VEDIC MATHEMATICS MANUAL

Transcription

1 VEDIC MATHEMATICS MANUAL

2 Material produced by: Resource persons/ Teachers in collaboration with IIVA

3 Introduction of Vedic Maths Benefits of Vedic Maths History of Vedic Maths Sutras & Sub Sutras Chapter - 3. Addition Chapter Subtraction (a) All from a last from 0 (b) Using Vinculum Chapter Digit Sum (casting out 9 s, 9-check method) Chapter Special Multiplication Methods a) Multiplication by and multiples of b) Multiplication by to 9 c) Multiplication by d) Multiplication by to 999 e) Base Method Multiplication f) If sum of unit digit is 0 and rest place digit are same. g) If sum of ten s place digits is 0 and ones place digit is same. h) Multiplication by 9 I) Multiplication by number ending with 9 i.e. 9 to 99 j) General Method k) Algebraic Expressions Chapter Squares and Square Roots Chapter Division

4 Mathematics is universally regarded as science of all sciences. In all early civilizations, the first expression of mathematical understanding appears in the form of counting systems. In Indian history MATHEMATICS had occupied a significant place. All earlier systems of numeration had certain drawbacks. It was Indian system that reached the western world through the Arabs and has now been accepted universally. India has always been a land of great mathematicians. Aryabhatta a man who knew infinity, Srinivasa Ramanujan etc are some mathematicians who revolutionized the world of mathematics. In Vedic period, records of mathematical activities are mostly to be found in Vedic texts associated with rituals. Vedic Mathematics is a term given to this ancient system of Mathematics which is supposed to be formulated over many centuries by ancient sages and rishis of India. It was rediscovered from Vedas between 9 and 98 by Jagadguru Swami Sri Bharti-Krishna Tirathji Maharaj. The term vedic mathematics refers to sixteen mathematical formulae or sutras and their corollaries or subsutras derived from the vedic system. In vedic system pupil are encouraged to be creative and use whatever method they like. All that the student has to do is to look for certain characteristics, spot them out, identify the particular type and apply the formulae which is applicable there to. Vedic Mathematics is useful in preparing students for competitive examination.

5 Mathematics is as old as human race itself. A human brain understands mathematics in a very little age. Little kids have number sense that can be easily judged in certain practical situations. Let's take an example, if you keep 5 toffees in one hand and toffees in other hand and ask child to choose either of the two. He/ She definitely go for hand with five toffees. This shows the child has number sense, he/ she knows the concept of more & less. The evidence of mathematics was available in all early civilizations also. During Harappa period decimal system was prevalent as weights corresponding to ratios of 0.05, 0., 0., 0.5,,, 5 etc and scales with decimal divisions were found. In Indian history during Vedic period, records of mathematical activities are mostly to be found in Vedic texts associated with ritual activities. Arithmetic's operations (GANIT) such as additions, subtractions, multiplications, fractions, squares, cubes and roots are enumerated in Narad- Vishnu Purana attributed to Ved Vyas (pre 000 B.C). Examples of Geometric knowledge (rehka ganit) are to be found in Sulva- Sutra of Baudhayana (800 B.C) and Apasthamba (600 B.C) which describe techniques for the construction of altars in use during Vedic Era. It is considered that all the information about Vedic period was stored in a book called VEDA. Later on it was divided into four Vedas namely RIGVEDA, YAJURVEDA, SAMAVEDA and ATHARVAVEDA. Indian Rishis and Sages used to teach everything in Gurukuls. Written Records of many things are not available. Vedic Mathematics is the name given to ancient system of mathematics which is supposedly is a part of Sthaptya Veda, an Upveda of Atharvaveda.

6 It is a gift to the world and was formulated over centuries by ancient sages & rishis of India. It was rediscovered from Vedas between 9 & 98 by Jagadguru Swami Sri Bharti-Krishna Tirathji Maharaj. Swamiji was the Shankracharya of Govardhan Math, Jagannath puri as well as Dwarka, Gujarat ( ). He was a great scholar and a brilliant student. He was M.A in seven subjects including English, Mathematics, Science, Sanskrit, History etc. He was a person who worked in this field and wrote 6 volumes on the sixteen sutras of vedic mathematics. These volumes got lost and only one volume containing all information is in print. The term Vedic Mathematics then onwards refers to sixteen sutras on which further derivations are being done. This work of Swamiji is considered to be first work done in field of vedic mathematics. The most striking features of Tirathji's work is its coherence. The whole system is inter-related and unified. The methods are simple and help students to give solution to various arithmetical problems mentally. Later on, in late 960's some mathematicians took interest in this system and started delivering lectures on this subject. There are number of books in print on the subject. Lots of people are working in the field of Vedic Mathematics. In India many organizations are working in this field to make the subject more interesting and help pupil remove the phobia of mathematics. 3

7 Vedic Mathematics provides pupil with the extra something which helps them to be different. It help them to play with numbers and thereby removes the phobia of Mathematics. It helps them to be more confident about the subject. Pupil start loving Mathematics once their interest is triggered and confidence level is increased. Advantages: It helps in reducing silly mistakes. It helps in solving problems 0-5 times faster. It helps in intelligent guessing. Reduces Phobia of Mathematics. It reduces the burden (Need to learn tables only up to 9). It increases concentration. Vedic Maths is a system of reasoning and mathematical working which is fast efficient and easy to learn and use. We all are aware that world is getting competitive day by day and in every competition there is a race against time only those people who have ability to calculate faster will able to win competitions as faster calculations will lead to time saving and saved time can be utilised to solve much difficult problems. Vedic mathematics provides answer in one line where as conventional method requires several steps. Its techniques simplifies multiplication, divisibility, squaring, cubing etc. It can even simplifies recurring decimals and Auxiliary fractions. The simplicity of techniques encourages flexibility and helps student to device newer techniques i.e. his/her own methods and not remain limited to same rigid approach. The simplicity of Vedic Mathematics encourage most calculations to be carried out without the use of pen and paper. Once the mind of students develops the understanding for system he/she begins to work closely 4

8 with numbers and becomes more creative. It is very easy to understand and practice. This process helps student to be confident about the subject mathematics and this removes his/her phobia of mathematics. By Applying Sutras we can save lot of time and effort in solving the problems, compared to the formal methods presently in use. Though the solutions appear like a magic, the application of the Sutras is perfectly logical and rational. This course of Vedic Mathematics seeks to present an integrated approach to learning Mathematics with interest and inquisitiveness, avoiding the monotony of accepting theories and logical proof of the Sutras is explained, which eliminates the misconception that the Sutras are wrongly explained. Application of the Sutras improves the computational skills of the learners ensuring both speed and accuracy. The knowledge of such methods enables the teachers to be more resourceful to enhance the students and strengthen their talent and knowledge. To specific problems application of sutras involves rational thinking, which, in turn, 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 the learning Mathematics at the school and college level in a way different from what is taught at present, but strictly embodying the principles of algebra for empirical accuracy. The presentation and explanation of concepts are in simple language to make it easy to understand. This course will build your confidence level as well as help you become self dependent. 5

9 ,dkf/kdsu iwosz.k (Ekâdhikena Pûrve?a) fuf[kya uor'pjea n'kr% (Nikhila? Navataœcarama? Daœata?) 3 Å/oZfr;ZXH;ke~ (Ûrdhva Tiryagbhyâ? 4 ijkor;z ;kst;sr~ (Parâvartya Yojayet) 5 'kwu;a lke;leqpp;s (Œûnya? Sâmyasamuccaye) 6 ¼vkuq#I;s½ 'kwu;eu;r~ (Ânurûpye Œûnayamanyat) 7 ladyuo;odyukh;ke~ (Sa?kalana Vyavakalanâbhyâ? ) 8 iwj.kkiwj.kkh;ke~ (Pûra?âpûra?âbhyâ? ) 9 pyudyukh;ke~ (Calana Kalanâbhyâ? ) 0 ;konwue~ (Yâvadûna? ) O;f"Vlef"V% (Vya??isama??i?) 'ks"kk.;m~dsu pjes.k (Œe?â?ya?kena Carame?a) 3 lksikur;};eur;e~ (Sopântyadvayamantya? ) 4,dU;wusu iwosz.k (Ekanyûnena Pûrvena) 5 xqf.krleqpp;% (Gu?itasamuccaya?) By One More than the One Before. All from 9 and the Last from 0. Vertically and Crosswise Transpose and Apply If the Samuccaya is the Same it is Zero If One is in Ratio the Other is Zero By Addition and by Subtraction By the Completion or Non-Completion Differential Calculus By the Deficiency Specific and General The Remainders by the Last Digit The Ultimate and Twice the Penultimate By One Less than the One Before The Product of the Sum 6 xq.kdleqpp;% (Gu?akasamuccaya?) All the Multipliers 6

