# FUNDAMENTAL OF VEDIC MATHEMATICS  Views:

Category: Entertainment

## Presentation Description

No description available.

## Presentation Transcript

### slide 2:

FUNDAMENTALS APPLICATIONS OF VEDIC MATHEMATICS 2014 State Council of Educational Research Training Varun Marg Defence Colony New Delhi-110024

### slide 3:

Chief Advisor Anita Satia Director SCERT Guidance Dr. Pratibha Sharma Joint Director SCERT Contributors Dr. Anil Kumar Teotia Sr. Lecturer DIET Dilshad Garden Neelam Kapoor Retired PGT Directorate of Education Chander Kanta Chabria PGT RPVV Tyagraj Nagar Lodhi Road Rekha Jolly TGT RPVV Vasant Kunj Dr. Satyavir Singh Principal SNI College Pilana Editor Dr. Anil Kumar Teotia Sr. Lecturer DIET Dilshad Garden Publication Officer Ms. Sapna Yadav Publication Team Navin Kumar Ms. Radha Jai Baghwan Published by : State Council of Educational Research Training New Delhi and printed at Educational Stores S-5 Bsr. Road Ind. Area Ghaziabad U.P.

### slide 4:

Vedic Mathematics introduces the wonderful applications to Arithmetical computations theory of numbers compound multiplications algebraic operations factorisations simple quadratic and higher order equations simultaneous quadratic equations partial fractions calculus squaring cubing square root cube root and coordinate geometry etc. Uses of Vedic Mathematics: It helps a person to solve mathematical problems 10-15 times faster It helps m Intelligent Guessing It reduces burden need to learn tables up to 9 only It is a magical tool to reduce scratch work and finger counting It increases concentration. It helps in reducing silly mistakes "Vedic Mathematics" is a system of reasoning and mathematical working based on ancient Indian teachings called Veda. It is fast efficient and easy to learn and use. Vedic mathematics which simplifies arithmetic and algebraic operations has increasingly found acceptance the world over. Experts suggest that it could be a handy tool for those who need to solve mathematical problems faster by the day. V edic Mathematics provides answer in one line where as conventional method requires several steps. It is an ancient technique which simplifies multiplication divisibility complex numbers squaring cubing square and cube roots. Even recurring decimals and auxiliary fractions can be handled by Vedic Mathematics. Vedic Mathematics forms part of Jyotish Shastra which is one of the six parts of Vedangas. The Jyotish Shastra or Astronomy is made up of three parts called Skandas. A Skanda means the big branch of a tree shooting out of the trunk. The basis of Vedic mathematics are the 16 sutras which attribute a set of qualities to a number or a group of numbers. The ancient Hindu scientists Rishis of Bharat in 16 Sutras Phrases and 120 words laid down simple steps for solving all mathematical problems in easy to follow 2 or 3 steps. V edic Mathematicsor one or two line methods can be used effectively for solving divisions reciprocals factorisation HCF squares and square roots cubes and cube roots algebraic equations multiple simultaneous equations quadratic equations cubic equations biquadratic equations higher degree equations differential calculus Partial fractions Integrations Pythogorus theoram Apollonius Theoram Analytical Conics and so on. Preface 3

### slide 5:

4 How fast your can solve a problem is very important. There is a race against time in all the competitions. Only those people having fast calculation ability will be able to win the race. Time saved can be used to solve more problems or used for difficult problems. This Manual is designed for Mathematics teachers of to understand Vedic System of Mathematics. The Chapters developed in this Manual will give teachers the depth of understanding of the Vedic methods for doing basic operations in Arithmetic and Algebra. Some important basic devices like Digit Sum the Vinculum are also explained along with independent Checking Methods. All the techniques are explained with examples. Also the relevant Sutras are indicated along with the problems. In Vedic System a manual approach is preferred. The simplicity of Vedic Mathematics encourages most calculations to be carried out without the use of paper and pen. The content developed in this manual will be applicable in the curriculum of VI-X classes. Methods like Shudh Method is applicable in statistics. This mental approach sharpens the mind improves memory and concentration and also encourages innovation. Since the Vedic Mathematics approach encourages flexibility the mathematics teachers encourage their students to device his/her own method and not remain limited to the same rigid approach which is boring as well as tedious. Once the mind of the student develops an understanding of system of mental mathematics it begins to work more closely with the numbers and become more creative. The students understand the numbers better. Vedic Mathematics is very flexible and creative and appeals to all group of people. It is very easy to understand and practice. I acknowledge a deep sense of gratitude to all the subject experts for their sincere efforts and expert advice in developing this manual which lead to qualitative and quantitative improvement in mathematics education and may this subject an interesting joyful and effective. Suggestions for further improvements are welcome so that in future this manual become more useful. —Anita Satia

### slide 6:

5 Contents Preface 03-04 Introduction 07-11 Chapter-1 Addition and Subtraction 12-24 1. Addition - Completing the whole 2. Addition from left to right 3. Addition of list of numbers - Shudh method 4. Subtraction - Base method 5. Subtraction - Completing the whole 6. Subtraction from left to right Chapter-2 Digit Sums Casting out 9s 9-Check Method 25-28 Chapter-3 11-Check method 29-31 Chapter-4 Special Multiplication methods 32-52 1. Base Method 2. Sub Base Method 3. Vinculum 4. Multiplication of complimentary numbers 5. Multiplication by numbers consisting of all 9s 6. Multiplication by 11 7. Multiplication by two-digit numbers from right to left 8. Multiplication by three and four-digit numbers from right to left. Chapter-5 Squaring and square Roots 53-57 Squaring 1. Squaring numbers ending in 5 2. Squaring Decimals and Fraction

### slide 7:

6 3. Squaring Numbers Near 50 4. Squaring numbers near a Base and Sub Base 5. General method of Squaring - from left to right 6. Number splitting to simplify Squaring Calculation 7. Algebraic Squaring Square Roots 1. Reverse squaring to find Square Root of Numbers ending in 25 2. Square root of perfect squares 3. General method of Square Roots Chapter-6 Division 58-64 1. Special methods of Division 2. Straight Division

### slide 8:

7 Introduction The “Vedic Mathematics” is called so because of its origin from Vedas. To be more specific it has originated from “Atharva V edas” the fourth V eda. “Atharva V eda” deals with the branches like Engineering Mathematics sculpture Medicine and all other sciences with which we are today aware of. The Sanskrit word V eda is derived from the root Vid meaning to know without limit. The word V eda covers all Veda-Sakhas known to humanity. The Veda is a repository of all knowledge fathomless ever revealing as it is delved deeper. Vedic mathematics which simplifies arithmetic and algebraic operations has increasingly found acceptance the world over. Experts suggest that it could be a handy tool for those who need to solve mathematical problems faster by the day. It is an ancient technique which simplifies multiplication divisibility complex numbers squaring cubing square roots and cube roots. Even recurring decimals and auxiliary fractions can be handled by Vedic mathematics. Vedic Mathematics forms part of Jyotish Shastra which is one of the six parts of Vedangas. The Jyotish Shastra or Astronomy is made up of three parts called Skandas. A Skanda means the big branch of a tree shooting out of the trunk. This subject was revived largely due to the efforts of Jagadguru Swami Bharathi Krishna Tirtha Ji of Govardhan Peeth Puri Jaganath 1884-1960. Having researched the subject for years even his efforts would have gone in vain but for the enterprise of some disciples who took down notes during his last days. The basis of Vedic mathematics are the 16 sutras which attribute a set of qualities to a number or a group of numbers. The ancient Hindu scientists Rishis of Bharat in 16 Sutras Phrases and 120 words laid down simple steps for solving all mathematical problems in easy to follow 2 or 3 steps. Vedic Mental or one or two line methods can be used effectively for solving divisions reciprocals factorisation HCF squares and square roots cubes and cube roots algebraic equations multiple simultaneous equations quadratic equations cubic equations biquadratic equations higher degree equations differential calculus Partial fractions Integrations Pythogorus Theoram Apollonius Theoram Analytical Conics and so on. V edic scholars did not use figures for big numbers in their numerical notation. Instead they preferred to use the Sanskrit alphabets with each alphabet constituting a number. Several mantras in fact denote numbers that includes the famed Gayatri Mantra which adds to 108 when decoded. How fast you can solve a problem is very important. There is a race against time in all the competitions. Only those people having fast calculation ability will be able to win the race. Time saved can be used to solve more problems or used for difficult problems. Given the initial training in modern maths in today’s schools students will be able to comprehend the logic of Vedic mathematics after they have reached the 8th standard. It will be of interest to everyone but more so to younger students keen to make their mark in competitive entrance exams. India’s past could well help them make it in today’s world. It is amazing how with the help of 16 Sutras and 13 sub-sutras the Vedic seers were able to mentally calculate complex mathematical problems.

### slide 9:

8 Sixteen Sutras S.N. Sutras Meaning 1. dkf/kdsu iwosZ.k One more than the previous one Ekadhikena Purvena also a corollary 2. fufkya uorpjea nkr All from 9 and last from 10 Nikhilam Navatascaramam Dasatah 3. Å/oZfrZXHke Criss-cross Vertically and cross-wise Urdhva-tiryagbhyam 4. ijkoRZ kstsr Transpose and adjust Transpose and apply Paravartya Yojayet 5. kwUa lkEleqPps When the samuchchaya is the same the samuch- Sunyam Samyasamuccaye chaya is zero i.e it should be equated to zero 6. ¼vkuq:Is½ kwUeUr If one is in ratio the other one is zero Anurupye Sunyamanyat 7. ladyuOodyukHke By addition and by subtraction Sankalana-vyavakalanabhyam also a corollary 8. iwj.kkiwj.kkHke By the completion or non-completion Puranapuranabhyam 9. pyudyukHke By Calculus Calana-Kalanabhyam 10- konwue By the deficiency Yavadunam 11. Of"Vlef"V Specific and General Use the average Vyastisamastih 12. ks"kk.dsu pjes.k The remainders by the last digit Sesanyankena Caramena 13. lksikUReURe The ultimate twice the penultimate Sopantyadvayamantyam 14. dUwusu iwosZ.k By one less than the previous one Ekanyunena Purvena 15. xqf.krleqPp The product of the sum of coefficients in the factors Gunitasamuccdyah The whole product 16. xq.kdleqPp Set of Multipliers Gunakasamuccayah

### slide 10:

9 Thirteen Sub-Sutras S.N. Sutras Meaning 1. vkuq:is.k Anurupyena 2. fk"rs ks"klaK Sityate Sesasanfitah 3. vk|ek|sukUReURsu Adyamadyenantyainantyena 4. dsoyS lIrda xq.kr Kevalalh Saptakan Gunyat 5. os"Vue Vestanam 6. konwra rkonwue Yavadunam Tavadunam 7. konwua rkonwuhd`RoxZ p kstsr Yavadunam Taradunikrtya Varganca Yojayet 8- vRksnZkds·fi Antyayordasakept 9- vURksjsn Antyayoteva 10- leqPpxqf.kr Samuccayaguaitah 11- yksiLFkkiukHke Lopanasthapandbhyam 12- foyksdue Vilokanam 13- xqf.krleqPp leqPpxqf.kr Gunitasamuccayah Samuccayagunitah Proportionately The remainder remains constant The first by the first and last by the last In case of 7 our multiplicand should be 143 By osculation Lessen by the Deficiency Whatever the extent of its deficiency lessen it still to that very extent and also set up the square of that deficiency. Whose last digits together total 10 and whose previous part is exactly the same Only the last terms The sum of the coefficients in the product By alternate elimination and retention By observation The product of sum of the coefficients in the factors is equal to the sum of the coefficients in the product.

