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GOVIND RAJU ☦✧ Vedam Classes✧☦ INTRODUCTION TO VEDIC MATHEMATICS The book on Vedic Mathematics written by Jagadguru Swami Sri Bharati Krisna Tirthaji Maharaja (Sankaracharya of Govardhana Matha, Puri, Orissa, India He was born in March 1884 to highly learned and pious parents. His father Sri P Narasimha Shastri was in service as a Tahsildar at Tinnivelly (Madras Presidency) and later retired as a Deputy Collector. His uncle, Sri Chandrasekhar Shastri was the principal of the Maharajas College, Vizianagaram and his great grandfather was Justice C. Ranganath Shastri of the Madras High Court. Born Venkatraman he grew up to be a brilliant student and invariably won the first place in all the subjects in all classes throughout his educational career. During his school days, he was a student of National College Trichanapalli; Church Missionary Society College, Tinnivelli and Hindu College Tinnivelly in Tamil Nadu. He passed his matriculation examination from the Madras University in 1899 topping the list as usual. His extraordinary proficiency in Sanskrit earned him the title “Saraswati” from the Madras Sanskrit Association in July 1899. After winning the highest place in the B.A examination Sri Venkataraman appeared for the M.A. examination of the American College of Sciences, Rochester, New York from the Bombay center in 1903. His subject of examination was Sanskrit, Philosophy, English, Mathematics, History and Science. He had a superb retentive memory. In 1911 he could not anymore resist his burning desire for spiritual knowledge, practice and attainment and therefore, tearing himself off suddenly from the work of teaching, he went back to Sri Satcidananda Sivabhinava Nrisimha Bharati Swami at Sringeri. He spent the next eight years in the profoundest study of the most advanced Vedanta Philosophy and practice of the Brahmasadhana. After several years in 1921 he was installed on the pontifical throne of Sharada Peetha Sankaracharya and later in 1925 he became the pontifical head of Sri Govardhan Math Puri where he served the remainder of his life spreading the holy spiritual teachings of Sanatana Dharma. In 1957, when he decided finally to undertake a tour of the USA he rewrote from his memory the present volume of Vedic Mathematics giving an introductory account of the sixteen formulae reconstructed by him What is Vedic Mathematics? The term Vedic Mathematics refers to the unique technique of Calculations based on a set of 16 sutras or aphorisms or formulae and their upa-sutras or corollaries derived from these sutras. Any mathematical problems (algebra, arithmetic, geometry or trigonometry) can be solved mentally with these sutras. Vedic Mathematics is more coherent than modern mathematics. The sixteen sutras and their corollaries are as follows: 1. Ekadhikina Purvena -By one more than the previous one (Cor: Anurupyena) 2. Nikhilam Navatashcaramam Dashatah -All from 9 and the last from 10. 3. Urdhva-Tiryagbyham-Vertically and crosswise. 4. Paraavartya Yojayet-Transpose and adjust. 5. Shunyam Saamyasamuccaye-When the sum is the same, that sum is zero. 1 GOVIND RAJU ☦✧ Vedam Classes✧☦ 6. (Anurupye) Shunyamanyat-If one is in ratio, the other is zero. 7. Sankalana-vyavakalanabhyam-By addition and by subtraction. 8. Puranapuranabyham-By the completion or non-completion. 9. Chalana-Kalanabyham-Differences and Similarities. 10. Yaavadunam-Whatever the extent of its deficiency. 11. Vyashtisamanstih-Part and Whole. 12. Shesanyankena Charamena-The remainders by the last digit. 13. Sopaantyadvayamantyam-The ultimate and twice the penultimate. 