27 | -63 | 0

x2    x1   x0

27x2 – 63x + 0

27x2 – 63x

7. List of Formulas and their Meaning

Formula Number.  Formula Name in English   (Formula Name in Sanskrit)

Formula Meaning in English

1. Ekādhikena Pūrvena (एकाधिकेन पूर्वेण)

One more than the Previous one.

2. Nikhilam Navataścaramam Daśatah (निखिलं नवतश्चरमं दशतः)

All from 9 and the last from 10

3. Ūrdhva – tiryagbhyām (ऊर्ध्वतिर्यग्भ्याम्)

Vertically and Crosswise

4. Parāvartya Yojayet (परावर्त्य योजयेत्)

Transpose and Apply

5. Sūnyam Samyasamuccaye (शून्यं साम्यसमुच्चये)

If the Samuccay (समुच्चय or समूह or Set) is same, it is ZERO.

6. Ānurūpye Śūnyamanyat (आनुरूप्ये शून्यमन्यत्)

If one is in Ratio, the other is ZERO

7. Sankalana – vyavakalanābhyām (संकलनव्यवकलनाभ्याम्)

8. Puranāpuranābhyām (पूरणापूरणाभ्याम्)

By the Completion or Non Completion

9. Calanā kalanābhyām (चलनकलनाभ्याम्)

By Calculus

By the Deficiency

11. Vyastisamastih (व्यष्टिसमष्टिः)

Individuality & Totality (Part & Whole)

12. Śesānyankena Caramena (शेषाण्यङ्केन चरमेण)

The Remainders by the last Digit

The Ultimate and twice the Penultimate

14. Ekanyūnena Pūrvena (एकन्यूनेन पूर्वेण)

By one less than the previous one.

15. Gunitasamuccayah (गुणितसमुच्चयः)

The sum of the product is equal to the product of the sum.

16. Gunakasamuccayah (गुणकसमुच्चयः)

All the Multipliers.

8. List of Corollaries (Sub Formulas) and their Meaning

Corollary Number.  Corollary Name in English (Corollary Name in Sanskrit)

Corollary Meaning in English

1. Ānurūpyena (आनुरूप्येण)

Proportionately

2. Śisyate Śesamjnah (शिष्यते शेषसंज्ञः)

The Remainder Remains Constant

3. Ādyamādyenantyamantyena (आद्यं आद्येन् अन्त्यम् अन्त्येन)

The First by the First and the Last by the Last

4. Kevalaih Saptakam Gunỹat (केवलैः सप्तकं गुण्यात्)

For 7 the Multiplier is 143

5. Vestanam (वेष्टनम्)

By Osculation

Lessen by the Deficiency

Lessen by the Deficiency and set up the Square of that Deficiency

8. Antyayordasake’ pi (अन्त्ययोर्द्दशकेऽपि)

When final Digits Added up gives10

9. Antyayoreva (अन्त्ययोरेव)

Only the Last Terms

10. Samuccayagunitah (समुच्चयगुणितः)

The Sum of the Products

11. Lopanasthāpanabhyām (लोपनस्थापनाभ्यां)

By Alternative Elimination & Retention

12. Vilokanam (विलोकनं)

By Mere Observation

13. Gunitasamuccayah Samuccayagunitah (गुणितसमुच्चयः समुच्चयगुणितः)

The Sum of the Product is equal to the Product of the Sum.

14. Dhvajāṅka (ध्वजांक)

On the Flag

15. Śūddha (शुद्धः)

Purification.

16. Dvaṅdvayoga (द्वन्द्वयोग)

Duplex Combination Process

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