### slide 11:

10 In the text the words Sutra aphorism formula is used synonymously. So are also the words Upa- sutra Sub-sutra Sub-formula corollary used. The Sutras apply to and cover almost every branch of Mathematics. They apply even to complex problems involving a large number of mathematical operations. Application of the Sutras saves a lot of time and effort in solving the problems compared to the formal methods presently in vogue. Though the solutions appear like magic the application of the Sutras is perfectly logical and rational. The computation made on the computers follows in a way the principles underlying the Sutras. The Sutras provide not only methods of calculation but also ways of thinking for their application. This course on Vedic Mathematics seeks to present an integrated approach to learning Mathematics with keenness of observation and inquisitiveness avoiding the monotony of accepting theories and working from them mechanically. The explanations offered make the processes clear to the learners. The logical proof of the Sutras is detailed which eliminates the misconception that the Sutras are a jugglery. Application of the Sutras improves the computational skills of the learners in a wide area of problems ensuring both speed and accuracy strictly based on rational and logical reasoning. The knowledge of such methods enables the teachers to be more resourceful to mould the students and improve their talent and creativity. Application of the Sutras to specific problems involves rational thinking which in the process helps improve intuition that is the bottom - line of the mastery of the mathematical geniuses of the past and the present such as Aryabhatta Bhaskaracharya Srinivasa Ramanujan etc. This course makes use of the Sutras and Sub-Sutras stated above for presentation of their application for learning Mathematics at the secondary school level in a way different from what is taught at present but strictly embodying the principles of algebra for empirical accuracy. The innovation in the presentation is the algebraic proof for every elucidation of the Sutra or the Sub-Sutra concerned. Terms and Operations a Ekadhika means ‘one more’ e.g: Ekadhika of 0 is 1 Ekadhika of 1 is 2 Ekadhika of 8 is 9 Ekadhika of 23 is 24 Ekadhika of 364 is 365 b Ekanyuna means ‘one less’ e.g: Ekanyuna of 1 2 3 ..... 8 ..... 14 .....69 ......is 0 1 2 ..... 7 ......13 .... 68 ...... c Purak means ‘complement’ e.g: Purak of 1 2 3 ..... 8 9 from 10 is 9 8 7..... 2 1 d Rekhank means ‘a digit with a bar on its top’. In other words it is a negative number. e.g: A bar on 7 is written as 7. It is called Rekhank 7 or bar 7. We treat Purak as a Rekhank. e.g: 7 is 3 and 3 is 7 At some instances we write negative numbers also with a bar on the top of the numbers as –4 can be shown as 4 . –21 can be shown as 21 .

### slide 12:

11 e 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. e.g: Beejank of 27 is 2 + 7 9. Beejank of 348 is 3 + 4 + 8 15 further 1 + 5 6. i.e. 6 is Beejank. Easy way of finding Beejank: Beejank is unaffected if 9 is added to or subtracted from the number. This nature of 9 helps in finding Beejank very quickly by cancelling 9 or the digits adding to 9 from the number. e.g. 1: 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. f Vinculum: The numbers which by presentation contain both positive and negative digits are called vinculum numbers. Conversion of general numbers into vinculum numbers We obtain them by converting the digits which are 5 and above 5 or less than 5 without changing the value of that number. Consider a number say 8. Note that it is greater than 5. Use it complement purak- rekhank from 10. It is 2 in this case and add 1 to the left i.e. tens place of 8. Thus 8 08 12 The number 1 contains both positive and negative digits. i.e.1 and 2. Here 2 is in unit place hence it is –2 and value of 1 at tens place is 10. Thus 12 10 – 2 8 Conveniently we can think and write in the following way General Number Conversion Vinculum number 6 10 – 4 14 97 100 – 3 103 289 300 – 11 311 etc.

### slide 13:

12 Chapter -1 Addition and subtraction Addition is the most basic operation and adding number 1 to the previous number generates all the numbers. The Sutra “By one more than the previous one describes the generation of numbers from unity. 0 + 1 1 1 + 1 2 2 + 1 3 3 + 1 4 4 + 1 5 5 + 1 6 6 + 1 7 7 + 1 8 8 + 1 9 9 + 1 10...... Completing the whole method The VEDIC Sutra ‘By the Deficiency’ relates our natural ability to see how much something differs from wholeness. 7 close to 10 The ten Point Circle 8 close to 10 9 close to 10 171819 are close to 20 27 28 29 are close to 30 37 38 39 are close to 40 47 48 49 are close to 50 57 58 59 are close to 60 67 68 69 are close to 70 77 78 79 are close to 80 87 88 89 are close to 90 97 98 99 are close to 100 ............... and so on We can easily say that 9 is close to 10 19 is close to 20 etc. We can use this closeness to find addition and subtraction. The ten Point Circle Rule : By completion non-completion Five number pairs 1+ 9 2 + 8 3 + 7 4 + 6 5 + 5 Use these number pairs to make groups of 10 when adding numbers. 10 1 2 3 4 5 6 7 9 8 19 18 17 16 15 14 13 12 11 20 10 1 2 3 4 5 6 7 9 8

### slide 14:

13 Example : 24 + 26 20 + 4 + 20 + 6 20 + 20 + 10 50 Below a multiple of ten Rule : By the deficiency 49 is close to 50 and is 1 short. 38 is close to 40 and is 2 short. Example : 59 + 4 59 + 1 + 3 60 + 3 63 59 is close to 60 and 1 short 50 59 + 4 is 60 Example : 38 + 24 38 + 2 + 22 40 + 22 62 or 38 + 24 40 + 24 – 2 64 – 2 62 38 is close and is 2 sheet so 38 + 24 is 2 short from 40 + 24 hence 38 + 24 40 + 24 – 2 64 – 2 62 Example Add 39 + 6 39 is close to 40 and is 1 less then it. So we take 1 from the 6 to make up 40 and then we have 5 more to add on which gives 45 Add 29 + 18 + 3 29 + 18 + 1 + 2 As 3 1 + 2 and 29 + 1 30 18 + 2 20 30 + 20 50 Note we break 3 into 1 + 2 because 29 need 1 to become 30 and 18 need 2 become 20 Add 39 + 8 + 1 + 4 39 + 8 + 1 + 2 + 2 40 + 10 + 2 52 Sum of Ten The ten point circle illustrates the pairs of numbers whose sum is 10. Remember : There are eight unique groups of three number that sum to 10 for example 1 + 2 + 7 10 1 + 2 + 7 10 Can you find the other seven groups of three number summing to 10 as one example given for you 2 + 3 + 5 10

### slide 15:

14 Adding a list of numbers Rule : By completion or non-completion Look for number pairs that make a multiple of 10 7 + 6 + 3 + 4 The list can be sequentially added as follows : 7 + 6 13 then 13 + 3 16 then 16 + 4 20 Or You could look for number pairs that make multiples of 10. 7 + 3 is 10 and 6 + 4 is 10 hence 10 + 10 is 20. Similarily : 48 + 16 + 61 + 32 48 + 32 + 16 + 1 + 60 80 + 77 157 10 10 10 or 7 + 8 + 9 + 2 + 3 + 5 + 3 + 1 + 2 + 3 + 7 + 9 10 10 10 + 10 + 10 + 10 + 10 + 9 59 PRACTICE PROBLEMS Add by using completing the whole method 1. 39 + 8 + 1 + 5 2. 18 + 3 +2 + 17 3. 9 + 41 + 11 +2 4. 47 + 7 + 33 23 5. 23 + 26 + 27 + 34 6. 22 + 36 + 44 + 18 7. 33 + 35 + 27 + 25 8. 18 + 13 + 14 + 23 9. 3 + 9 + 8 + 5 + 7 + 1+ 2 10. 37 + 25 + 33 11. 43 + 8 + 19 + 11 12. 42 + 15 + 8 +4 13. 24 + 7 + 8 + 6 +13 14. 16 +43 + 14 +7 15. 13 + 38 +27 ADDITION Completing the whole method class VI commutative associative property 1. 39 + 17 + 11 + 13 2. 16 + 23 + 24 + 7 3. 12 + 51 + 9 + 18 4. 35 + 12 +55

### slide 16:

15 5. 123 + 118 + 27 6. 35 + 15 + 16 + 25 7. 58 + 41 + 12 + 9 8. 223 + 112 + 27 9. 24 + 106 + 508 + 12 10. 506 + 222 + 278 Adding from left to right The conventional methods of mathematics teachers use to do calculation from right and working towards the left. In Vedic mathematics we can do addition from left to right which is more useful easier and sometimes quicker. Add from left to right 1. 23 2. 2 3 4 + 15 + 5 2 4 38 7 5 8 3. 1 5 4. 2 3 5 3 8 5 2 6 4 3 7 5 1 Add 1 Add 1 53 761 The method: This is easy enough to do mentally we add the first column and increase this by 1 if there is carry coming over from the second column. Then we tag the last figure of the second column onto this Mental math Add from left to right 1 6 6 2 5 4 6 3 5 3 4 4 1 4 5 7 + 5 5 + 6 7 1 + 7 1 7 + 2 8 5 7 5 4 5 6 3 1 2 4 6 5 7 7 4 5 8 1 4 3 2 + 7 6 + 7 6 1 2 4 6 + 2 7 + 8 6 6 8 9 8 5 10 5 3 7 11 4 5 6 12 2 6 4 8 + 2 3 + 7 1 8 + 1 2 7 + 8 3 6 5 13 1 3 4 5 14 5 4 6 15 7 8 8 5 16 3 7 8 + 5 8 3 6 + 4 5 6 1 + 1 5 4 3 + 4 8 17 3 5 6 7 1 18 2 4 6 8 + 1 2 3 4 5 + 1 2 3