10 vkuq:i;s.k (Ânurûpye?a) f'k";rs 'ksôlak% (Œi?yate Œe?as? jña?) vk ek s ukur;eur;su (Âdyamâdyenântyamantyena) dsoys% lirda xq.;kr~ (Kevalai? Saptaka? Gu?yât) os"vue~ (Ve??ana? ) Proportionately The Remainder Remains Constant The First by the First and the Last by the Last For 7 the Multiplier is 43 By Osculation ;konwua rkonwue~ (Yâvadûna? Tâvadûna? ) ;konwua rkonwuhd`r; oxz p ;kst;sr~ (Yâvadûna? Tâvadûnîk?tya Vargañca Yojayet) vur;;ksnz kds fi (Antyayordaœake'pi) Lesson by the Deficiency Whatever the Deficiency, less on by that amount and set up the square of the Deficiency. Last Totalling 0 9. vur;;ksjso (Antyayoreva) Only the Last Terms leqpp;xqf.kr% (Samuccayagu?ita?) yksilfkkiukh;ke~ (Lopanasthâpanâbhyâ? ) foyksdue~ (Vilokana? ) xqf.krleqpp;% leqpp;xqf.kr% (Gu?itasmuccaya? Samuccayagu?ita?) The sum of the Products By Alternate Elimination and Retention By Mere Observation The Product of the Sum is the Sum of the Products. 7

11 Terms and Operations (a) Ekadhika means one more e.g : Ekadhika of 0 is Ekadhika of is Ekadhika of 7 is 8 Ekadhika of 7 is 8 Ekadhika of 45 is 46 (b) Ekanyuna means one less e.g : Ekanyuna of is 0 Ekanyuna of is Ekanyuna of 4 is 3 Ekanyuna of 98 is 97 Ekanyuna of 75 is 74 (c) Purak means complement e.g: Purak of is 9 Purak of is 8 Purak of 3 is 7 Purak of 4 is 6 Purak of 5 is 5 and Vice Versa (d) Rekhank means a digit with a bar on its top. In other words it is a negative number. e.g : A bar on 8 is written as 8. It is called Rekhank 8 or bar 8. We treat Purak as a Rekhank. e.g : 8 is and is 8 At some instances we write negative numbers also with a bar on the top of the numbers as - can be shown as. -45 can be shown as 45. (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 4 is + 4 = 6 Beejank of 648 is = 8, further + 8 = 9 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. 8

12 e.g. : Find the Beejank of 6743 As above we have to follow But a quick look gives 6 & 3 ; &7 are to be ignored because 6+3=9, +7=9. Hence remaining is the beejank of 6743 (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 digit which are 5 and above 5 or less than 5 without changing the value of that number. Consider a number say 7.(Note that it is greater than 5). Use its complement (Purak-rekhank) from 0. It is 3 in this case and add to the left ( i.e. tens place) of 7. Thus, 7 = 07 = 3 The number contains both positive and negative digits. i.e. and 3. Here 3 is in units place, hence it is - 3 and value of at tens place is 0. T O 3 3 x = -3 + x 0 = 0 Conveniently we can think and write in the following way General Number Conversion Vinculum number etc. 7 9

13 Chapter- Addition is the most basic operation and adding number 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 + = + = + = = = = = = = = 0... Completing the whole method The VEDIC Sutra By the Deficiency relates our natural ability to see how much something differs from wholeness. 7close to 0 8 close to 0 9 close to 0 7, 8, 9, are close to 0 7, 8, 9 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 and so on We can use this closeness to find addition and subtraction The Ten Point Circle 9 Rule : By completion non-completion Five number pairs Use these number pairs to make groups of 0 when adding numbers. Example : = = Adding a list of numbers Below a multiple of ten Rule : By the deficiency 49 is close to 50 and is short 38 is close to 40 and is short

14 e.g. : = = = 53 (49 is close to 50 and short of 50, 49+ = is 50) e.g. : = = 30 + = 4 or = = 4 (8 is close and is short too of 30) So, 44 - = 4 e.g. : Add =? 49 is close to 50 and is less then it. So we take from the 6 to make up 50 and then we have 5 more to add on which gives 55 Add (As 3 = + and 39 + = 40, 8 + =30) = 70 [Note: we break 3 into + because 39 need to become 40 and 8 need to become 30] Add =6 Sum of Ten The ten point circle illustrates the pairs of numbers whose sum is 0. Remember : There are eight unique groups of three number that sum to0 for example = 0 Can you find the other seven groups of three number summing to 0 as one example given for you? Rule : By completion or non-completion Look for number pairs that make a multiple of The list can be sequentially added as follows : = 3 then = 6 then = 0 or you could look for number pairs that make multiples of is 0 and is 0 hence is 0 Similarly : =(78 + ) + ( ) = = 67 Example = = 45 This method can be easily used in commutative and associative property.

15 Practice Sheet = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

16 Chapter- Subtraction Sutra applied is Nikhila? Navataœcarama? Daœata? This kind of subtraction is done when there are number of zeros in minuend. eg. :,00,000 Minuend -7,354 Subtrahend ,646 difference We subtract unit digit of subtrahend from 0 and remaining from 9 till the highest place of subtrahend here for eg. till ten thousand place. From the remaining digit of places of minuend subtract Here for eg, remaining place is just lakhs and digit is so - = 0 so the difference is 7646 eg. : Here subtract unit place digit of subtrahend from 0 (0-3=7) And remaining digits i.e. 4 & 3 from 9 (9-4=5; 9-3=6) From the remaining digit of places of minuend. Here for eg. remaining places are thousand and ten thousand subtract and digits are 5 and 0 So 50 - = 49 So the difference is

17 Practice Sheet 00-37= = = = = = = = = = = = = = = = = = = = 00-57= = = = = = = = = = = = = = = = = = = = 4

18 What is vinculum? Vinculum : The numbers which by presentation contain both, positive and negative digits are called vinculum numbers. Use of Vinculum in Vedic Maths : It is used to represent negative digit in Vedic Maths. It helps to show mixed digits i.e.+ ve in one place and-ve in the other. Advantages of using vinculum : () It gives us flexibility, we use the vinculum when it suits us. () Large numbers like 6, 7, 8, 9 can be converted to smaller digits. (3) Figures tend to cancel each other or can be made to cancel. (4) 0 and occur twice as frequently as they otherwise would. (5) Borrowing can be avoided Converting from positive to vinculum form : Sutras : (Nikhila? Navataœcarama? Daœata?) & (Ekâdhikena Pûrve?a) (All from 9 and last from 0) and (One more than the previous one). 9= (i.e. 9-0), 8 =, 7 = 3, 6 = 4, 9 =, 9 = 3 8 = 3, 36 = 44, 38 = 4 Step to convert from positive to vinculum form : () Find out the digits that are to be converted. () Apply all from 9 and last from 0 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. Consider a number say 7.(Note that it is greater than 5). Use it complement (Purak-rekhank) from 0. It is 3 in this case and add to the left ( i.e. tens place) of 7. Thus, 7 = 07 = 3 The number 3 contains both positive and negative digits i.e. and 3. Here 3 is in unit place hence it is - 3 and value of at tens place is 0. T O 3 3 x = -3 + x 0 = 0 7 Conveniently we can think and write in the following way General Number Conversion Vinculum number etc. 5

19 Change units digit into a vinculum. X means negative X, for the purpose of subtraction and division in Vedic Mathematics. If you want to vinculate unit digit of a number then subtract that number from 0 and put the bar on number obtained (i.e find its Purak-Rekhank). Then after that add to tens place digit. eg. : 58 = 6 (8-0 = ) (5+ =6) 87 = 93 (7-0 = ) 3 (8+ = 9) 394 = 406 ) (4-0 = ) 6 (39+ = 40) If you want to vinculute all digit except first then subtract 0 from unit place digit and 9 from remaining except the highest place digit. Add to highest place digit. eg. : = (3-0=7) (7-9=) (6-9=3) (8-9=) (3+=4 ) eg. : 907 = (7-0=3) (0-9=9) (9-9=0) or (9+ =30) (+=3) Zero is neither positive nor negative it can be written without bar in both cases Steps to convert from vinculum to positive form : () Find out the digits that are to be converted i.e. digits with a bar on top. () Apply all from 9 and the last from 0 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 If you want to devinculate unit place. Subtract unit place from 0 and write the number obtained without bar and subtract from remaining digits. eg. : 5 = 49 (0- = 9) (5- =4) eg.: 3 = 8 (0- = 8) 6