14. Ekanyunena Purvena-By one less than the previous one. 15. Gunitasamuchyah-The product of the sum is equal to the sum of the product 16. Gunakasamuchyah-The factors of the sum is equal to the sum of the factors Addition of numbers In vedic mathematics during addition the numbers does not exceed 18 as we add the numbers to more then 9 we put a dot and the remainder is carried forward. i.e. if we add 6 and 8 we get 14 so we put a dot on the digit and 4 is taken for further addition Example :- Lets add 36,57,84,39,24 . Add 36 . 57 36 84 57 . 39 84 ____24_____ 39 0 Now the dots are added first and the same ____24____ th process is repeated for the 10 place and the th If we start adding the unit place from down to up dots on the 10 place are add and the value is th we have 4 plus 9 equals 13. So put a dot on 9 placed on 100 place and 3 is taken for further addition with 4 gives 7 again added with seven to give 14 . so put a dot on 7 and 4 is taken away for further addition with 6 which comes to 10, so put a dot on 6 and 0 is written as the addition for the unit place Nikhilam Navatashcaramam Dashatah -All from 9 and the last from 10 The difference between the number and the base is termed as deviation. Deviation may be positive or negative. Positive deviation is written without the positive sign and the negative deviation, is written using Rekhank (a bar on the number). Now observe the following table. Number Base Number – Base Deficiency _ 8 10 8 – 10 -2or 2 97 100 97 – 100 -03 or 03 112 100 112 – 100 12 2 GOVIND RAJU ☦✧ Vedam Classes✧☦ Subtraction For subtraction the above sutra All from 9 and the last from 10 is of best use . For example subtract 1876 from 3241 So starting from the right 6 is more then 1 so we take the complement from 10 since it is the last digit i.e. 4 and add it to 1 that comes to 5 th Now for the tenth column, the 9 compliment of 7 (i.e 2) is added to 4 that comes to 6 th Now for the hundredth column, the 9 compliment of 8(i.e. 1) is added to 2 that comes to 3 To end with the subtraction 3 is subtracted form 1 the comes to 2 and is further reduced by 1 (i.e. 3 – 1 – 1 = 1 ) that comes out to 1 3241 1876 1365 Vinculum Vinculum means deficiency. In vedic mathematics vinculum has a greater role. As vedic maths. Says that any digit/number can be expressed as number containing less then or equal to 5. i.e. 7 can be expressed as 10-3 or 1 3 and it is read as 1 vinculum 3. th As 1 is in the 10 place it means 10 with 3 deficiency which equals 7 hear we the nikhilam sutra i.e. all by nine and last by 10 and 1 is added to the extreme. Example : Express a given number in vinculum form 1) 97 = It can be written as 100 – 3 or 103 2) 378 = 422 3) 52862 = 53142 In short increase the first by 1 and use all by 9 and last by 10 Devinculate It is just a reverse of vinculum to bring back a number to normal form. As it is reverse we subtract 1 from the last and apply nikhilam sutra (all by 9 and last by 10 ) Example : Devinculate the following number. 1) 314 = 286 2) 532897231 = 527296771 3 GOVIND RAJU ☦✧ Vedam Classes✧☦ Multiplying a number by 11. To multiply any 2-figure number by 11 we just put the total of the two figures between the 2 figures. 1. 26 x 11 = 286 Notice that the outer figures in 286 are the 26 being multiplied.And the middle figure is just 2 and 6 added up. 2. So 72 x 11 = 792 3. 