### slide 17:

16 Shudh method for a list of number Shudh means pure. The pure numbers are the single digit numbers i.e. 0 1 2 3…9. In Shudh method of addition we drop the 1 at the tens place and carry only the single digit forward. Example: Find 2 + 7 + 8 + 9 + 6 + 4 2 • 7 • 8 9 • 6 4 36 We start adding from bottom to top because that is how our eyes naturally move but it is not necessary we can start from top to bottom. As soon as we come across a two-digit number we put a dot instead of one and carry only the single digit forward for further addition. We put down the single digit 6 in this case that we get in the end. For the first digit we add all the dots 3 in this case and write it. Adding two or three digit numbers list . 23.4 We start from the bottom of the right most columns and get a single digit 6 at the unit 6.5.8 place. There are two dots so we add two to the first number 4 of .81.8 the second column and proceed as before. The one dot of this 46 column is added to the next and in the end we just put 1 down 1756 for one dot as the first digit of the answer. Shudh method • 52 6 • 9 • 4•5 4 3 4 • 6 • 8 1 7 5 2 • 8 23 8 4 43 Add the following by Shudh method 1. 5 2. 37 3. 345 7 64 367 6 89 289 826 + 167 4 + 71 + 9

### slide 18:

17 4. 3126 5. 468 6. 235 1245 937 579 4682 386 864 + 5193 654 + 179 7. 59 8. 49 9. 98 63 63 83 75 78 78 82 85 62 + 91 + 97 + 44 10. 37 11. 2461 12. 9721 79 4685 2135 52 6203 5678 88 1234 207 + 91 + 5432 + 1237 Number Spliting Method Quick mental calculations can be performed more easily if the numbers are split into more manageable parts. For example : Split into two more manageable sums + 3642 36 42 Note : The split allows us to add 36 + 24 2439 + 24 39 and 42 + 39 both of which can be done 60 81 mentally Remember : Think about where to place the split line. Its often best to avoid number carries over the line. For example : 342 3 42 34 2 + 587 5 87 58 7 2 29 92 9 carry 1 No carry is required A carry of 1 over the line is required

### slide 19:

18 SUBTRACTION Sutra: All from 9 and the Last from 10 The Concept of Base Numbers made up of only 1’s and 0’s are known as a Base. Examples of a Base are 10 100 1000 1 .01….etc The base method is used for subtracting multiplying or dividing numbers. Like 98 898 78999 etc that are close to base. Applying the formula “All form 9 and Last form 10” to any number especially the big one’s reduces it to its smaller Counterpart that can be easily used for calculations involving the big digits like 7 8 and 9. Applying the formula “All from 9 and the last from 10” Example: Apply ‘All from 9 Last from 10’ to Subtract 789 from 1000 7 8 9 ↓ ↓ ↓ Here all from 9 last from 10 means subtract 78 8 from 9 and 9 from 10 so weget 211 2 1 1 We get 211 because we take 7 and 8 from 9 and 9 from 10. from 10000 from 100 from 100 from 100000 2772 54 97 10804 ↓↓↓↓ ↓↓ ↓↓ ↓↓↓↓↓ 7228 46 03 89196 If you look carefully at the pairs of numbers in the above numbers you may notice that in every case the total of two numbers is a base number 10 100 1000 etc. This gives us an easy way to subtract from base numbers like 10 100 1000……. Subtracting from a Base Example: - 1000 – 784 216 Just apply ‘All from 9 and the Last from 10’ to 784 difference of 7 from 9 is 2 8 from 9 is 1 4 from 10 is 6 so we get 216 after subtraction. When subtracting a number from a power of 10 subtract all digits from 9 and last from 10. 1000 from 9 from 10 – 276 724 276 ↓↓↓ 724

### slide 20:

19 Subtracting from a Multiple of a Base Sutra: ‘All from 9 and the last from 10’ and ‘One less than the one before’ Example: 600 – 87 We have 600 instead of 100. The 6 is reduced by one to 5 and the All from 9 and last from 10 is applied to 87 to give 13. Infact 87 will come from one of those six hundred so that 500 will be left. ∴ 600 – 87 513 Note : First subtract form 100 then add 500 as 500 + 13 513 Example: Find 5000 – 234 5 is reduced to one to get 4 and the formula converts 234 to 766 ∴ 5000-2344766 Example: 1000 – 408 592 Example:100 – 89 11 Example:1000 – 470 530 Remember apply the formula just to 47 here. If the number ends in zero use the last non-zero number non-zero number as the last number for example. 10000 from 9 from 10 – 4250 5750 4250 ↓↓↓ 5750 Hence 1000 – 4250 5750 Adding Zeroes In all the above sums you may have noticed that the number of zeros in the first number is the same as the numbers of digits in the number being subtracted. Example: 1000 – 53 here 1000 has 3 zeros and 53 has two digits. We can solve this by writing 1000 – 053 947 We put on the extra zero in front of 53 and then apply the formula to 053. Example: 10000 – 68 Here we need to add two zeros. 10000 – 0068 9932

### slide 21:

20 Practice Problems Subtract from left to right 1 86 – 27 2 71 – 34 3 93 – 36 4 55 – 37 5 874 – 567 6 804 – 438 7 793 – 627 8 5495 – 3887 9 9275 – 1627 10 874– 579 11 926 – 624 12 854– 57 13 8476 – 6278 14 9436 – 3438 Subtract the following mentally 1 55 – 29 2 82 – 558 3 1000 – 909 4 10000 – 9987 5 10000 – 72 6 50000 – 5445 7 70000 – 9023 8 30000 – 387 9 46678 – 22939 10 555 – 294 11 8118 – 1771 12 61016 – 27896 Example: Find 9000 – 5432 Sutra: ‘One more than the previous one’ and ‘all from 9 and the Last from the 10’ Considering the thousands 9 will be reduced by 6 one more than 5 because we are taking more than 5 thousand away ‘All from 9 and the last from 10’ is than applied to 432 to give 568 9000 – 5432 3568 Similary—7000 – 3884 3116 3 7 – 4 4 is one more than 3 and 116 4000 – 3884 by all from a and the last from 10 If the number is less digits then append zero the start : 1000 from 9 from 10 – 425 9575 042 5 ↓↓↓ ↓ 957 5

### slide 22:

21 When subtracting form a multiple ofa power of 10 just decrement the first digit by 1 then subtract remaining digits : 4000 from 9 from 10 – 257 3743 257 ↓↓↓ 4 – 1 → 3 753 Look at one more example : Money: A great application of "all from 9 and last from 10" is money. Change can be calculated by applying this sutra mentally for example : 10.00 from 9 from 10 – 4.25 5.75 4.25 5.75 This is helpful because most our rupee notes are multiple of 10s. PRACTICE PROBLEMS Subtract base method 1 1000 – 666 2 10000 – 3632 3 100 – 54 4 100000 – 16134 5 1000000 – 123456 6 1000 – 840 7 1000 – 88 8 10000 – 568 9 1000 – 61 10 100000 – 5542 11 10000 – 561 12 10000 – 670 Subtract multiple of base 1 600 – 72 2 90000 – 8479 3 9000 – 758 4 4000 – 2543 5 7000 – 89 6 300000 – 239 7 1 – 0.6081 8 5 – 0.99 Subtracting Near a base Rule : By completion or non completion. when subtracting a number close to a multiple of 10. Just subtract from the multiple of 10 and correct the answer accordingly.

### slide 23:

22 Example : 53 – 29 29 is just close to 30 just 1 short so subtaract 30 from 53 making 23 then add 1 to make 24. 53 – 29 53 – 30 + 1 23 + 1 24 Similarily 45 – 18 45 – 20 + 2 25 + 2 27 18 is near to 20 just 2 short Use the base method of calculating To find balance Q. Suppose you buy a vegetable for Rs. 8.53 and you buy with a Rs. 10 note. How much change would you expect to get Ans. You just apply “All from 9 and the last from 10” to 853 to get 1.47. Q. What change would expect from Rs. 20 when paying Rs. 2.56 Ans.The change you expect to get is Rs. 17.44 because Rs. 2.56 from Rs.10 is Rs. 7.44 and there is Rs. 10 to add to this. Practice Problem Q1. Rs. 10 – Rs. 3.45 Q2. Rs. 10 – Rs. 7.61 Q3. Rs. 1000 – Rs. 436.82 Q4. Rs. 100 – Rs. 39.08 Subtracting number just below the base Example: find 55 – 29 Subtraction of numbers using "complete the whole" Step 1: 20 is the sub base close to 19 19 is 1 below 20 Step 2: take 20 from 55 to get 35 Step 3: Add 1 back on 55 – 19 36 Example 61 – 38 38 is near to 40 40 – 38 2 61 – 40 21 61 – 38 21 + 2 23

### slide 24:

23 Example 44 – 19 19 + 1 20 44 – 20 24 44 – 19 24 – 1 23 Example 88 – 49 49+150 88 – 50 38 88 – 49 38 + 1 39 Example 55 – 17 17 + 3 20 55 – 20 35 55 – 17 35 + 3 38 Number spliting Method As you have use this method in addtion the same can be done for subtraction also : + 3642 36 42 Note : The split allows on to add 36 – 24 2439 + 24 39 and 42 – 39 both of which can be done 12 03 mentally General Method of subtraction Subtraction from left to right In this section we show a very easy method of subtracting numbers from left to right that we have probably not seen before. We start from the left subtract and write it down if the subtraction in the next column can be done. If it cannot be done you put down one less and carry 1 and then subtract in the second column. Subtraction from left to right. Example: Find Find 83 – 37 78 – 56 8 3 7 8 – 3 7 – 5 6 4 6 2 2

### slide 25:

24 Left to right 3 4 5 5 1 13 1 2 1 1 3 0 1 1 – 4 9 – 2 8 9 –2 0 4 0 2 0 3 2 1 9 7 6 3 0 1 7 3 5 1 5 6 1 7 –2 0 1 –1 1 8 2 8 1 0 0 2 3 7 3 9 Starting from the left we subtract in each column 3-12 but before we put 2 down we check that in next column the top number is larger. In this case 5 is larger than 1 so we put 2 down In the next column we have 5-14 but looking in the third column we see the top number is not larger than the bottom 5 is less than 8 so instead putting 4 down we put 3 and the other 1 is placed as the flag as shown so that 5 becomes 15 so now we have 15-87. Checking in the next column we can put this down because 6 is greater than 2. In the fourth column we have 6-24 but looking at the next column 7 is smaller than 8 we put down only 3 and put the other flag with 7 as shown finally in the last column 17-89.