20 If you want to devinculate, all digits except the first, then subtract units digit from 0 and remaining except highest place digit from 9. Write all numbers without bars. Now subtract from the highest place digit. eg. : = (0-8=) (9-9=0) (9-3=6) (9-=8) If you want to devinculate a number which is without bar at tens place again a number with bar on hundreds place and thousands place without bar. In such case subtract units place number from 0 and subtract from tens place number again subtract hundred place number from 0 and subtract from thousands place. eg. : 6 3 = 5708 (0-=8) (-=0) (0-3=7) (6-=5) Do subtraction at every place and get the respective answer with sign (Minuendsubtrahend). For every negative digit put bar on it. Thus, the answer will come in mixed form that is positive & negative digits together. Now devinculate the answer. eg. : (8-= ) (3-= ) (9-3=6 ) (7-=6 ) Devinculate = 5397 (7 as it is) (0-=9) (9-6=3) (6-=5) Here s the answer =

21 Practice Sheet 39 = 37=. 48 = 48= 97= 77 =. 376 = 76 = 3. 8= 3. 88= = = 4. 8= 4. 99= 5. 58= 5. 8 = 6. 79= 6. 9= 7. 58= = 8. 43= 8. 78= 8. 7= 9. 98= 0. 5= = 47= 37 =. 3= 4 = = = 4. 5= 4. 4 = 5. = 5. 4 = 6. 73= = = = 8. 4 = = 9. 8 = = = 0. 47= = = = = = = = = = = = = = = 5= 3 3=. 63= 3. 53= 3. 4= = 4. 74= = 5. 84= = 6. 63= = 7. 50= 37. 6= 8. 43= = = = = = 8

22 Practice Sheet 4-787= 543-7= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 9

23 Chapter-3 Digit Sum The word digit means a single figure of a number. The numbers,, 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 4 is +4+ = 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 0. The digit sum of 57 is = + = 3 greater than 9, so need to add again Hence, the digit sum of 57 is 3 The digit sum of 687 is = + = 3 Hence, the digit sum of 687 is 3 Keep finding the digit sum of the result, until it is less than 0 0 and 9 are equivalent Look and understand some more examples: To find the digit sum of 8, add and 8, i. e. +8 = 9. So the digit sum of 8 is 9. And the digit sum of 34 is 9 because +3+4 = 9 Following table shows the digit sum of the following numbers: Sometimes, two steps are needed to find the digit sum. So, for the digit sum of 9, we add +9 = Since is digit number, we again find digit sum, i. e. add + = So, for the digit sum of 9, we can write 9 = +9 = = + = Similarly digit sum of 49 = 4+9 = 3 = +3 = 4 So the digit sum of 49 is 4. Number 4 Digit sum +4 = 5 Single digit sum is = = = = 0

24 Adding 9 to a number does not affect its digit sum So, 5, 59, 95, 959 all have same digit sum of 5. For example, to find out the digit sum of 4939, we can cast out 9s and just add 3 and 4. So the digit sum is 7 or using = 5 +5 = 7 There is another way of casting out the 9s from number when you are finding its digit sum. Casting out 9s and digit totaling 9 comes under the Sutra When samuccaya is the same, it is zero. Observe in 465, 4 & 5 addition is 9, they are cast out and the digit sum becomes 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 or Nine Point Circle 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. Digit sum can be used to check whether the answers are correct. Example: Solve 3 + and check the answer using the digit sums 3 = digit sum of 3 is +3 = 5 + = digit sum of is + = 3 44 = digit sum of 44 is 4+4 = 8 If the sum has been done correctly, the digit sum of the question and answer should also be same, i. e. 8.

25 Digit sum of 44 = 8, so according to this check method, the answer is correct. There are 4 steps to use digit sum to check the answers: Do the question.. Write down the digit sums of the numbers in the question. 3. Add the digit sum 4. Check whether the two answers are same in digit sums. If the digit sums are same, then the answer is correct. Add 78 and 9 and check the answer We get 397 for the answer.. We find the digit sum of 78 and 9 which are 8 & respectively. 3. Adding 8 and gives 0, digits sum of 0 = +0 = 4. Digit sum of 397 is = 9 +9 = 0 = +0 = This confirms that the answer is correct. CAUTION: Check the following sum: =8 +8= = =3 +3=4 3 Here, an estimation can help you, to find the result more accurate. If by mistake, you write 400 in place of 490, then also it will show that the result is correct. The check is 9+4 = 3 = 4, which is same as the digit sum of the answer which confirms the answer. However, if we check the addition of the original number, we will find that it is correct. This shows that the digit sum does not always find errors. It usually works but not always. Note : The difference of 9 and its multiples in the answer make errors. So, keep in mind a rough estimation.

26 Digit sum puzzles The digit sums of a two digit number is 8 and figures are the same, what is the number?. The digit sums of a two digit number is 9 and the first figure is twice the second, what is it? 3. Give three digit numbers that have a digit sum of A two digit number has a digit sum of 5 and the figures are the same. What is the number? 5. Use casting out 9s to find the digit sum of the following numbers: Numbers Add the following and check your answer using digit sum check: () = () = (3) = (4) = (5) = (6) = (7) = (8) = 3

27 Sub Sutra applied is Ânurûpye?a Chapter-4 Multiplication Application of sutra is very easy, we just have to add digits in pairs. Only addition is required and no need to learn the table of We call this rule as Naught Sandwich. Naught means zero. I) Step- Make sandwich of multiplicand between single zeroe. Step x Add the numbers in pairs from right to left (0+4=4) (+4=5) x x (+3=4) (+3=5) x x (0+=) x Here s your answer 34 x =5454 ii) When sum of any two digits pair of multiplicand is a two digit number. Step & step are same as above but when the sum of two digits of multiplicand add up to two digits, then carry the ten's place number and add to the next digit (0+4=4) (+4=6) x x (8+=0) (8++=) x x (0++=3) x Here s your answer 84 x =

28 For two digit number Example : 43 x Short cut method for two digit number 43 x Write ones place of digit on ones place. Tens place digit on hundreds place x x Now add the two digit of multiplicand & write the sum of digits at tens place 43 (4 +3 = 7) x 473 So, 43 x = 473 b. Multiplication by to 99 ( Multiples of ) Multiples mean numbers appearing in table of Step- Find out the rank of multiple of (eg. is nd multiple or 88 is 8th multiple) and multiply the multiplicand with the number of rank, then apply the rule for Example- 45 x 45 (45x=90) (0+0 = 0) x x 45x = (9+0 = 9) (0+9 = 9) x x Ans. 990 Example 45 x (45x3=35) (5+0 = 0) x 33 x 45x 3 = (3+5 = 8) (+3 = 4) x x (0+ = ) x 485 Ans

29 Practice Sheet 45 x = 8. 9 x = x =. 99 x = x = x = x = x = x = x = 33 x = x = 5. 8 x =. 86 x = x = 6. 7 x = x = x = x = x = 4 65 x = x = x = x = 9. 8 x = x = x = x = x = x = 39 x = x = x = 48 x = x = x = 3. 3 x = x = x = x = 3 68 x = x = x = x = x = 6. 6 x = x = x = 7. 8 x = x = 35 x = x 33 = x 77 =. 68 x 44 = x 66 = x 88 = 3. 5 x 55 = x 77 = x = x 77 = 94 x = x 66 = 5. 4 x 33 =. 67 x 88 = x 77 = x 99 = 3. 9 x 55 = x 66 = 7. 6 x 66 = x 99 = 4 3 x 77 = x 77 = x 77 = x 44 = x 44 = x 44 = x 88 = x 88 = x 66 = x 99 = 5 x = 8. 4 x 88 = x 55 = 65 x 44 = x = x 44 = x 55 = x 33 = x 33 = 4. 8 x 99 = 3 53 x 55 = x 77 = x 66 = x 66 = x = 6. 4 x 44 = x 44 = x 55 = x 55 = x = 6

30 Sutra applied is Sopântyadvayamantya? Using the sutra which says that the ultimate and twice the penultimate, we can solve the multiplication by Using same pattern, we can solve the multiplication by of 3, Step- Sandwich the multiplicand between single zero x Step- Multiply every number of multiplicand by from right to left and add it to the number following it (x5+0=0) (x3=6+5=+=) x x (x =4+3=7+=8) (x=+=4) x x (0x=0 +=) x x x Here s your answer 35 x = 480 Note : If the answer after product and addition, in any step is digit number, then 0s place digit is carry forward to next higher place digit For any number between to 9 same procedure will apply: In step multiplication of number will change. The given multiplicand would be multiplied by 3 in case of 3 ; 4 is case of 4 and so on till 9. eg. 35x x5+0= (5x3=5+5=0) x5 x5 (0+) (5x=0+3=3) (5x=5+=7) x x x x5 (3+) x5 (7+) x x x x x (5x0=0+=) x Here s your answer 35 x 5 =855 7