234 x 11 = 2574 We put the 2 and the 4 at the ends. We add the first pair 2 + 3 = 5. and we add the last pair: 3 + 4 = 7. MULTIPLICATION (by Nikhilam Sutra) Let us first take up a very easy and simple illustrative example for the numbers which are nearer to the base of 10,100,100 such as 8,9,11,12,98,97,103,105,997,988,1003,1006 Numbers less then near to base 10 Suppose we have to multiply 9 by 7. (10) base (i) We should take 10 as Base for our calculations , 9-1 7-3 6/3 (ii) Put the two numbers 9 and 7 above and below on the left hand side (as shown in the working alongside here on the right-hand side margin) ( iii )Subtract each of them from the base (10) and write down the remainders (1 and 3) on the right- hand side with a connecting minus sign (-) between them, to show that the numbers to be multiplied are both of them less than 10. (iv) The product will have two parts, one on the left side and one on the right. A vertical dividing line may be drawn for the purpose of discriminate between the two parts. ( v,) Now, the left-hand-side digit (or the answer) can be arrived at in either by cross-subtract 9 by 3 or 7 by 1 that comes to 6. The answer is 63 Numbers less then near to base 100 Suppose we have to multiply 94 by 97. base (100) (i) We should take 100 as Base for our calculations , 94-06 97-03 91/18 4 GOVIND RAJU ☦✧ Vedam Classes✧☦ Same step as illustrated above The answer is 9118 Numbers less then near to base 1000 Suppose we have to multiply 984 by 997. base (100) (i) We should take 1000 as Base for our calculations , 984-016 997-003 981/048 Same step as illustrated above The answer is 981048 Numbers more then near to base 10 Suppose we have to multiply 14 by 12. (10) base (i) We should take 10 as Base for our calculations , 14+4 12+2 16/8 (ii) Put the two numbers 14 and 12 above and below on the left hand side (as shown in the working alongside here on the right-hand side margin) ( iii )Subtract each of them from the base (10) and write down the remainders (4 and 2) on the right- hand side with a connecting plus sign (+) between them, to show that the numbers to be multiplied are both of them less than 10. (iv) The product will have two parts, one on the left side and one on the right. A vertical dividing line may be drawn for the purpose of discriminate between the two parts. ( v,) Now, the left-hand-side digit (or the answer) can be arrived at in either by cross-adding 14 with 2 or 12 with 4 that comes to 16. The answer is 168 Numbers more then near to base 100 Suppose we have to multiply 104 by 108. base (100) (i) We should take 100 as Base for our calculations , 104+04 108+08 112/32 Same step as illustrated above The answer is 11232 5 GOVIND RAJU ☦✧ Vedam Classes✧☦ Numbers near to base Suppose we have to multiply 994 by 1003. base (100) (i) We should take 1000 as Base for our calculations , 994-006 1003+003 997/018 Same step as illustrated above With the help of devinculum the answer comes to 996982 Numbers away from the base Suppose we have to multiply 42 by 44. Both these numbers are so far away from the base100 that by our adopting that we shall get 58 and 56 as the deficiency from as our actual base, the base. And thus the consequent vertical multiplication of 58 by 56 would prove too cumbrous a process to be permissible under the Vedic system and will be positively inadmissible. 100 merely as a theoretical base and We therefore, accept take sub-multiple 50 (which is conveniently near 42 and 46) as our working basis, work the sum up accordingly and then do the proportionate multiplication or division, for getting the correct answer. Our chart will then take this shape :- - (i) We take 50 as our working base. 100/ 2 = 50 (ii) By cross-subtraction, we get 36 on the left hand side. 42 -08 (iii) As 50 is a half of 100, we therefore divide 36 by 2 and put 18 44 -06 down as the real left-hand-side portion of the answer. 2 )36 / 48 (iv) The right-hand-side portion (48) remains unaffected. 18 / 48 (v) The answer therefore is 1848. 