### slide 26:

25 Chapter 2 Digit sums casting out 9’s and 9’ check method The word digit means a single figure number: The numbers 1 2 3 4 5 6 7 8 9 0 are all digits. Big numbers can be reduced to single digit by adding the constituents. Digit Sums A digit sum is the sum of all the digits of a number and is found by adding all of the digits of a number The digit sum of 35 is 3 + 5 8 The digit sum of 142 is 1 + 4 + 2 7 Note : If the sum of the digits is greater than 9 then sum the digits of the result again until the result is less than 10. The digit of 57 is 5 + 7 12 → 1 + 2 3 greater than 9 so need to add again Hence the digit sum of 57 is 3. The digit sum of 687 is 6 + 8 + 7 21 → 2 + 4 3 Hence the digit sum of 687 is 3. ● Keep findig the digit sum of the result + unitl its less then 10 ● 0 and 9 are requivalent Look and undevstand some more example : To find the digit sum of 18 for the example we just add 1 and 8 i.e.1 + 8 9 so the digit sum of 18 is 9. And the digit sum of 234 is 9 because 2 + 3+ 4 9 Following table shows how to get the digit sum of the following members 15 6 12 3 42 6 17 8 21 3 45 9 300 3 1412 8 23 5 22 4 Sometimes two steps are needed to find a digit sum. So for the digit sum of 29 we add 2 + 9 11 but since 11 is a 2-digit number we add again 1+11 So for the digit sum of 29 we can write 29 2 + 9 11 1 + 1 1

### slide 27:

26 Similarity for 49 4 + 9 13 1 + 3 4 So the digit sum of 49 is 4. Number 14 Digit sum 1 + 4 5 Single digit 5 19 1 + 9 10 1 39 3+ 9 12 3 58 5 + 8 13 4 407 4 + 0 + 7 11 2 CASTING OUT NINE Adding 9 to a number does not affect its digit sum So 5 59 95 959 all have digit sum of 5. For example to find out the digit sum of 4939 we can cast out nines and just add up the 3 and 4 so digit sum is 7 or using the longer method we add all digit 4 + 9 + 3+ 9 25 2 + 57 There is another way of casting out the nines from number when you are finding its digit sum. Casting out of 9’s and digit totalling 9 comes under the Sutra when the samuccaya is the same it is zero. So in 465 as 4 and 5 total nine they are cast out and the digit sum is 6: when the total is the same as 9 it is zero can be cast out cancelling a common factor in a fraction is another example. Number Digit sum Nine Point Circle 1326 3 25271 8 9643 4 23674 4 128541 3 1275 6 6317892 9 or 0 Number at each point on the circle have the same digit sum. By casting out 9s finding a digit sum can be done more quickly and mentally. 1 2 3 4 5 6 7 9 8 17 26 16 25 15 24 14 23 13 22 12 11 10 18 27 19 20 21

### slide 29:

28 Practice Problems Digit sum Puzzles 1. The digit sums of a two digit number is 8 and figures are the same what is the number 2. The digit sum of a two digit number is 9 and the first figure is twice the second what is it 3. Give three two digit numbers that have a digit sum of 3. 4. A two digit number has a digit sum of 5 and the figures are the same. What is the number 5. Use casting out 9’s to find the digit sums of the numbers below. Number 465 274 3456 7819 86753 4017 59 6. Add the following and check your answer using digit sum check 1 66 + 77 2 57 + 34 3 94 + 89 4 304 + 233 5 787 + 132 6 389 + 414 7 5131 + 5432 8 456 + 654

### slide 30:

29 Chapter 3 Eleven Check Method We have already used the digit sum check that helps to show if a calculation is correct. This method works because adding the digit in a number gives the remainder of the number after division by 9. A similar method works by using remainders of numbers after division by 11 rather than 9 Alternate digit sum or Eleven-check Method Suppose we want another check for 2434 × 23 55982 it can be done in the following steps Step1: Alternately add and subtract starting from right moving towards left the digits of each numbers as described below Number Alternating signs Digit sum 2434 –2 + 4 – 3 + 4 3 23 –2 + 3 1 55982 +5 – 5 + 9 – 8 + 2 3 Step 2: Now multiply the Digit Sum to get the product 3 × 1 3 Since the Digit Sum of the product and the two numbers is the same the answer is correct as per 11 check method. Two digit and Negative number in the digit sum checking the sum of addition 4364 + 1616 Left to right 4364 1916 6280 Number Alternating signs Digit sum Single digit 4364 –4 + 3 – 6 + 4 –3 11-3 8 1916 –1 + 9 – 1 +6 1311+2 2 6280 –6 + 2 – 8 + 0 – 12 10 11 –12 –1 11 – 1 10 Step2: Apply the following rules to get a single positive digit for the number • Subtract the negative numbers below 11 from 11 to get its positive counterpart so – 3 11 –3 8 And –12 –12 +11 –1 11 – 1 10 • For the two digit number above 11 divide the number by 11 and get the remainder as the positive digit sum so 13 ÷ 13 gives remainder 2. Alternately adding and subtracting digit of 13 starting from right can obtain this same result.

### slide 31:

30 Step 3 : now add the Digit sums to get the sum 8 + 2 10 the answer is correct as per 11 check method. Two digits in the digit sum Check subtraction problem 2819174 – 839472 2819174 839472 1979702 Step 1: Alternatively add and subtract staring from right moving towards left the digit of each numbers as described below Number Alternating signs Digit sum Single digit 2819174 +2–8+1–9+1–7+4 –16–16+11 –5 11–56 839472 –8+3–9+4–7+2 –15–15+11 –4 11–47 1979702 +1–9+7–9+7–0+2 –1 11–110 Step 2: Apply the following rules to get a single positive digit for the number • The negative numbers below –11 are to be first divided by 11 to get the remainder. Than subtract the remainder from 11 to get its positive counterpart. So –16/11 Remainder is –5 and –5 11 – 5 6 similarly –15/11 Remainder –4 11 –4 7. • The negative number –1 11 – 1 10 Step3: Now subtract the Digit sums to get the answer 6 – 7 –1 10 the answer is correct as per 11- checked method. Practice Problems Get the digit sum and single digit for the following numbers. Numbers Alternative signs Digit sums Single digit 567 1536 93823 1978712 849391 82918 5949393 176780

### slide 32:

31 Using 11 check method check the following Addition problems: 1 37 + 47 84 2 55 + 28 83 3 47 + 25 72 4 29 + 36 65 5 526 + 125 651 6 1328 + 2326 3654 7 129 + 35644 35773 8 3425 + 7491 + 8834 19750 9 1423178 + 5467 + 123 + 34 1428802 10 1314 + 5345 + 65 +781 7505 Check the following subtraction problems: 1 63 – 28 35 2 813 – 345 468 3 695 – 368 372 4 3456 – 281 3175 5 7117 – 1771 5346 6 8008 – 3839 4165 7 6363 – 3388 2795 8 51015 – 27986 23029 9 14285 – 7148 7137 10 9630369 – 3690963 5939406

### slide 33:

32 Chapter- 4 Special Multiplication Methods Multiplication in considered as one of the most difficult of the four mathematical operations. Students are scared of multiplication as well as tables. Just by knowing tables up to 5 students can multiply bigger numbers easily by some special multiplication methods of Vedic Mathematics. We should learn and encourage children to look at the special properties of each problem in order to understand it and decide the best way to solve the problem. In this way we also enhance the analytical ability of a child. Various methods of solving the questions /problems keep away the monotonous and charge up student’s mind to try new ways and in turn sharpen their brains. Easy way for multiplication Sutra:Vertically and Cross wise : For speed and accuracy tables are considered to be very important. Also students think why to do lengthy calculations manually when we can do them faster by calculators. So friends/ teachers we have to take up this challenge and give our students something which is more interesting and also faster than a calculator. Of course it’s us the teachers/parents who do understand that more we use our brain more alert and active we will be for that is the only exercise we have for our brain. Example 1: 7 x8 Step 1: Here base is 10 7 – 3 7 is 3 below 10 also called deficiencies × 8 – 2 8 is 2 below 10 also called deficiencies Step 2: Cross subtract to get first figure or digit of the answer: 7 – 2 5 or 8 – 3 5 the two difference are always same. Step 3 : Multiply vertically i.e. –3 × –2 6 which is second part of the answer. So 7 – 3 8 – 2 i.e. 7 × 8 56 5 / 6 Example 2: To find 6 × 7 Step 1 : Here base is 10 6 – 4 6 is 4 less than 10 i.e. deficiencies 7 – 3 7 is 3 less than 10 i.e. deficiencies Step 2: Cross subtraction : 6 – 3 3 or 7 – 4 3 both same Step 3: – 3 × – 4 + 12 but 12 is 2 digit number so we carry this 1 over to 3 obtained in 2 step 6– 4 7 – 3 3 / 1 2 i.e. 6 × 7 42 Try these : 1 9 × 7 ii 8 × 9 iii 6 × 9 iv 8 × 6 v 7 × 7

### slide 34:

33 Second Method: Same Base Method : When both the numbers are more than the same base. This method is extension of the above method i.e. we are going to use same sutra here and applying it to larger numbers. Example 1: 12 × 14 Step 1: Here base is 10 12 + 2 12 is 2 more than 10 also called surplus 14 + 4 14 is 4 more than 10also called surplus Step 2: Cross add: 12 + 4 16 or 14 + 2 16both same which gives first part of answer 16 Step 3: Vertical multiplication: 2 × 4 8 So 12 + 2 14 +4 16 / 8So 12 × 14 168 14 + 2 12 + 4 Example 2:105x 107 Step1: Here base is 100 105 + 05 105 is 5 more than 100 or 5 is surplus 107 + 07 107 is 7 more than 100 or 7 is surplus Base here is 100 so we will write 05 in place of 5and 07 in place of 7 Step 2: Cross add: 105 + 7 112 or 107 + 5 112 which gives first part of the answer 112 Step 3: Vertical multiplication: 05 × 07 35 two digits are allowed As the base in this problem is 100 so two digits are allowed in the second part. So 105 × 107 11235 Example 3: 112 x 115 Step 1: Here base is 100 112 + 12 2 more than 100 i.e. 12 is surplus 115 + 15 15 more than 100 i.e. 15 is surplus Step 2: Cross add: 112 + 15 127 115 + 12 to get first part of answer i.e.127 Step 3: Vertical multiplication 12 × 15 Oh my godIt’s such a big number. How to get product of this Again use the same method to get the product. 12 + 2 15 + 5 12 + 5 15 + 2 17/ 1 0 17 + 1 / 0 180 i.e. 12 × 15 180 But only two digits are allowed here so 1 is added to 127 and we get 127 + 1 128 So 112 × 115 128 80