31 Practice Sheet 3 x = x 5 = x 5 =. 3 x 3 = 8 x 6 = x 7 = 3. 5 x 4 =. 38 x 7 = 4 78 x 6 = x 5 = x 8 = x 9 = x 6= x 9 = x 8 = x 7 = x = x = 7. 9 x 8 = 6. 5 x 4 = x 3 = x 9 = x 3 = x 4 = x = x 5 = x = x 3 = x 6 = x 6 = 74 x4 = x 7 = x 8 = 83 x 5 = 3 65 x 8 = x 3 = 3. 9 x 6 = x 9 = x 7 = x 6 = x 8 = x = x 9 = x 3 = 7. 8 x = x 5 = x 3 = x 7 = x 4 = x 4 = 8

32 Concept : Double Naught Sandwich Step-I Sandwich the mulitlplicand between two zeroes 348 x = x Step-II Add three digits in groups from right to left (0+0+8=8) (0+8+4=) x x (8+4+3=5) (+3+4=8) x (5+=6 x (8+) (0++3=4) (0+0+=) x x Here s your answer 348 x = 4968 Step- Find out the rank of multiple of and multiply it with the multiplicand (eg. is nd multiple & 444 is 4th multiple) x = 696 x Step- Now apply the concept of on it. Add three digits of multiplicand in groups (0+0+6=6) (0+6+9=5) x x (6+9+6=) (9+6+=7) x (+=) x (7+-9) (6++0=8) (0+0+=) x (8+) x Here s your answer 348 x =

33 Practice Sheet 4 x = x = x =. 9 x = x = x = x = x = x = 4. 8 x = 43 x = x = x =. 86 x = x = 6. 7 x = x = x = x = x = x = x = x = x = x = x = x = x = x = x = 34 x = x = x = 68 x = x = x = 3. x = x = x = x = x = x = x = x = x = 6. 6 x = x = x = x = x = 37 x = x 333 = x 777 =. 67 x 444 = x 666 = x 888 = 3. 8 x 555 = x 777 = x = x 777 = 94 x = x 666 = x 333 =. 67 x 888 = x 777 = x 999 = 3. 9 x 555 = x 666 = 7. 6 x 666 = x 999 = 4 3 x 777 = x 777 = x 777 = x 444 = x 444 = x 444 = x 888 = x 888 = x 666 = x 999 = 5 x = 8. 4 x 888 = x 555 = 65 x 444 = x = x 444 = x 555 = x 333 = x 333 = 4. 8 x 999 = 3 53 x 555 = x 777 = x 666 = x 666 = x = 6. 4 x 444 = x 444 = x 555 = x 555 = x = 30

34 Sutra applied is Nikhila? Navataœcarama? Daœata? and Sub Sutra Ânurûpye?a A Base is a number followed by zeros (0) eg: 0, 0, 00, 500, 000, 0000 A) Below Base-0 Step- Write down the number with the negative sign 9 - ( below base i.e. deficiency) x 9 - ( less than 0) Step- Multiply the deficiencies and write the product at one s place 9 - (- x -=) x Step-3 Add the numbers diagonally or cross addition and write the sum on ten s place (both addition will give same answer.) 9-9+ (-) =8 x Here s your answer 9x9= In case If the number obtained in step is double digit then we carry forward the subscript and add it to next place digit in next step after cross addition. 7-3 (-3 x -4= ) x Cross addition. 7-3 [7+(-4)+ =4] x6-4 [6+(-3)+=4] Here s your answer 7 x 6= B) Below Base 0-90 Step Write down the numbers with negative sign after subtracting the given number from 0,30, 40 & till 90 according to the base Add the numbers diagonally and write the answer on tens and hundreds places. eg. 9 - ( below base 0 i.e. deficiency) x8 - ( below base 0 i.e. deficiency)

35 Step Multiply the deficiencies and write the product on one s place. eg. 9 - (- x - =) x Step 3 : Add the number diagonally or cross addition (both additions will give the same number). Now multiply the sum with the first digit of base number i.e. by for 0 base, by 3 for 30 base by 4 for 40 base and so on. eg. 9 - [{9+ (-)} x ] x 8 - [{8+ (-)} x ] (=34) 34 Here s your answer 9 x 8 = 34 In case of step if product obtained is digit number then write ten s place digit as subscript and later add it, in step 3 after cross addition. eg: 7 x 6 Step- 7-3 Step [(7 + (-4)] x 6-4 x6-4 = (7-4) x = 3 x = 6 7 = 6+ = Here s your answer 7 x 6= 7 C) Below Base-00 Step- Write down the deficiency with -ve sign (0 below base 00 i.e. deficiency) x 98-0 (0 below base 00 i.e. deficiency) Step- Multiply the deficiencies and write down the product on one s and ten s place x Step-3 Add the number diagonally or cross addition and write the sum on hundred s and thousand s places (Both additions will give same result) 99 - [99+(-)= 97] x 98 - [98+ (-)=97] Here s your answer 99 x 98= 970 3

36 In case the product obtained in step is three digit number then write the hundred s place number as subscript.. After addition in step 3, add the number written as subscript to obtain the answer. eg. 90 x 98 Step Step i) -0 x(-) =0 X 98 - X 98 - ii) 90+(-) = [78+= 79] Here s your answer 99 x 98= 790 D) Below Base Step- Write down the deficiency with -ve sign ( ) X x ( ) Step- Multiply the deficiencies numbers and write the product on one s & ten s place as base is of (-x-5)= (-4 x-6) =4 X x Step-3 Add the numbers diagonally or cross addition. Multiply the sum with base highest place digits for 00, 3 for 300 etc and write the answer on hundred thousand and ten thousand place [98+ (-5) =93] [596+ (-6)=590] X (93 x 3) Base of 300 X (590 x 6)B ase of Here s your answer 98 X 95 = X 594 = In Case in step if the product is 3 digit number then write the hundred s place digit as subscript and after performing step-3 add the number. Now write it on hundreds, thousand and & ten thousands place. eg. 90 x 89 Step Step (I) [( -0) x (-) = 0)] x89 - x89 - (ii) 89+(-0) = (iii) 79x3 = (iv) = Here s your answer 90 x 89 =

37 E) Above Base 0 Step Write down the surplus with + ve sign after eg. : + ( above base 0 i.e. surplus) x + ( above base 0 i.e. surplus) Step- Multiply the surplus numbers and write the product on one s place. + ( x ) x Step-3 Add the numbers diagonally or cross addition and write the answer on tens and hundreds place. + + = 3 x + + = Here s your answer x = 3 In case number obtained in step is two digit number then write the ten s place digit as subscript. After performing step 3 add the subscript digit and write the answer on tens and hundreds place. eg. : 4 x 4 Step Step (I) 4x4 = x4 + 4 (ii) 4+4 = iii) 8+ = Here s your answer 4 x 4 = 96 F) Above Base 0-90 Step- Write down the surplus with +ve sign. eg: + (I above base 0 i.e. surplus ) 4 + x 3 +3 (3 above base 0 i.e. 3 surplus ) x Step- Multiply the surplus numbers and write down the product at one s place. eg: + (X 3=3 ) 4 + (X =4) x 3 +3 x Step-3 Add the number diagonally or cross addition, then multiply the sum by bases (, 3, 4 etc) and write the answer at tens and hundreds place. + (3+=4) 4 + (4+=44) x 3 +3 (4x=48) X 4 + (44x4=76) Here s your answer (a) x 3 = 483 (b) 4 x 44 =

38 In case the product obtained in step is two digit number then write the ten s place number as subscript. After Step 3 Add the subscript digit to answer and write then on ten s & hundreds place. eg. 5 x (i) 5x4 = 0 x 4 +4 (ii) 5 +4 = (iii) 9 x = (iv) 58 + = Here s your answer 5 x 4 = 600 G) Above Base 00 Step- Write down the surplus [in two digit] with +ve sign. eg: 0 +0 (I above base 00 i.e. surplus ) x (3 above base 00 i.e. 3 surplus ) Step- Multiply the surplus numbers and write down the product at one s and ten s place as base is of (0x03= 03) x Step-3 Add the numbers diagonally cross addition and write down answer at hundred, thousand and tens thousand place (0+3 = 04) x Here s your answer 0 x 03 = 0403 In case the product obtained in step is three digit number then write the hundred place digit as subscript and after obtaining answer in step 3 add that subscript digit. eg. 0 x (I) 0 x = 0 x + (ii) 0+ = (iii) + = 0 Here s your answer 0 x = 0 35