6 GOVIND RAJU ☦✧ Vedam Classes✧☦ MULTIPLICATION (Urdhva-Tiryak Sutra) The Criss-Cross system of multiplication. The traditional system of multiplication taught to students is universal, it is applicable to all types of numbers. In Vedic Mathematics a similar system is there, but it helps the student to get the result much faster. This system of multiplication is given by the Urdhva-Tiryak Sutra, which means „vertically and cross- wise‟. STEP 1: multiply the digits in the unit place. STEP 2: cross multiply and add the products (2x2)+(3x1). STEP 3: now multiply the remaining digits. This system is self consistent, in the sense that it is applied to all kind of multiplications, once the crossing technique have been understood. 7 GOVIND RAJU ☦✧ Vedam Classes✧☦ SQUAR OF NUMBERS Squaring numbers near a base. This method takes advantage of the sub-sutra whatever the deficiency lessen by that amount, and set up the square of the deficiency. Duplex of numbers. One of the byproducts of the criss-cross method, is a general rule for squaring numbers. This is obvious as squaring a number is nothing less than a multiplication of the number by itself. There is an extension and general way to square numbers, that make use of the duplex D(…) function. By definition the duplex of a one digit number is its square 2 2 2 D(3)=3 =9 D(5)=5 =25 D(7)=7 =49 The duplex of a two digit number is twice their product: D(32)=3x2x2=12 D(15)=1x5x2=10 D(36)=3x6x2=36 The duplex of a three digit number is twice the product of the outers digits plus the square of the middle one: 2 2 2 D(321)=3x1x2+2 =10 D(135)=1x5x2+3 =19 D(336)=3x6x2+3 =45 Squaring numbers. Squaring two digit numbers. In this case we have three duplex to calculate: 2 43 = D(4)=16 D(43)=24 D(3)=9 2 43 = 1 8 2 4 9 2 56 = D(5)=25 D(56)=60 D(6)=36 2 56 = 3 1 1 6 3 3 6 Squaring three digit numbers. In this case we have five duplex to calculate: 2 123 = D(1)=1 D(12)=4 D(123)=10 D(23)=12 D(3)=9 2 123 = 1 5 1 1 1 2 9 Square numbers only have digit sums of 1, 4, 7, 9 and they only end in 1, 4, 5, 6, 9 ,0 8 GOVIND RAJU ☦✧ Vedam Classes✧☦ DIVISION The format of vedic base division. We will use a diagram where the base and the difference from the base of the divided is in evidence. The base is the nearest multiple of 10 to the divisor. The number of digits in the RH side is equal to the zeros of the base. The base difference can be positive or negative, when the divisor is less than or exceeds the base, respectively Base division: Divide 617 by 95, take base as 100 and put the deficit 05 below 95 , draw the lines as stated above after 2 digit as 100 is our base, now bring the 6 down and then 6 multiplied by 05 gives 30 and place just below the next digit, and add to get the reminder, Base division greater then base: Divide 1296 by 113, take base as 100 and put the surplus - 13 below 113, draw the lines as stated above after 2 digit as 100 is our base, now bring the 1 down and then 1 multiplied by -1 and -3 gives -1and -3 and place just below the next digit, next to next digit and add the second column to get the quotient and repeat the cycle to get the remainder, x EXP 2 9 GOVIND RAJU ☦✧ Vedam Classes✧☦ Straight division. Straight division is the general division method by which any number of any size can be divided in one line. Sri Bharati Krishna Tirthaji, the man who rediscovered the Vedic system, called this the “crowning gem of Vedic Mathematics”. It comes under the Urdhva Sutra-vertically and cross-wise and the subsutra on the flag. We want to divide 209 by 52. THIS IS THE FLAG We put the 2 of 52 on the flag! 2 2 0 09 5 4 1 The steps are: 5 into 20 goes 4 remainder 0 4 multiplied by 2 gives 8, which taken from 9 gives the remainder 1. Example 2 We want to divide 321 by 63. 3 3 2 21 6 5 6 6 into 32 goes 5 remainder 2 5 multiplied by 3 gives 15, which taken from 21 gives the remainder 6. Decimalizing the remainder. Divide 40342 by 73 to 5 decimal place! 3 4 0 5 3 3 4 5 2401010306020 7 5 5 2 6 3 0 1 3 7 Nothing more simple, just add as many zero‟s you need after the number… and you‟ve got it! 10 GOVIND RAJU ☦✧ Vedam Classes✧☦ Larger divisors. Divide 2250255 by 721! 