### slide 35:

34 Try these: i12 × 14 ii 14 × 17 iii 17 × 19 iv 19 × 11 v 11 × 16 vi 112 × 113 vii 113 × 117 viii 117 × 111 ix 105 × 109 x 109 × 102 xi 105 × 108 xii 108 × 102 xiii 102 × 112 xiv 112 × 119 xv 102 × 115 Both numbers less than the same base: Same sutra applied to bigger numbers which are less than the same base. Example1: 99 × 98 Step 1: Check the base: Here base is 100 so we are allowed to have two digits on the right hand side. ∴ 99 – 01 1 less than 100 i.e. 01 deficiency 98 – 02 2 less than 100 i.e. 0 2 deficiency Step 2: Cross – subtract: 99 – 02 97 98 – 01 both same so first part of answer is 97 Step3: Multiply vertically – 01 × – 02 02 As base is 100 so two digits are allowed in second part So 99 × 98 9702 Example 2 : 89 × 88 Step1: Here base is 100 So 89 – 11 i.e. deficiency 11 88 – 12 i.e. deficiency 12 Step2: Cross subtract: 89 – 12 77 88 – 11both same So first part of answer can be 77 Step 3:Multiply vertically – 11 × – 12 Again to multiply 11 × 12 apply same rule 11 + 1 10 + 1 12 + 2 10 + 2 11 + 2 13 12 + 1 / 1 × 2 12 so 11 × 12 1 32 as only two digits are allowed on right hand side so add 1to L.H.S. So L.H.S. 77 + 1 78 Hence 89 × 88 7832 Example 3: 988 × 999 Step 1: As the numbers are near 1000 so the base here is 1000 and hence three digits allowed on the right hand side 988 – 012 012 less than 1000 i.e. deficiency 0 12 999 – 001 001 less than 1000 i.e. deficiency 00 1 Step 2: Cross – subtraction: 988 – 001 987 999 – 012 987 So first part of answer can be 987 Step 3: Multiply vertically: –012 xs – 001 012 three digits allowed ∴ 988 × 999 987012 How to check whether the solution is correct or not by 9 – check method.

### slide 36:

35 Example 1: 99 × 98 9702 Using 9 – check method. As 99 0 Product L.H.S. 0 × 8 0 taking 9 0 98 8 R.H.S. 9702 7 2 9 0 + 9702 9 both are same As both the sides are equal answer may be correct. Example 2: 89 × 88 7832 89 8 88 8 + 8 16 1 + 6 7 add the digits L.H.S. 8 × 7 56 5 + 6 11 2 1 + 1 R.H.S. 7832 8 + 3 11 1 + 1 2 As both the sides are equal so answer is correct Example 3: 988 × 999 987012 988 8 + 8 16 1 + 6 7 999 0 As 0 × 7 0 LHS 987012 0 As 7 + 2 9 0 8 + 1 9 0 also 9 0 ∴ RHS 0 As LHS RHS So answer is correct. Try These: i 97 × 99 ii 89 × 89 iii 94 × 97 iv 89 × 92 v 93 × 95 vi 987 × 998 vii 997 × 988 viii 988 × 996 ix 983 × 998 x 877 × 996 xi 993 × 994 xii 789 × 993 xiii 9999 × 998 xiv 7897 × 9997 xv 8987 × 9996. Multiplying bigger numbers close to a base: number less than base Example 1: 87798 x 99995 Step1: Base here is 100000 so five digits are allowed in R.H.S. 87798 – 12202 12202 less than 100000 deficiency is 12202 99995 – 00005 00005 less than100000 deficiency is 5 Step 2: Cross – subtraction: 87798 -00005 87793 Also 99995 – 12202 87793 both same So first part of answer can be 87793 Step 2 : Multiply vertically: –12202 × – 00005 + 61010 ∴ 87798 × 99995 8779361010

### slide 37:

36 Checking: 87798 total 8 + 7 + 7 + 8 30 3 single digit 99995 total 5 LHS 3 x 5 15 total 1 + 5 6 RHS product 8779361010 total 15 1 + 5 6 L.H.S R.H.S. So correct answer Example 2 : 88777 × 99997 Step 1: Base have is 100000 so five digits are allowed in R.H.S. 88777 – 11223 i.e. deficiency is 11223 99997 – 00003 i.e. deficiency is 3 Step 2: Cross subtraction: 88777 – 00003 88774 99997 – 11223 So first part of answer is 88774 Step 3: Multiply vertically: – 11223 × – 00003 + 33669 ∴ 88777 × 99997 8877433669 Checking: 88777 total 8 + 8 + 7 + 7 + 7 37 + 10 1 99997 total 7 ∴ LHS 1 × 7 7 RHS 8877433669 8 + 8 + 7 + 7 + 4 34 3 + 4 7 i.e. LHS RHS So correct answer Try These: i 999995 × 739984 ii 99837 × 99995 iii 99998 × 77338 iv 98456 × 99993 v 99994 × 84321 Multiply bigger number close to base numbers more than base Example 1: 10021 × 10003 Step 1: Here base is 10000 so four digits are allowed 10021 + 0021 Surplus 10003 + 0003 Surplus Step 2: Cross – addition 10021 + 0003 10024 10003 + 0021 both same ∴ First part of the answer may be 10024 Step 3: Multiply vertically: 10021 × 0003 0063 which form second part of the answer ∴ 10021 × 10002 100240063

### slide 38:

37 Checking: 10021 1+ 2 + 1 + 1 4 10003 1 + 3 4 ∴ LHS 4 × 4 16 1 + 6 7 RHS 100240063 1 + 2 + 4 7 As LHS RHS So answer is correct Example 2: 11123 × 10003 Step 1: Here base is 10000 so four digits are allowed in RHS 11123 + 1123 surplus 10003 + 0003 surplus Step 2: Cross – addition: 11123 + 0003 11126 10003 + 1123 both equal ∴ First part of answer is 11126 Step 3: Multiply vertically: 1123 × 0003 3369 which form second part of answer ∴ 11123 × 10003 111263369 Checking: 11123 1 + 1 + 1 + 2 + 3 8 10003 1 + 3 4 and 4 × 8 32 3 + 2 5 ∴ LHS 5 R.H.S 111263369 1 + 1 + 1 + 2 5 As L.H.S R.H.S So answer is correct Try These: i 10004 × 11113 ii 12345 × 111523 iii 11237 × 10002 iv 100002 × 111523 v 10233 × 10005 Numbers near different base: Both numbers below base Example 1: 98 × 9 Step 1: 98 Here base is 100 deficiency 02 9 Base is 10 deficiency 1 ∴ 98 – 02 Numbers of digits permitted on R.H.S is 1 digits in lower base Step 2: Cross subtraction: 98 -1 88 It is important to line the numbers as shown because 1 is not subtracted from 8 as usual but from 9 so as to get 88 as first part of answer. Step 3: Vertical multiplication: -02 x -1 2 one digits allowed ∴ Second part 2 ∴ 98 × 9 882

### slide 39:

38 Checking: Through 9 – check method 98 8 9 0 LHS 98 × 9 8 × 0 0 RHS 882 8 + 8 + 2 18 1 + 8 9 0 As LHS RHS So correct answer Example 2: 993 × 97 Step 1: 993 base is 1000 and deficiency is 007 97 base is 100 and deficiency is 03 ∴ 993 – 007 digits in lower base 2 So 2 digits are permitted on × 97 – 03 RHS or second part of answer Step 2: Cross subtraction: 993 – 03 963 Again line the number as shown because 03 is subtracted from 99 and not from 93 so as to get 963 which from first part of the answer. Step 3: Vertical multiplication: –007 – –03 21 only two digits are allowed in the second part of answer So second part 21 ∴ 993 × 97 96321 Checking: through 9 – check method 993 3 97 7 ∴ L.H.S. 3 × 7 21 2 + 1 3 R.H.S. 96321 2 + 1 3 As LHS RHS so answer is correct Example 3 : 9996 base is 10000 and deficiency is 0004 988 base is 1000 and deficiency is 012 ∴ 9996 – 0004 digits in the lower base are 3 so3digits × 988 – 012 permitted on RHS or second part of answer Step 2 : Cross – subtraction: 9996 – 012 9876 Well again take care to line the numbers while subtraction so as to get 9876 as the first part of the answer. Step3 : Vertical multiplication: –0004 × –012 048

### slide 40:

39 Remember three digits are permitted in the second part i.e. second part of answer 048 ∴ 9996 × 988 9876048 Checking:9 – check method 9996 6 988 8 + 8 + 16 1 + 6 7 ∴ LHS 6 × 7 42 4 + 2 6 RHS 9876045 8 + 7 15 1 + 5 6 As LHS RHS so answer is correct When both the numbers are above base Example 1: 105 × 12 Step 1: 105 base is 100 and surplus is 5 12 base is 10 and surplus is 2 ∴ 105 + 05 digits in the lower base is 1 so 1 digit is permitted in the second part of answer 12 + 2 Step 2: Cross – addition: 105 + 2 125 again take care to line the numbers properly so as to get 125 ∴ First part of answer may be 125 Step 3: Vertical multiplication : 05 × 2 10 but only 1 digit is permitted in the second part so 1 is shifted to first part and added to 125 so as to get 126 ∴ 105 × 12 1260 Checking: 105 1 + 5 6 12 1 + 2 3 ∴ LHS 6 × 3 18 1 + 8 9 0 ∴ RHS 1260 1 + 2 + 6 90 Example 2: 1122 × 104 Step1: 1122 – base is 1000 and surplus is 122 104 – base is 100 and surplus is 4 ∴ 1122 + 122 104 + 04 digits in lower base are 2 so 2-digits are permitted in the second part of answer Step 2: Cross – addition 1122 + 04 again take care to line the nos. properly so as to get 1162 1162

### slide 41:

40 ∴ First part of answer may be 1162 Step 3: Vertical multiplication: 122 × 04 4 88 But only 2 – digits are permitted in the second part so 4 is shifted to first part and added to 1162 to get 1166 1162 + 4 1166 ∴ 1122 × 104 116688 Can be visualised as: 1122 + 122 104 + 04 1162 / ← 4 88 116688 + 4 / Checking: 1122 1 + 1 + 2 + 2 + 6 104 1 + 4 5 ∴ LHS 6 × 5 30 3 RHS 116688 6 + 6 12 1 + 2 3 As LHS RHS So answer is correct Example 3: 10007 × 1003 Now doing the question directly 10007 + 0007 base 10000 × 1003 + 003 base 1000 10037 / 021 three digits per method in this part ∴ 10007 × 10003 10037021 Checking : 10007 1 + 7 8 1003 1 + 3 4 ∴ LHS 8 × 4 32 3 + 2 5 RHS 10037 021 1 + 3 + 1 5 As LHS RHS so answer is correct Try These: i 1015 × 103 ii 99888 × 91 iii 100034 × 102 iv 993 × 97 v 9988 × 98 vi 9995 × 96 vii 1005 × 103 viii 10025 × 1004 ix 102 × 10013 x 99994 × 95 VINCULUM: “Vinculum” is the minus sign put on top of a number e.g. 5 41 63 etc. which means –5 40 – 1 60 – 3 respectively Advantages of using vinculum: 1 It gives us flexibility we use the vinculum when it suits us . 2 Large numbers like 6 7 8 9 can be avoided. 3 Figures tend to cancel each other or can be made to cancel. 4 0 and 1 occur twice as frequently as they otherwise would.