39 H) Above Base Step- Write down the surplus with +ve sign. eg (05-00=05) ( =07) x (06-00=06) x ( =08) Step- Multiply the surplus numbers and write the product on ones and tens place (5x6 =30) (7x8 =56) x x Step-3 Add the number diagonally or cross addition and multiply the sum with respective bases (,3,4,...9) (i) 05 +6= (i) (507+8=55) x (ii) x= 4 x (ii) (55x5=575) Here s your answer 05 x 06 = x 508 = In case the answer obtained in step- is 3 digit number then add hundred s place digit to answer obtained in step-3. eg (i) 0 x =0 x 3 + (ii) 30 + = (iii) 3 x 3 = (iv) = Here is your answer 30 x 3 = 9670 In case of Base above or below 000 same procedure will follow except after multiplication in step answer will be written at three places since in thousand three zeros are there. Similarly for bases above or below 0,000 answer of multiplication will be written at four places. eg. : i) (-00) x (-00) = 00 x ii) (-00) = i) 005 x 0 = 50 x ii) =

40 Step- Write the number with the + ve sign which is surplus and write the number with -ve sign which is deficit. eg. : (05 above base 00 then 00 i.e. 5 surplus ) x95-05 ((05 below base 00 then 00 i.e. 5 deficiency) ) Step- Multiply the surplus and deficit. Since the result of multiplication is -ve number. Write the product using vinculum eg.: [(+5) x (-5) = 5)] x as base is 00 so we write product at two places. Step-3 Add the number diagonally or cross addition and write the answer at ones & tens places. eg.: (05+(-05) = 00) x95-05 ( =00) Step-3 Devinculate the number 005 (by using all from 9 and last from 0) = 9975 Here is your answer 05 x 95 = 9975 Example : 03 x 98 Step : Here, Base is (03 above base i.e. 3 surplus) (i) 03 x -0 = (0 below base i.e. deficiency) (ii) 03 + (-0) = Now Devinculate 006 =

41 eg : 04 x Step Vertical Multiplication (multiplication of surplus) for 04, base is 00 & surplus is 4 for, base is 0 & surplus is The digits in the lower base is one, so digits permitted in second part of answer [i.e. product of surplus] is only one x Step : Do cross - addition [only of higher base number and surplus of lower base number]. 04 Checking : +0 (one zero is attached with surplus as number of digits in higher base is one more) 4 (again take care to line the numbers properly, so as to get 4 i.e. First part of answer) X = + 4 = 5, = + = LHS = 5 x = 0 = + 0 = RHS = 44 = = 0 = +0 = : LHS = RHS, so answer is correct eg 0006 x 00 Now doing the question directly = (base 0000) x = (base 000) 0 (Three digits are permitted as per our method for second part) For first part of the answer we will do cross addition So the required Answer to question 0006x 00= Checking = 0006 = + 6 = 7, 00 = + = 3 LHS = 7 x 3 = = + = 3 RHS = 0060 = = = + = 3 AS LHS = RHS so answer is correct 38

42 eg. 97 x Here base is 00 and deficiency is Base is 0 and deficiency is Step : Multiply the deficiencies and write the answer in ones place as the number of digits permitted after multiplication is equal to number of digits in the lower base Step : Cross subtraction from higher base number The subtraction is done after placing the number of extra zeros which are in higher base deficiency starting from ones place, while writing subtrahend. Here in this case becomes 0 after writing extra zero so the answer to question 97 x 9= 873 eg. 998 x Here base is 000 and deficiency is Here base is 00 and deficiency is x Step: Multiply the deficiencies and write the answer on ones and tens place as number of digits will be equal to the lower base digits x Step : Cross subtraction will be done from higher base number after placing a zeroes equal to the higher base = x / 04 so answer to question 998 x 98 = Checking : - (through the check method) 99 =, 98 = 8 LHS = x8 = 6 =+6 = 7 RHS = 976 = +6 = 7 As LHS = RHS, so answer is correct 39

43 Practice Sheet + Below Base X 6= 9 X 5= 6 X 5= 8 X 8= 9 X 4= 6. 8 X 3= 7. 7 X = 8. 8 X 6= 9. 9 X = 0. 7 X 7= 8 X 6= 9 X 4= 3. 8 X 3= 4. 7 X 5= 5. 5 X 4= 6. 6 X 3= Below Base X 6= 9 X 5= 6 X 5= 8 X 8= 9 X 4= 8 X 3= 7 X = 8 X 6= 9 X = 7 X 7= 8 X 6= 9 X 4= 8 X 3= 7 X 5= 5 X 4= 6 X 3= Below Base X 3= 8 X 5= 7 X 6= 8 X 8= 9 X 4= 8 X 3= 7 X = 8 X 6= 9 X = 7 X 7= 8 X 6= 6 X 4= 3. 8 X 3= 4. 7 X 5= 5. 5 X 4= 6. 6 X 3= Below Base -40 Below Base -50 Below Base X 3= 38 X 35= 37 X 34= 38 X 38= 39 X 34= 38 X 33= 37 X 3= 38 X 36= 39 X 3= 37 X 37= 38 X 36= 36 X 36= 38 X 37= 37 X 36= 35 X 3= 36 X 3= X 43= 48 X 45= 47 X 44= 48 X 43= 49 X 4= 45 X 43= 47 X 4= 48 X 46= 49 X 4= 47 X 47= 48 X 46= 46 X 46= 48 X 47= 47 X 46= 45 X 4= 46 X 4= X 56= 59 X 58= 58 X 58= 53 X 5= 57 X 55= 56 X 54= 58 X 5= 54 X 5= 58 X 5= 59 X 58= 54 X 54= 53 X 5= 57 X 56= 56 X 56= 59 X 5= 57 X 53= 40

44 Practice Sheet Below Base X 66= 69 X 68= 68 X 63= 66 X 6= 67 X 66= 67 X 64= 68 X 6= 64 X 6= 68 X 68= 67 X 6= 6 X 6= 65 X 64= 67 X 66= 66 X 6= 69 X 67= 69 X 63= Below Base -00 Below Base X 77= 79 X 7= 78 X 75= 77 X 74= 76 X 77= 75 X 74= 77 X 7= 74 X 7= 78 X 73= 77 X 7= 7 X 7= 76 X 73= 79 X 77= 74 X 73= 79 X 78= 79 X 73= Below Base -00 Below Base X 84= 88 X 8= 84 X 8= 89 X 83= 88 X 8= 8 X 8= 86 X 83= 89 X 88= 84 X 83= 89 X 89= 89 X 83= 89 X 88= 89 X 8= 89 X 85= 88 X 84= 86 X 88= Below Base X 96= 99 X 95= 96 X 95= 98 X 98= 99 X 94= 98 X 93= 97 X 9= 98 X 96= 99 X 9= 97 X 97= 98 X 96= 99 X 94= 98 X 93= 97 X 95= 95 X 94= 96 X 93= X 96= 96 X 95= 95 X 95= 98 X 98= 95 X 94= 98 X 93= 97 X 9= 98 X 96= 99 X 9= 97 X 97= 98 X 96= 97 X 94= 98 X 93= 97 X 95= 95 X 94= 96 X 93= X 96= 96 X 85= 96 X 9= 98 X 98= 95 X 94= 98 X 93= 97 X 9= 98 X 96= 99 X 9= 97 X 97= 98 X 96= 97 X 94= 98 X 93= 97 X 95= 89 X 94= 96 X 93= 4

45 Practice Sheet Below Base X 396= 396 X 385= 396 X 39= 398 X 398= 395 X 394= 398 X 393= 397 X 393= 398 X 396= 399 X 393= 397 X 397= 398 X 384= 397 X 379= 398 X 39= 399 X 395= 397 X 394= 396 X 393= Below Base-700 Below Base X 49= 499 X 495= 497 X 494= 496 X 494= 496 X 49= 498 X 498= 495 X 494= 499 X 494= 497 X 497= 498 X 484= 497 X 479= 498 X 494= 497 X 494= 498 X 496= 487 X 496= 496 X 485= Below Base -800 Below Base X 59= 599 X 595= 597 X 595= 597 X 595= 596 X 590= 598 X 59= 595 X 595= 59 X 58= 597 X 597= 598 X 585= 597 X 579= 598 X 595= 597 X 595= 598 X 596= 587 X 596= 596 X 585= Below Base X 69= 699 X 696= 697 X 696= 697 X 696= 696 X 690= 698 X 69= 696 X 696= 69 X 68= 697 X 697= 698 X 686= 697 X 679= 698 X 696= 697 X 696= 698 X 696= 687 X 696= 696 X 686= X 79= 799 X 797= 797 X 797= 797 X 799= 797 X 790= 798 X 79= 797 X 798= 79 X 78= 797 X 797= 798 X 787= 797 X 779= 798 X 797= 79 X 79= 798 X 797= 787 X 790= 797 X 787= X 89= 899 X 898= 898 X 898= 898 X 895= 898 X 890= 898 X 89= 895 X 896= 89 X 88= 89 X 895= 898 X 888= 898 X 889= 896 X 89= 893 X 893= 894 X 895= 888 X 898= 898 X 888= 4