21 2 2 1 5 1 0 1 2 0 5 5 7 3 1 2 1 0 14 Dividing by 3 or 4 figure numbers is an easy extension of the same technique, which involves putting al but 1 or 2 figures on the flag! Since there are 2 figures on the flag, the decimal point goes two figures from the right, as shown. We start as usual dividing 22 by 7, we get 3 remainder 1 Next we multiply 3 by 2 (only the first figure of the flag), get 6, taken from 15 and divided by 7 gives 2 remainder 2. From now on we start cross multiplying two figure numbers, scaling one digit after the other. The first cross multiplication is: We take 5 from 20, get 15, divided by 7 gives 2 remainder 1 Next we cross multiply 21 by 12, take the result from 12, divide by 7, gives 1 remainder 0. The procedure become cyclic again, we go on cross multiplying, and subtracting as usual… 11 GOVIND RAJU ☦✧ Vedam Classes✧☦ General Square Roots. If we have a number which we want the square root, there are two things we can immediately get from the number: the number of digits before decimal point of the square root the first digit of the square root We will see the general squaring, as a square of a perfect square will be recognized when the remainder will be naught. Find the square root of 38. 2 There is clearly one figure before the point, and it is 6. The divisor will be 6x2=12. 6 =36, so there is a remainder of 2 from 38. Then 20:12=1 remainder 8. From now on, before dividing, we deduce the duplex of all the numbers after the first. 2 So 80-D(1)=80-1 =79, divided by 12 gives 6 remainder 7. Then 70-D(16)=70-(1x6x2)=58, divided by 12 gives 4 remainder 10. 2 Now 100-D(164)=100-(1x4x2+6 )=56, divided by 12 gives 4 remainder 8. Then 80-D(1644)=80-(1x4x2+6x4x2)=24, divided by 12 gives 2 remainder 0. Example 2 Find the square root of 293764. There are clearly three figures before the point, the first is 5, so the divisor will be 10. The first 2 remainder is 29-5 =4, so 43:10=4 rem. 3. Now again with the duplex‟s. 37-D(4)=21, divided by 10 gives 2 rem. 1, and we put down the decimal point. Then 16-D(42)=16-16=0, divided by 10 gives 0 remainder 0. The 4-D(420)=4-4=0, divided by 10 gives 0 remainder 0. From here on, all the duplex will be null, the remainder will be naught, so this is a perfect square! 12 GOVIND RAJU ☦✧ Vedam Classes✧☦ Negative difference When there is a negative difference, we treat it in the usual way, we reduce the previous figure and take a full remainder, or just put down a barred number, and then remove it later (I prefer the first option). The square root has a one figure number, it‟s 4, the divisor is 4x2=8 and the first remainder is 3. 32:8=4 rem. 0. You can see that the next step you will get a negative number, you can just say 0- D(4)=16, 16:8=2 remainder 0. But this will be too much cumbersome, as the duplex will have negative numbers too. So I prefer to take 1 from the previous, and say 32:8=3 rem. 8. So 80-D(3)=71:8=8 rem. 7. Now 70-D(38)=70-48=22:8=2 rem. 6. An then 60-D(382)=60-76…. Again we take one from the previous… So 140-D(381)=140-70=70:8=8 rem. 6, but…. And so on…. 13 GOVIND RAJU ☦✧ Vedam Classes✧☦ Instant calculations: Method for multiplying numbers where the first figures are the same and the last figures add up to 10. e.g. i)We multiply 3x4=12; ii)We multiply the last figures; Quick way to square numbers that end by 5 BY ONE MORE THAN THE ONE BEFORE. e.g. i)The first number (7)multiplied by the number”one more”,which is 8; ii)Is always25; Adding time A simple way to add hours and minutes together: Let's add 1 hr and 35 minutes and 3 hr 55 minutes together. i)135+355=490 ii) No matter what the hours and minutes are, just add the 40 time constant to the sub total. =>490+40=530 (finaly 5h.30min) 14