### slide 42:

41 Converting from positive to negative form or from normal to vinculum form: Sutras: All from 9 the last from 10 and one more than the previous one 9 1 1 i.e. 10 – 1 8 12 7 13 6 14 19 21 29 3 1 28 32 36 44 40 – 4 38 42 Steps to convert from positive to vinculum form: 1 Find out the digits that are to be converted i.e. 5 and above. 2 Apply “all from 9 and last from 10” on those digits. 3 To end the conversions “add one to the previous digit”. 4 Repeat this as many times in the same number as necessary. Numbers with several conversions: 159 241 i.e. 200 – 41 168 232 i.e. 200 – 32 237 243 i.e. 240 – 7 1286 1314 i.e. 1300 – 14 2387129 24 1313 1 here only the large digits are be changed From vinculum back to normal form: Sutras: “All from 9 and last from ten” and “one less than then one before”. 1 1 09 10 – 1 13 07 10 – 3 24 16 20 – 4 241 200 – 41 159 162 160 – 2 158 222 200 – 22 178 1314 1300 – 14 1286 2413131 2387129 can be done in part as 13 1 130 – 1 129 and 2413 2400 – 13 2387 ∴ 2413131 2387129. Steps to convert from vinculum to positive form: 1 Find out the digits that are to be converted i.e. digits with a bar on top. 2 Apply “all from 9 and the last from 10” on those digits 3 To end the conversion apply “one less than the previous digit” 4 Repeat this as many times in the same number as necessary Try These: Convert the following to their vinculum form: i 91 ii 4427 iii 183 iv 19326 v 2745 vi 7648 vii 81513 viii 763468 ix 73655167 x 83252327 Try These: From vinculum back to normal form. i 14 i 21 iii 23 iv 231 v 172 vi 1413 vii 2312132 viii 241231 ix 632233 1 x 14142323

### slide 43:

42 When one number is above and the other below the base Example1: 102 × 97 Step 1: Here base is 100 102 + 02 02 above base i.e. 2 surplus 97 – 03 03 below base i.e. 3 deficiency Step 2: Divide the answer in two parts as 102 / + 02 97 / – 03 Step 3: Right hand side of the answer is + 02 × – 03 – 06 06 Step 4: Left hand side of the answer is 102 – 3 99 97 + 02 same both ways ∴ 102 × 97 9906 9894 i.e. 9900 – 6 9894 Checking: 102 1 + 2 3 9 7 7 ∴ L.H.S. 3 × 7 21 1 + 2 3 ∴ R.H.S 9894 8 + 4 12 1 + 2 3 As L.H.S. R.H.S. So answer is correct Example 2 : 1002 × 997 1002 + 002 006 1000 – 6 994 and 1 carried from 999 to 999 reduces to 998 997 – 003 999 006 ∴ 1002 × 997 998 994 When base is not same: Example1: 988 × 12 988 – 012 base is 1000 deficiency 12 12 + 2 base is 10 surplus is 2 1 digit allowed in R.H.S. 1188 – 2 024 1186 24 ∴ 988 × 12 1186 4 11856 because 4 10 – 4 6 Checking: 9 88 8 + 8 16 1 + 6 7 12 1 + 2 3 ∴ LHS 7 × 3 21 2 + 1 3 R.H.S 11856 1 + 5 + 6 12 1 + 2 3 As LHS RHS So answer is correct Example 2: 1012 × 98 1012 1012 + 012 base is 1000 12 surplus +ve sign – 02 98 – 02 base is 100 2 deficiency –ve sign 992 992 24 As 012 × – 02 – 24 2 digits allowed in RHS of

### slide 44:

43 Answer ∴ 1012 × 98 99224 99176 As 992200 – 24 99176 Checking:1012 1 + 1 + 2 4 98 8 LHS 4 × 8 32 3 + 2 5 RHS 99176 1 + 7 + 6 14 1 + 4 5 As RHS LHS so answer is correct Try These: i 1015 × 89 ii 103 × 97 iii 1005 × 96 iv 1234 × 92 v 1223 × 92 vi 1051 × 9 vii 9899 × 87 viii 9998 × 103 ix 998 × 96 x 1005 × 107 Sub – base method: Till now we have all the numbers which are either less than or more than base numbers. i.e.10 100 1000 10000 etc. now we will consider the numbers which are nearer to the multiple of 10 100 10000 etc. i.e. 50 600 7000 etc. these are called sub-base. Example: 213 × 202 Step1: Here the sub base is 200 obtained by multiplying base 100 by 2 Step 2: R. H. S. and L.H.S. of answer is obtained using base- method. 213 + 13 202 + 02 215 13 × 02 26 Step 3: Multiply L.H.S. of answer by 2 to get 215 × 2 430 ∴ 213 × 202 43026 ∴ Example 2: 497 × 493 Step1: The Sub-base here is 500 obtained by multiplying base 100 by 5. Step2: The right hand and left hand sides of the answer are obtained by using base method. Step3: Multiplying the left hand side of the answer by 5. 497 –03 493 –07 Same 497–07 490 21 493 – 03 490 490 × 5 2450 ∴ 497 × 493 245021

### slide 45:

44 Example 3: 206 × 197 Sub-base here is 200 so multiply L.H.S. by 2 206 + 06 197 – 03 206 – 3 203 –18 197 + 06 203 × 2 18 406 ∴ 206 × 197 40618 40582 Example 4: 212 × 188 Sub – base here is 200 212 + 12 188 -12 200 – 12 200 144 188 + 12 200 × 2 400 –1 399 ∴ 212 × 188 399 44 39856 Checking:11 – check method + – + 2 1 2 2 + 2 – 1 3 + – + 1 8 8 1 – 8 + 8 1 L.H.S. 3 × 1 3 + – + – + R.H.S. 3 9 8 5 6 3 As L.H.S R.H.S. So answer is correct. Try these 1 42 × 43 2 61 × 63 3 8004 × 8012 4 397 × 398 5 583 × 593 6 7005 × 6998 7 499 × 502 8 3012 × 3001 9 3122 × 2997 10 2999 × 2998 Doubling and Making halves Sometimes while doing calculations we observe that we can calculate easily by multiplying the number by 2 than the larger number which is again a multiple of 2. This procedure in called doubling:

### slide 46:

45 35 × 4 35 × 2 + 2 × 35 70 + 70 140 26 × 8 26 × 2 + 26 × 2 + 26 × 2 + 26 × 2 52 + 52 + 52 + 52 52 × 2 + 52 × 2 104 × 2 208 53 × 4 53 × 2 + 53 × 2 106 × 2 212 Sometimes situation is reverse and we observe that it is easier to find half of the number than calculating 5 times or multiples of 5. This process is called Making halves: 4. 1 87 × 5 87 × 5 × 2/2 870/2 435 2 27 × 50 27 × 50 × 2/2 2700/2 1350 3 82 × 25 82 × 25 × 4/4 8200/4 2050 Try These: 1 18 × 4 2 14 × 18 3 16 × 7 4 16 × 12 5 52 × 8 6 68 × 5 7 36 × 5 8 46 × 50 9 85 × 25 10 223 × 50 11 1235 × 20 12 256 × 125 13 85 × 4 14 102 × 8 15 521 × 25 Multiplication of Complimentary numbers : Sutra: By one more than the previous one. This special type of multiplication is for multiplying numbers whose first digitsfigure are same and whose last digitsfiguresadd up to 10100 etc. Example 1: 45 × 45 Step I: 5 × 5 25 which form R.H.S. part of answer Step II: 4 × next consecutive number

### slide 47:

46 i.e. 4 ×5 20 which form L.H.S. part of answer ∴ 45 × 45 2025 Example 2: 95 × 95 9 × 10 90/25 → 5 2 i.e. 95 × 95 9025 Example 3: 42 × 48 4 × 5 20/16 → 8 × 2 ∴ 42 × 48 2016 Example 4: 304 × 306 30 × 31 930/24 → 4 × 6 ∴ 304 × 306 93024 Try These: 1 63 × 67 2 52 × 58 3 237 × 233 4 65 × 65 5 124 × 126 6 51 × 59 7 762 × 768 8 633 × 637 9 334 × 336 10 95 × 95 Multiplication by numbers consisting of all 9’s : Sutras: ‘By one less than the previous one’ and ‘All from 9 and the last from 10’ When number of 9’s in the multiplier is same as the number of digits in the multiplicand. Example 1 : 765 × 999 Step I : The number being multiplied by 9’s is first reduced by 1 i.e. 765 – 1 764 This is first part of the answer Step II : “All from 9 and the last from 10” is applied to 765 to get 235 which is the second part of the answer. ∴ 765 × 999 764235 When 9’s in the multiplier are more than multiplicand Example II : 1863 × 99999 Step I : Here 1863 has 4 digits and 99999 have 5-digits we suppose 1863 to be as 01863. Reduce this by one to get 1862 which form the first part of answer.