46 Practice Sheet Above Base X 3= 5 X 4= 6 X 5= 9 X 6= 8 X 8= 3 X x3= X 9= 4 X = X = X 4= 5 X 8= 7 X 7= X 6= 3 X 6= 4 X 6= 5 X 9= Above Base -40 Above Base X = X 4= 5 X 8= 7 X 7= X 6= 3 X 5= X 9= 4 X = 3 X = 4 X 6= 5 X 9= 9 X 6= 8 X 8= X 3= 5 X 4= 6 X 5= Above Base -50 Above Base X 38= 37 X 37= 39 X 36= 38 X 38= 3 X 36= 3 X 3= 3 X 34= 34 X 3= 33 X 38= 34 X 36= 35 X 39= 35 X 34= 36 X 35= 3 X 33= 33 X 33= 3 X 39= Above Base X 43= 44 X 46= 45 X 48= 47 X 47= 4 X 43= 43 X 49= 46 X 46= 4 X 4= 4 X 49= 48 X 48= 45 X 49= 45 X 44= 46 X 45= 4 X 46= 44 X 4= 49 X 46= X 54= 55 X 5= 59 X 56= 57 X 57= 5 X 53= 5 X 59= 58 X 58= 5 X 5= 53 X 53= 55 X 56= 54 X 59= 55 X 55= 56 X 55= 5 X 56= 56 X 56= 55 X 58= X 64= 66 X 6= 69 X 66= 67 X 67= 65 X 63= 6 X 69= 68 X 68= 6 X 6= 63 X 63= 66 X 66= 64 X 69= 67 X 68= 6 X 64= 6 X 66= 67 X 65= 6 X 68= 43

47 Practice Sheet Above Base X 77= 77 X 75= 7 X 78= 77 X 77= 73 X 73= 77 X 76= 74 X 79= 7 X 7= 73 X 74= 77 X 7= 79 X 76= 76 X 78= 7 X 74= 75 X 73= 7 X 79= 78 X 78= Above Base X 07= 08 X 05= 0 X = 0 X 0= 4 X 0= 07 X 06= 04 X 06= 08 X 0= 03 X 04= 6 X 0= 0 X 08= 06 X 0= 0 X 04= 05 X 03= 0 X 05= X 05= Above Base X 87=. 88 X 85= 3. 8 X 88= X 8= X 83= X 86= X 89= 8. 8 X 8= X 84= X 8= 89 X 86= 86 X 88= X 84= 85 X 83= 8 X 89= 88 X 88= Above Base X 06= 08 X 0= 03 X 04= 36 X 0= 5 X 03= 07 X 06= 0 X 08= 06 X = 0 X 04= 0 X 7= 0 X 05= 05 X 03= 07 X 05= X 09= 0 X = 0 X 0= Above Base X 97= 98 X 95= 9 X 99= 99 X 9= 94 X 93= 97 X 96= 94 X 99= 98 X 9= 93 X 94= 97 X 9= 99 X 98= 96 X 99= 9 X 94= 95 X 93= 9 X 95= 99 X 99= Above Base X 303= 34 X 303= 308 X 303= 303 X 304= 36 X 30= 307 X 306= 304 X 306= 30 X 307= 308 X 305= 36 X 30= 305 X 303= 306 X 303= 30 X 333= 306 X 309= 30 X 305= 3 X 305= 44

48 Practice Sheet Above Base X 404= 44 X 40= 407 X 406= 404 X 406= 446 X 40= 404 X 408= 409 X 404= 404 X 404= 403 X 404= 405 X 408= 40 X 405= 40 X 405= 40 X 407= 408 X 405= 40 X 4= 403 X 4= Above Base-700 Above Base X 505= 503 X 504= 556 X 50= 505 X 505= 54 X 50= 507 X 506= 504 X 506= 505 X 504= 505 X 503= 50 X 505= 505 X 508= 506 X 505= 50 X 507= 508 X 505= 505 X 5= 55 X 505= Above Base -800 Above Base X 65= 603 X 604= 606 X 605= 609 X 604= 64 X 60= 607 X 606= 60 X 607= 64 X 606= 604 X 606= 66 X 60= 66 X 603= 6 X 606= 64 X 607= 605 X 63= 60 X 605= 6 X 60= Above Bas X 777= 707 X 707= 704 X 706= 708 X 707= 74 X 70= 707 X 706= 70 X 707= 708 X 705= 703 X 704= 776 X 70= 705 X 703= 70 X 705= 707 X 704= 707 X 708= 706 X 707= 7 X 705= X 808= 809 X 804= 807 X 806= 804 X 806= 80 X 87= 80 X 807= 804 X 805= 808 X 808= 83 X 804= 86 X 80= 808 X 808= 80 X 8= 804 X 808= 805 X 803= 809 X 805= 8 X 80= X 90= 907 X 906= 909 X 999= 909 X 909= 903 X 904= 96 X 90= 904 X 906= 908 X 909= 90 X 907= 909 X 904= 905 X 903= 906 X 909= 908 X 905= 909 X 908= 90 X 905= 9 X 905= 45

49 Practice Sheet WHERE ONE NUMBER IS ABOVE BASE AND ANOTHER NUMBER IS BELOW BASE NUMBERS NEAR DIFFERENT BASE BOTH THE NUMBERS ARE ABOVE BASE NUMBERS NEAR DIFFERENT BASE BOTH NUMBERS ARE BELOW BASE X 9= 04 X = 97 X 9= X 8= 05 X = 98 X 8= 3 5 X 7= 3 08 X 4= 3 96 X 7= 4 9 X 9= X 3= X 9= 5 6 X 6= X 5= X 7= 6 X 9= X = X 8= 7 3 X 8= 7 00 X 0= X 98= 8 5 X 49= X 0= X 97= 9 59 X 6= X 05= X 95= 0 89 X 9= X 0= X 9= 08 X 95= 000 X = 9995 X 9= 0 X 99= 000 X 3= 9996 X 8= 3 08 X 9= X = X 7= 4 0 X 9= X 5= X 6= 5 03 X 95= X 4= X 99= 6 98 X 09= X 00= X 96= 7 0 X 99= X 005= X 94= X 96= X 00= X 97= X 399= X 00= X 98= 0 98 X 30= X 00= X 95= 699 X 50= 000 X 03= 9999 X 998= 005 X 995= 000 X 0= 9999 X 999= X 9998= X 0= X 995= X 9994= X 08= X 99= X 9999= X 05= X 99= 46

50 Sub sutra applied is Antyayordaœake'pi Step- Multiply the units digit and write the answer on one s and ten s place eg: 67 7x3= eg. 9 x 9 = 09 x63 x Here in this case product of & 9 is 9 so we write it as 09 Step- Multiply the tens place number with its successor eg. 67 6x7= 4 eg. 9 9x 0 = 90 x63 x Here are the answer 67 x 63 = 4 9 x 99 = 9009 Note : In case of triple digit number, when unit place add to 0 and tens and hundreds place digits are same, then tens and hundred s place number are together multiplied by its successor. eg. : 06 (i) 6 x 4 = 4 x04 (ii) 0 x = Here,s is the answer 06 x 04 = 004 Step Multiply the units digits and write the answer on one s and ten s place. eg. : (x = 04) eg. 77 (7x7 = 49) x8 x Step. Multiply the tens place digits and add units place digit to it and write the answer on hundred and thousand place. eg. : (8x +) eg. 77 7x3+7 = x8 6+ =8 x37 +7 = x 8 = x 37 =

51 Practice Sheet 67 x 63 = 8. 6 x 68 = x 67 =. 9 x 99 = x 36 = x 98 = 3. 8 x = x 73 = x 03 = x 47 = 93 x 97 = x 36 = x 65 =. x 9 = x 6 = x 4 = x 57 = x 06 = x 8 = x 75 = 4 x 8 = x 64 = x 8 = x 55 = x 3 = x 6 = x 6 = 0. 7 x 3 = x 74 = x 3 = 45 x 45 = 8. 4 x 6 = x 3 = 8 x 89 = 9. 4 x 48 = x 5= x 46 = x 3 = x 08 = 4. 5 x 58 = 3 5 x 59 = 48. x 9 = 5. 9 x = x 65 = x 5 = 6. 8 x 89 = x 7 = x 4 = x 46 = x 86 = 5 x 85 = 9. 5 x 5 = x 33 =. 9 x 99 = x 43 = x 74 = 3. 8 x = 5 x 5 = x 53 = x 33 =. 44 x 64 = x 44 = x 56 = 3. 3 x 7 = 4 55 x 55 = x 37 = 4. x 8 = x 65 = x 4 = 5. x 9 = 43. x 8 = x 75 = x 74 = 44. x 9 = x 65 = x 63 = x 76 = 0. 8 x = 8. 3 x 7 = x 6 = 86 x 6 = 9. 4 x 84 = x 67 = 89 x 9 = x 85 = x 75 = 3. 4 x 6 = 3 83 x 3 = x 95 = x 46 = 3. 4 x 6 = x 7 = x 55 = x 57 = 6. 6 x 4 = x 48 = x 36 = x 38 = 8. 7 x 3 = x 4 = 48