### slide 48:

47 Step II: Apply ‘All from 9 and last from 10’ to 01863 gives 98137which form the last part of answer ∴ 1863 x 99999 186298137 When 9’s in the multiplier are less than multiplicand Example 3 : 537 x 99 Step I: Mark off two figures on the right of 537 as 5/37 one more than the L.H.S. of it i.e. 5+1 is to be subtracted from the whole number 537 – 6 531this forms first part of the answer Step II: Now applying “all from 9 last from 10” to R.H.S. part of 5/37 to get 63 100 – 37 63 ∴ 537 x 99 53163 Try these 1 254 × 999 2 7654 × 9999 3 879 × 99 4 898 × 9999 5 423 × 9999 6 876 × 99 7 1768 × 999 8 4263 × 9999 9 30421 × 999 10 123 × 99999 Multiplication by 11 Example 1: 23 × 11 Step 1 : Write the digit on L.H.S. of the number first. Here the number is 23 so 2 is written first. Step 2 : Add the two digits of the given number and write it in between. Here 2 + 3 5 Step 3 : Now write the second digit on extreme right. Here the digit is 3. So 23 × 11 253 OR 23 × 11 2 / 2+3 / 3 253 Here base is 10 so only 2 digits can be added at a time Example 2: 243 × 11 Step 1: Mark the first second and last digit of given number First digit 2 second digit 4 last digit 3 Now first and last digits of the number 243 form the first and last digits of the answer. Step 2: For second digit from left add first two digits of the number i.e. 2 + 4 6 Step 3: For third digit add second and last digits of the number i.e. 3 + 4 7 So 243 × 11 2673 OR 243 × 11 2 / 2 + 4 / 4 + 3 / 3 2673 Similarly we can multiply any bigger number by 11 easily. Example 3: 42431 × 11

### slide 49:

48 42431 × 11 4 / 4 + 2 / 2 + 4 / 4 + 3 / 3 + 1 / 1 466741 If we have to multiply the given number by 111 Example 1: 189 × 111 Step 1: Mark the first second and last digit of given number First digit 1 second digit 8 last digit 9 Now first and last digits of the number 189 may form the first and last digits of the answer Step 2: For second digit from left add first two digits of the number i.e. 1 + 8 9 Step 3: For third digit add first second and last digits of the number to get 1 + 8 + 9 18 multiplying by 111 so three digits are added at a time Step 4: For fourth digit from left add second and last digit to get 8 + 9 17 As we cannot have two digits at one place so 1 is shifted and added to the next digit so as to get 189 × 111 20979 OR 1 1 + 8 9 1 + 8 + 9 8 + 9 9 9 + 1 18 1 7 1 + 1 2 1 0 18 + 1 1 9 ∴189 × 111 20979 Example 2 : 2891 × 111 2 2 + 8 2 + 8 + 9 8 + 9 + 1 9 + 1 10 + 2 19 + 1 18 + 1 1 0 1 1 2 2 0 1 9 2891 × 111 320901 Try These: 1 107 × 11 2 15 × 11 3 16 × 111 4 112 × 111 5 72 × 11 6 69 × 111 7 12345 × 11 8 2345 × 111 9 272 × 11 10 6231 × 111. Note: This method can be extended to number of any size and to multiplying by 1111 11111 etc. This multiplication is useful in percentage also. If we want to increase a member by 10 we multiply it by 1.1

### slide 50:

49 General Method of Multiplication. Sutra: Vertically and cross-wise. Till now we have learned various methods of multiplication but these are all special cases wherenumbers should satisfy certain conditions like near base or sub base complimentary to each other etc. Now we are going to learn about a general method of multiplication by which we can multiply any two numbers in a line. Vertically and cross-wise sutra can be used for multiplying any number. For different figure numbers the sutra works as follows: Two digit – multiplication Example: Multiply 21 and 23 Step1: Vertical one at a time 2 1 1 × 3 3 3 2 3 Step2: Cross –wise two at a time 2 1 2 × 3 + 2 × 1 8 2 3 8 3 Step3: Vertical one at a time 2 1 2 × 2 4 2 3 4 8 3 ∴ 21 × 23 483 Multiplication with carry: Example: Multiply 42 and 26 Step1: Vertical 42 2 × 6 12 12 26 Step2: Cross-wise 4 2 4 × 6 + 2 × 2 2 8 1 2 2 6 24 + 4 28 Step3: Vertical 42 4 × 2 8 8 8 2 26 + 2 1 10 2 9 ∴ 42 × 26 1092 Three digit multiplication: Example: 212 × 112 | | ↓ | | ↓ | | ↓ | | ↓

### slide 51:

50 Step1: Vertical one at a time 212 2 × 2 112 4 4 Step2: Cross-wise 2 1 2 2 × 1 + 2 × 1 4 4 two at a time 1 1 2 2 + 2 4 Step3: Vertical and cross-wise 2 1 2 three at a time 1 1 2 2 × 2 + 2 × 1 + 1 × 1 4 + 2 + 1 7 7 4 4 Step4: cross wise 2 1 2 2 × 1 + 1 × 1 3 7 4 4 Two at a time 1 1 2 2 + 1 3 Step 5: vertical one at a time 2 1 2 2 × 1 2 2 3 7 4 4 1 1 2 ∴ 212 × 112 23744 Three digits Multiplication with carry: Example: 816 × 223 8 1 6 8 × 2 8 × 2 + 2 × 1 8 × 3 + 6 × 2 + 1 × 2 3 × 1 + 2 × 6 6 × 3 18 16 + 2 18 24 + 12 + 2 3 + 12 15 38 2 2 3 16 16 + 2 18 + 3 38 + 1 15 + 1 1 8 2 1 3 9 1 6 ∴ 816 × 223 181968 Checking by 11 – check method + – + – + 8 1 6 14 – 1 1 3 3 – 1 2 + - + 2 2 3 3 ∴ L.H.S. 3 × 2 6 | | ↓ | | ↓ ↑ | | ↓ ↑ | | | | | ↓

### slide 52:

51 - + - + - + - + 1 8 1 9 6 8 1 7 7 – 1 6 As L.H.S. R.H.S. ∴ Answer is correct Try These: 1 342 × 514 2 1412 × 4235 3 321 × 53 4 2121 × 2112 5 302 × 415 6 1312 × 3112 7 5123 × 5012 8 20354 × 131 9 7232 × 125 10 3434 × 4321 Number Split Method As you have earlier used this method for addition and subtraction the same may be done for multiplication also. For example : 263 26 3 Note : The split allows us to add 36 + 24 × 2 × 2 ×2 and 42 + 39 both of which can be done 52 6 mentally Multiplication of algebraic expressions: Sutra: Vertically and cross-wise Example1 : x + 3 x + 4 x + 3 x × x 4x + 3x 4 × 3 x 2 7x 12 x + 4 x 2 7 × 12 x 2 + 7x + 12 Example2: 2x + 5 3x + 2 2x + 5 2x × 3x4x + 15x 10 6x 2 19x 3x + 2 6x 2 +19x + 10 | | | | ↓ | | | | ↓ | | | | ↓ | | | | ↓

### slide 53:

52 Example3: x 2 + 2x + 5 x 2 – 3x + 1 x 2 + 2x – 5 x 4 –3x 3 +2x 3 x 2 + 5x 2 –6x 2 2x –15x 5 × 1 x 2 – 3x + 1 –x 3 0 – 13x 5 x 4 – x 3 – 13x + 5 x 4 –x 3 0 –13x 5 Try These: 1 2x – 1 3x + 2 2 2x + 1 x 2 + 3x – 5 3 5x+ 5 7x – 6 4 x + 5 x 2 – 2x + 3 5 x – 4 x 2 + 2x + 3 6 x 2 + 4x – 5 x + 5 7 x 3 – 5 x 2 + 3 8 x 2 – 2x + 8 x 4 – 2 9 x 2 – 7x + 4 x 3 – 1 10 x 3 – 5x 2 + 2 x 2 + 1 | | ↓ | | ↓

### slide 54:

53 Chapter 5 Squaring and square Roots Square of numbers ending in 5 : Sutra: ‘By one more than previous one” Example: 75 × 75 or 75 2 As explained earlier in the chapter of multiplication we simply multiply 7 by the next number i.e. 8 to get 56 which forms first part of answer and the last part is simply 25 5 2 . So 75 × 75 5625 This method is applicable to numbers of any size. Example: 605 2 60 × 61 3660 and 5 2 25 ∴ 605 2 366025 Square of numbers with decimals ending in 5 Example : 7.5 2 7 × 8 56 0.5 2 0.25 7.5 2 56.25 Similar to above example but with decimal Squaring numbers above 50: Example: 52 2 Step1: First part is calculated as 5 2 + 2 25 + 2 27 Step2: Last part is calculated as 2 2 04 two digits ∴ 52 2 2704 Squaring numbers below 50 Example : 48 2 Step1: First part of answer calculated as: 5 2 – 2 25 – 2 23 Step2: second part is calculated as : 2 2 04 ∴ 48 2 2304 Squaring numbers near base : Example : 1004 2 Step1: For first part add 1004and 04 to get 1008 Step2: For second part4 2 16 016 asbase is 1000 a three digit no. ∴ 1004 2 1008016

### slide 55:

54 Squaring numbers near sub - base: Example 302 2 Step1: For first part 3 302 + 02 3 × 304 912 Here sub – base is 300 so multiply by 3 Step2: For second part 2 2 04 ∴ 302 2 91204 General method of squaring: The Duplex Sutra: “Single digit square pair multiply and double” we will use the term duplex` D’ as follows: For 1 figureor digit Duplex is its squaree.g. D4 4 2 16 For2 digitsDuplex is twice of the product e.g. D34 2 3 x 4 24 For 3 digit number: e.g. 341 2 D3 3 2 9 D 34 2 3 × 4 24 D 341 2 3 × 1 + 4 2 6 + 16 22 9 4 2 8 1 D 41 2 4 × 1 8 2 2 D 1 1 2 1 116281 ∴ 341 2 116281 Algebraic Squaring : Above method is applicable for squaring algebraic expressions: Example: x + 5 2 D x x 2 Dx + 5 2 x × 5 10x D 5 5 2 25 ∴ x + 5 2 x 2 + 10x + 25 Example: x – 3y 2 D x x 2 Dx – 3y 2 x × – 3y – 6xy D–3y –3y 2 9y 2 ∴ x – 3y 2 x 2 – 6xy + 9y 2 Try these: I 85 2 II 8 2 1 2 III 10.5 2 IV 8050 2 V 58 2 VI 52 2 VII 42 2 VIII 46 2 IX 98 2 X 106 2 XI 118 2 XII x + 2 2 XIII y – 3 2 XIV 2x – 3 2 XV 3y – 5 2

### slide 56:

55 SQUARE ROOTS: General method: As 1 2 1 2 2 4 3 2 9 4 2 1 6 5 2 2 5 6 2 3 6 7 2 4 9 8 2 6 4 9 2 81 i.e. square numbers only have digits 145690 at the units place or at the end Also in 16 digit sum 1 + 6 7 25 2 + 5 7 36 3 + 6 9 49 4 + 9 13 13 1 + 3 4 64 6 + 4 10 1 + 0 1 81 8 + 1 9 i.e. square number only have digit sums of 1 4 7 and 9. This means that square numbers cannot have certain digit sums and they cannot end with certain figures or digits using above information which of the following are not square numbers: 1 4539 2 6889 3 104976 4 27478 5 12345 Note: If a number has a valid digit sum and a valid last figure that does not mean that it is a square number. If 75379 is not a perfect square in spite of the fact that its digit sum is 4 and last figure is 9. Square Root of Perfect Squares: Example1: √5184 Step 1: Pair the numbers from right to left 5184 two pairs Therefore answer is 2 digit numbers 7 2 49 and 8 2 64 49 is less than 51 Therefore first digit of square root is 7. Look at last digit which is 4 As 2 2 4 and 8 2 64 both end with 4 Therefore the answer could be 72 or 78 As we know 75 2 5625 greater than 5184 Therefore √5184 is below 75 Therefore √5184 72 Example 2: √9216 Step 1: Pair the numbers from right to left 9216two pairs Therefore answer is 2 digit numbers 9 2 81 and 10 2 100 81 is less than 92 Therefore first digit of square root is 9. Look at last digit which is 6 As 4 2 16 and 6 2 36 both end with 6 Therefore the answer could be 94 or 96

### slide 57:

56 As we know 95 2 9025 less than 9216 Therefore √9216 is above 95 Therefore √9216 96 General method Example 1 : √2809 Step1: Form the pairs from right to left which decide the number of digits in the square root. Here 2 pairs therefore 2 - digits in thesquare root Step 2: Now √28 nearest squares is 25 So first digit is 5 from left Step3: As 28 – 25 3 is reminder which forms 30 with the next digit 0. Step 4: Multiply 2 with 5 to get 10 which is divisor 10 √2809 30 Now 3 × 10 30 30 Q R 10 3 0 Step 5: As 3 2 9 and 9 – 9 last digit of the number 0 ∴ 2809 is a perfect square and √2809 53 Example 2:3249 Step1: Form the pairs form right to left which decided the number of digits in the square root. Here 2 pairs therefore 2 digits in the square root. Step2: Now 32 25 5 2 so the first digit in 5 from left Step 3: 32 – 25 7 is remainder which form 74 with the next digit 4 5 7 Step 4: Multiply 2 with 5 to get 10 which is divisor 10√3249 Now 74 Q R 7 4 107 4 Step5: 7 2 49 and 49 – 49 0 remainder is 4 which together with9 form49 ∴ 3249 is a perfect square and √3249 57 Example 3: 54 75 6 Step1: Form the pairs from right to left therefore the square root of 54756 has 3-digits. Step2: 5 4 2 2 i.e. nearest square is 2 2 4 So first digit is 2 from left Step3: As 5 – 4 1 is remainder which form 14 with the next digit 4.

### slide 58:

57 Step4: Multiply 2 with 2 to get 4 which is divisor 2 4 5 1 4 2 75 6 Now 14 Q R 4 3 2 Step 5: Start with remainder and next digit we get 27. Find 27 – 3 2 27 – 9 18 square of quotient 234 Step 6: 18 Q R 4 5 1 4 2 7 2 5 1 6 4 4 2 Now 25 – 3 × 4 × 2 25 – 24 1 1 Q R 4 0 1 16 – 4 2 16 – 16 0 ∴ 54756 is a perfect square and so √5 4 7 5 6 234 Try These: 1. 2116 2. 784 3. 6724 4. 4489 5. 9604 6. 3249 7. 34856 8. 1444 9. 103041 10. 97344

### slide 59:

58 CHAPTER-6 DIVISION Defining the Division terms There are 16 balls to be distributed among 4 people How much each one will get is a problems of division. Let us use this example to understand the terms used in division. Divisor: —Represent number of people we want to distribute them or the number that we want to divide by. Here the divisor is 4. Dividend: -Represents number of balls to be divided 16 in this case. Quotient:Represents the number of balls in each part 4 is this case. Remainder:What remains after dividing in equal parts 0 in this case The remainder theorem follows from the division example above and is expressed mathematically as follows. Divided Divisor × Quotient + Remainder The remainder theorem can be used to check the Division sums in Vedic Mathematics as described in the following sections. Different methods are used for dividing numbers based on whether the divisor is single digit numbers below a base above a base or no special case. Special methods of Division. Number splitting Simple Division of Divisor with single digits can be done using this method. Example:The number 682 can be split into 6/82 and we get 3/41 because 6 and 82 are both easy to halve Therefore 682/2 341 Example : 3648/2 becomes 36/48/2 18/24 1824 Example:1599/3 we notice that 15 and 99 can be separately by 3 so 15/99/3 5/33 533

### slide 60:

59 Example: 618/6 can also be mentally done 6/18/6 103 note the 0 here Because the 18 takes up two places Example: 1435/7 14/35/7 2/05 205 Example: 27483/3 becomes 27/48/3/3 9/16/1 9161 Practice Problem Divided mentally Numbers Splitting 1 2656 2 2726 3 31899 4 61266 5 32139 6 22636 7 4812 8 64818 9 840168 10 5103545 Division by 9 As we have seen before that the number 9 is special and there is very easy way to divide by 9. Example : Find 25 ÷ 9 25/9 gives 2 remainder 7 The first figure of 25 is the answer And adding the figures of 25 gives the remainders 2 + 5 7 so 25 ÷ 9 2 remainder 7. It is easy to see why this works because every 10 contains 9 with 1 left over so 2 tens contains 2 times with 2 left over. The answer is the same as the remainders 2. And that is why we add 2 to 5 to get remainder. It can happen that there is another nine in the remainder like in the next example Example: Find 66 ÷ 9 66/9 gives 6 + 6 12 or 7 or 3 We get 6 as quotient and remainder 12 and there is another nine in the remainder of 12 so we add the one extra nine to the 6 which becomes 7 and remainder is reduced to 3 take 9 from 12 We can also

### slide 61:

60 get the final remainder 3 by adding the digits in 12. The unique property of number nine that it is one unit below ten leads to many of the very easy Vedic Methods. This method can easily be extended to longer numbers. Example: 3401 ÷ 9 377 remainder 8 Step 1: The 3 at the beginning of 3401 is brought straight into the answer. 93401 3 Step 2: This 3 is add to 4 in 3401 and 7 is put down 93401 37 Step 3: This 7 is then added to the 0 in 3401 and 7 is put down. 93401 377 Step 4: This 7 is then added to give the remainder 9 340/1 377/8 Divided the following by 9 1 951 2 934 3 917 4 944 5 960 6 926 7 946 8 964 9 988 10 996 Longer numbers in the divisor The method can be easily extended to longer numbers. Suppose we want to divide the number 21 3423 by 99. This is very similar to division by 9 but because 99 has two 9’s we can get the answer in two digits at a time. Think of the number split into pairs. 21/34/23 where the last pair is part of the remainder.

### slide 62:

61 Step 1: Then put down 21 as the first part of the answer 9921/34/23 21 Step 2: Then add 21 to the 34 and put down 55 as next part 9921/34/23 21/55 Step 3: Finally add the 55 to the last pair and put down 78 as the remainder 9921/34/23 21/55/78 So the answer is 2155 remainder 78 Example: 12/314 ÷ 98 1237 Step 1: This is the same as before but because 98 is 2 below 100 we double the last part of the answer before adding it to the next part of the sum. So we begin as before by bringing 12 down into the answer. 98 12/13/14 12 Step 2: Then we double 12 add 24 to 13 to get 37 98 12/13/14 12/37 Step 3: Finally double 37 added 37 × 2 74 to 14 9812/13/14 12/37/88 1237 remainder 88. It is similarly easy to divide by numbers near other base numbers 100 1000 etc. Example: Suppose we want to divide 236 by 88 which is close to 100. We need to know how many times 88 can be taken from 235 and what the remainder is Step 1: We separate the two figures on the right because 88 is close to 100 Which has 2 zeros 88 2/36 Step 2: Then since 88 is 12 below 100 we put 12 below 88 as shown 88 2/36 Step 3: We bring down the initial 2 into the answer 88 2/36 12 2

### slide 63:

62 Step 4: This 2 is multiplies Haggled 12 and the 22 is placed under the 36 as Shown 88 2/36 12 2 / 24 Step 5: We then simply add up the last two columns. 88 2/36 12 2 r 60 In a similar way we can divide by numbers like 97 and 999. Practice problems Divide the following using base method 1 121416 by 99 2 213141 by 99 3 332211 by 99 4 282828 by 99 5 363432 by 99 6 11221122 by 98 7 3456 by 98 Sutra: Transpose and Apply A very similar method allows us to divide numbers which are close to but above a base number. Example: 1479 ÷ 123 12 remainder 13 Step 1: 123 is 23 more than base 100 Step 2: Divide 1479 in two columns therefore of 2digit each Step 3: Write 14 down Step 4: Multiply 1 by 23 and write it below next two digits. Add in the Second column and put down 2. Step 5: Add multiply this 2 the 2 3 and put 46 then add up last two Columns 123 14 78 23 23 46 12/02

### slide 64:

63 Straight Division The general division method also called Straight division allows us to divide numbers of any size by numbers of any sine in one line Sri BharatiKrsnaTirthaji called this “the cowing gem of V edic Mathematics” Sutra: - ‘vertically and crosswise’ and ‘on the flag’ Example: Divide 234 by 54 The division 54 is written with 4 raised up on the flag and a vertical line is drawn one figure from the right hand end to separate the answer 4 from the remainder 28 23 3 4 5 4 20 16 428 Step 1: 5 into 20 goes 4 remained 3 as shown Step 2: Answer 4 multiplied by the flagged 4 gives 16 and this 16 taken from 34 leaves the remainder 28 as shown Example: Divide: 507 by 72 50 1 7 7 2 49 14 7 3 Step 1: 7 into 50 goes 7 remainder 1 as shown Step 2: 7 times the flagged 2 gives 14 which we take from 17 to have remainder of 3 Split Method Split method can be done for division also. For example : 6234 ÷ 2 62 34 ÷ 2 ÷2 31 17 The split may require more parts. 30155 ÷ 5 30 15 5 ÷5 ÷5 ÷5 6 03 1 6031 244506 ÷ 3 24 45 06 ÷3 ÷3 ÷3 8 15 02 81502

### slide 65:

64 Practice Question Divide the following using straight division 1 209 ÷ s52 2 621 ÷ 63 3 503 ÷ 72 4 103 ÷ 43 5 74 ÷ 23 6 504 ÷ 72 7 444 ÷ 63 8 543 ÷ 82 9 567 ÷ 93 10 97 ÷ 28 11 184 ÷ 47 12 210 ÷ 53 13 373 ÷ 63 14 353 ÷ 52 15 333 ÷ 44 16 267 ÷ 37 17 357 ÷ 59 18 353 ÷ 59 19 12233 ÷ 53 Books for Reference 1. Sri BharatiKrsnaTirthaji “Vedic Mathematics” published by MotilalBanarsidass 1965. ISBN 81- 208-0163-6. 2. Williams K.R. “Discover Vedic Mathematics.” Vedic Mathematics Research Group 1984. ISBN 1- 869932-01-3 3. Williams K.R. and M. Gaskell “The Cosmic Calculator”. MotilalBanarsidass 2002.ISBN 81-208- 1871-7. 4. Nicholas A.P. Williams J. Pickles. “Vertically and Crosswise”. Inspiration Books 1984. ISBN 1- 902517-03-2. 