52 Step- Multiply the number by 0 eg. : 53 53x0 = 530 x Step- Subtract the multiplicand from the answer obtained in step = 477 x Here is your answer 53 x 9 = 477 Step- : Multiply the multiplicand by 0,30,... as the case may be. eg. : ( 53 x 0) ( 53x 50) x9 x Step- :Subtract the multiplicand from the answer obtained in step- eg. : 53 (060-53) 53 (650-53) x9 x x 9 = x 49 =

53 Practice Sheet 5 X 9 = X 9 = X 9 =. 5 X 9 = X 9 = X 39 = X 9 = X 39 = X 49 = X 9 = X 49 = X 59 = 5. 8 X 9 = 4 9 X 59 = X 69 = X 9 = X 69 = X 79 = X 9 = X 79 = X 89 = X 9 = X 89 = X 99 = X 9 = X 99 = 8 57 X 9 = X 9 = X 9 = X 9 = 6 X 9 = X 9 = 83. X 9 = 99 X 9 = 48. X 9 = X 39 = 3. 7 X 9 = 49. X 9 = X 49 = 4. 8 X 9 = X 39 = X 59 = X 9 = 5 5 X 9 = X 69 = X 9 = 5. 5 X 9 = X 79 = X 9 = X 9 = X 89 = X 39 = X 39 = X 99 = 9. X 9 = X 49 = 9 9 X 9 = X 9 = X 59 = X 9 = 6 X49 = X 69 = X 9 =. 37 X 59 = X 79 = X 39 = X 69 = X 89 = X 49 = 4. 5 X 79 = X 99 = X 59 = 5. 8 X 89 = 6 6 X 89 = X 69 = 6. 8 X 99 = X 79 = 98. X 79 = X 9 = X 69 = 99. X 89 = 8. 7 X 9 = X 59 = 00 0 X 99 = X 39 = X 49 = X 49 = X 39 = 3 57 X 59 = X 9 = X 69 = X 9 = 33. X 79 = 69. X 9 = X 89 = X 99 = X 99 = 7 6 X 9 = X 9 = X 9 = 50

54 Sutra applied is Ûrdhva Tiryagbhyâ? Step : Divide the answer block in three parts (Blocks are always less than total no. of digits in a question) Step : Multiply the two digits in right column and write the answer in last block i.e. units place digit with units place digit. Step 3: Cross multiply the unit place digit with the tens place digits to obtain two products. Add the two products and write the answer in the middle block. Step 4: Again vertically multiply the digits in left column i.e. tens place digit with tens place digit. and write the answer in left block. eg.: x = 54 ) In this first we multiply 4 x=4 and write the answer in the last block. ) We multiply cross wise 4 x=4, x= and add 4 and and write answer 5 in middle block. 3) Multiply vertically the tens place digits x= and write in the first block. 4) If each block has only one digit then these digits make the answer. 7 x Eg. 7 x 43 Same steps as explained in above example will be followed. If number of digit in each block is more than one, then the higher place digits will be carried and added in next block. 9 Ans. :

55 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 4 is + 4 = 6 Beejank of 648 is = 8, further + 8 = 9 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. : Find the Beejank of 6743 As above we have to follow But a quick look gives 6 & 3 ; &7 are to be ignored. [This method is called casting out 9s] because 6+3=9, +7=9. Hence remaining is the beejank of eg. : 4 x 3 4 x =768 Cross Checking 4 x 3 = 768 (+4) x (3+) = x 5 = 30 = = 3 3 = 3 LHS= RHS Hence the product is correct. This digit cross check method can be used in any kind of calculation to cross check your answer. This is also explained earlier. 5

56 Practice Sheet 54 X 7 = 6 X 34 = 4 98 X 59 =. 7 X 58 =. 37 X 9 = X 64 = X 89 = X 9 = X 87 = X 34 = 4. 5 X 63 = X 39 = 5. 8 X 46 = 5. 8 X 49 = X 4 = X 56 = 6. 8 X 87 = X 57 = X 69 = X 96 = X 63 = 8. 5 X 79 = 8. 7 X 56 = 48. X 78 = 9. 7 X 84 = X 89 = X 89 = 0. 8 X 9 = X 55 = X 59 = 6 X 8 = 3 57 X 58 = 94 X 73 = X 9 = 3. 7 X 56 = 33. X 96 = 4. X 59 = X 3 = X 48 = X 75 = X 35 = X 58 = X 8 = X 6 = X 9 = X 74 = 9. X 9 = X 8 = 0. 6 X 96 = X 98 = 53

57 I) Binomial x Binomial (x+3y) (3x+4y) Step : Step : Divide the answer block in three parts (Blocks are always less than total no. of digits in a question) Multiply the two digits in right column and write the answer in last block Step 3: Step 4: Cross multiply the first digit column with the second digit of second column, and second digit of the first column with the first digit of the second column. Then add the two products. Write the answer in second block. Again vertically multiply the digits in left column and write the answer in starting block. Step 5: The answer in each block will br written separately & like term will be added (if any). eg:- (3x+y) (4x+5y) 3x + y x4x + 5y x X 4x 3x X 5y = 5xy y x 5y =x 4x X y = 8xy = 0y =3xy Ans. x +3xy+0y II) Trinomial x Trinomial (x +x+5) (3x +x+) I) Step: In 3 digit * 3 digit multiplication, we make 5 blocks (Here there are 6 digits in all, so less than 6 is 5 blocks). Step: Step3: Vertically multiply the right hand side digits and write the answer in the first block from right. Cross multiply the digits of the two columns from right. Add both the products and write the answer in second block starting from right to left. 54

58 Step4: Cross multiply the digits in three columns i.e., multiply extreme right digits with extreme left of different blocks and multiply vertically the middle digits. Now add the three products and write the answer in third block. Step5 : Multiply the digits of two columns from the left. Add the two products and write the answer in the fourth block. Step6 : Vertically multiply the digits of the column of the left side and write the answer in th 5 block. Step7 : The answers in each block are written separately & like terms are added (if any). eg. (x +x+5) (3x +x+) x +x+5 3x +x+ 3 x +3x x Xx=4x x X= x x +x =x 5x 3 3x Xx= 6x 3x X 5= 5x x X x= 4x xx 5 = 0x 4 3 =6x =0x = x = x = =6x +0x +x +x+5 (7x -3x-) (3x+) (for Trinomial x Binomial multiplication put 0 in Binomial to make it trinomial) 7x -3x- 0+ 3x+ 7x X 3x 7x X = 4x -3x X - 0X-3x 0 X = 0 3x X x 3 =x -3x X 3x = -9x -6x+(-6x) -4 = -x 5x 3 Ans. x + 5x -x-4 55

59 Practice Sheet (x+5) (x-) (3x +x+5) (x +x-). (3p-) (9p+7). (3p+6q-) (9p-4q+7) 3. (x +) (x -) 3. (x +y -) (4x -3y +7) 4. (5k-4) (4k+6) 4. (7m-3n-4) (m+n+8) 5. (9m-4n) (m+n) 5. (a+3b-) (a-b+4) 6. (ab-3c) (7ab+c) (x +3x -7) (5x -5x -7) 7. (x-3y) (3x-9y) 7. (x -3x+5) (8x -9x+8) 8. (3a-7b) (5a+9b) 8. (3a-7b+7) (5a+9b-7) 9. (9y-4z) (7y+8z) 9. (x+9y-4z) (-5x+7y+8z) 0. (7x-y) (8x+3y) 30. (8x-5y+3) (5x+3y-) (x+6) (x+8) 3 (x +3x-5) (6x -9x-) (9p-6) (8p+4) 3. (p-3q+) (6p-7q-37) 3. (x +5) (x -5) 33. (x -7x+) (x -4) 4. (8x+9) (3x-4) 34. (7m+-4n-4) (6n+6) 5. (6m-7n) (m+n) 35. (9a-4b+3c) (7b+3c) 6. (8ab-9c) (4ab+c) 36. (7a -3b +3) (6b -4) 7. (5x-7y) (3x-4y) 37. (x-35y) (3x-y) 8. (a-9b) (6a+7b) 38. (7a -7b) (6a +9b) 9. (y-7z) (3y+4z) 39. (3x-5y-4z) (8x-9y+z) 0 (8x-3y) (4x+y) 40. (x-3y) (5x+y) 56

60 Chapter-5 Squares and Squares Roots For square of number ending with 5, same procedure discussed earlier, where the sum of the ones place digits is 0 and tens place digits are same is followed 5 = 65 Ist Multiply 5 with 5 to get 5 (second half) nd Multiply by 3 ( successor) to get 6 (first half) For square of number starting with 5, same procedure as discussed earlier where the sum of tens place digits is 0 and ones place digits are same is followed 5 = 704 Step Multiply by and write the answer as 04 [ x = 04] (second half) Step- Multiply 5 by 5 and add the common number [5 x 5 + =7] (first half) Above Base (0-90) I) If we are to find square of Determine the nearest base for (i. e. 0) and find its difference from the base i. e. +The answer comes in two parts. For first part, add the difference from base to the given number (+=). For second part, square the difference and put answer = +/ = /= 4 =4+4/4 = 8/6= 9/6=96 + In the above example, the square of 4, (6) is in double digit, so put as subscript and add it to 8 and put 6 in second part. II) For finding squares where base is between 0 to 90, the whole procedure is same. Lastly, we will multiply the respective base rank with the answer of first part.( for 0, 3 for for 90) e. g. 63 =6(63+3)/3 = 66x6/9=396/9=

61 Above Base (00-900) I) If we are to find square of 0 Same procedure for above base is applied. Only the placement in second part changes, answer 0 is written, as the base has two zeroes. 0 = 0+/ = 0/0=00 II) In case the base is between 00 to 900, the whole procedure remains same. Just multiply the respective base rank with the answer of first part.( for 00, 3 for for 900) e. g. 603 =6(603+3)/3 = 606x6/09=3636/09= Below Base (0-90) I) If we are to find square of 9. Same method as for above base is applied, the only difference is the deficiency of base is now subtracted from the number and answer is written in the first part. Now square the deficiency 9-0= - For second part, square the difference from of number and the base. i. e. - 9 = 9-/(-) = 8 II) In case where base is between 0 to 90, the whole procedure is same. Lastly, we will multiply the respective base rank with the answer of first part. ( for 0, 3 for for 90) e. g. 48 =5(48-)/(-) = 30/4= 304 (Here base will be 50) Below Base (00-900) I) If we are to find square of 98. Same procedure discussed above is applied. Now the placement in second part changes, as the base has two zeroes. 98 = 98-/(-) = 96/04=9604 II) In case the base is between 00 to 900, the whole procedure remains same. Just multiply the respective base rank with the answer of first part.( for 00, 3 for for 900) e. g. 399 =4(399-)/(-) = 398x4/0= 59/0=

62 Practice Sheet 5 = 50 =. 55 = 506 = 3. 5 = = = = = 5. 5 = = = = = = = = 9. 5 = = = 5 = 45 =. 5 = 55 = = = = = = = = 6. 5 = = = = = 9. 5 = 9. 5 = 0. 5 = = 59

63 The Duplex Sub Sutra: follows:. We will use the term Duplex, D' as For Single digit, its duplex will be its square, eg. D (4) = 4 = 6 For Double digit, its duplex will be twice of their product. For eg. D (35) = X3X5 = 30 For Three digit, D (34), duplex will be : 4 + X3X = = 46 Question: Find the square of 3 For finding the square, we will find the duplex: D(3) = 3 = 9 D(3) = x x 3 = D(3) = + x x 3 = 0 D() = x x = 4 D() = = So the answer becomes 59 Algebraic Squaring: Above method is applicable for squaring algebraic expressions also:- Example: (x+5) D(5) = 5 = 5 D(x, 5) = X x X 5 = 0x D(x) = x So square of x+5 is x + 0x +5 Example: (x 3y) D(-3y) = (-3y) = 9y D(x, -3y) = X x X -3y = -6xy D(x) = x So square of x-3y is x - 6xy +9y 60

64 Practice Sheet BASE-0 BASE-60 BASE-00 BASE = 63 = 04 = 70 =. 7 =. 67 =. =. 753 = BASE-0 BASE-70 BASE-300 BASE = 73 = 303 = 84 =. 5 =. 75 =. 33 =. 803 = BASE-30 BASE-80 BASE-400 BASE = 8 = 4 = 946 =. 39 =. 84 =. 46 =. 986 = BASE-40 BASE-90 BASE = 9 = 54 =. 48 =. 97 =. 55 = BASE-50 BASE-00 BASE = 0 = 663 =. 5 =. 09 =. 67 = BASE-0 BASE-60 BASE-00 BASE = 58 = 79 = 698 =. 7 =. 56 =. 9 =. 668 = BASE-0 BASE-70 BASE-300 BASE = 69 = = 779 =. 9 =. 6 =. 97 =. 795 = BASE-30 BASE-80 BASE-400 BASE = 76 = 359 = 875 =. 8 =. 79 =. 399 =. 89 = BASE-40 BASE-90 BASE = 85 = 445 =. 35 =. 89 =. 489 BASE-50 BASE-00 BASE = 94 = 545 =. 46 =. 98 =. 598 = 6

65 Practice Sheet = 5 = 4 (x+). 43 =. 36 = 4. (a+b) 3. 7 = 3. 3 = 43. (3x+4y) 4. 6 = 4. 4 = 44. (8c+4d) 5. 3 = 5. 5 = = 6. 6 = 7. 5 = 7. 5 = 47. (9c+4d) 8. 6 = 8. 7 = 48. (m+3) = = 0. 3 = = 49. (m+n) 6 = 3 36 = 50. (5z+8k) 7 = = = = = = 5. 4 = = = = 7. 9 = = 8. 4 = = = = = = (3x+y) (7a+3b) 6

66 General Method: As =, = 4, 3 = 9, 4 = (6), 5 = (5), 6 = 3(6), 7 = 4(9), 8 = 6(4), 9 = 8(), i. e., square numbers only have digits, 4, 5, 6, 9, 0 at the units place (or at the end) Also in 6, digit sum = +6 = 7, 5 = +5 = 7, 36 = 3+6 = 9, 49 = 4+9 = 3 = 4, 64 = 6+4 = 0 =, 8 = 8+ = 9, i. e. square number only have digit sums of, 4, 7 & 9. This means that square numbers cannot have certain digit sums and they cannot end with certain figures (or digits). Using above information state which of the following are not square numbers: () 4539 () 6889 (3) (4) 7478 (5) 345 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 is not a perfect square inspite of the fact that its digit sum is 4 and last figure is 9. Example : 584 Step : Pair the numbers in two, from right to left i. e Therefore, answer will come in digits Step : For the first digit, look where 5 lies in squares table, and consider Step 3: the lower number. Here, 5 lies in between 49 and 64. So, we will consider 7 at tens place, as 49 is square of 7. Look at the last digit of number, 4. 4 lies in = 04 and 8 = 64. So, the ones place digit is either or 8 We can say the answer is either 7 or 78. Step 4: Now find the square of 75 and compare with the number (given in Example : 96 the question). If the number is lesser, then consider the lower value and if number is more than the square, then consider the higher value. Here 584<565, therefore we will consider the lower value that is 7 Step : Pair the numbers in two, from right to left i. e. 9 6 Therefore, answer will come in digits Step : The first digit is 9, as 9 lies between 9 and 0, and we will take lower value, 9 as 8 is square of 9. Step 3: The last digit of number, 6. 6 lies in 4 = 6 and 6 = 36. So, the ones place digit is either 4 or 6 The answer is either 94 or 96. Step 4: Now, find the square of 95 and compare with the number (given in the question). Here 96>905, therefore we will consider the higher value that is 96 63

67 64 Practice Sheet

68 Chapter-6 Division For Division by 9 we need to know the table of 9 but in vedic mathematics we can easily calculate Quotient and Remainder with help of just addition Write first place number as it is then add the number diagonally to the second number and repeat this addition till the unit place of dividend you will get the desired answer. If sum is 3 9 it will be written as 9 3 (leaving unit place separately to show the remainder. a) The initial is brought straight down as the first digit of quotient b) This is then added to the 3 in 3 and 5 is next quotient digit. c) This 5 is then added to the in the 3 and 6 is next quotient digit a) b) c) d) This 6 is then added to to give the remainder, 7 Here the answer 3 9 = 56 (Q) & 7 Type-II (I) If the digit of dividend add upto two digit number then tens place number is carried and added to the next number eg. : d) (ii) (iii) If we add + 8 then result obtained is 0 then we will write tens place no. of 0 i.e. below as a subscript as shown above and second digit i.e. zero just along with. Now add 0 to the next number i.e = 4. Write as subscript and four in answer row. Now add 4 to next number 3. Write 7 after the slash as 3 is the last number of dividend (iv) Add the carry forward in the quotient ( +=3) (0+ =) 65