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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Sutras, EKĀDHIKENA PŪRVE•A, The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the previous, one”., i) Squares of numbers ending in 5 :, Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the, example 252., Here the number is 25. We have to find out the square of the number. For the, number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than, the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure, 'to multiply the previous digit 2 by one more than itself, that is, by 3'. It becomes, the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand, side) of the result is 52, that is, 25., Thus 252 = 2 X 3 / 25 = 625., In the same way,, 352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;, 652= 6 X 7 / 25 = 4225;, 1052= 10 X 11/25 = 11025;, 1352= 13 X 14/25 = 18225;, , Apply the formula to find the squares of the numbers 15, 45, 85, 125,, 175 and verify the answers., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (1 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Algebraic proof:, a), , Consider (ax + b)2 Ξ a2. x2 + 2abx + b2., This identity for x = 10, , and b = 5 becomes, , (10a + 5) 2 = a2 . 102 + 2. 10a . 5 + 52, = a2 . 102 + a. 102 + 52, = (a 2+ a ) . 102 + 52, = a (a + 1) . 10 2 + 25., Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the, values a = 1, 2, 3, -------,9 respectively. In such a case the number (10a + 5)2 is, of the form whose L.H.S is a (a + 1) and R.H.S is 25, that is, a (a + 1) / 25., Thus any such two digit number gives the result in the same fashion., Example:, , 45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10, and b = 5. giving the answer a (a+1) / 25, that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025., , b) Any three digit number is of the form ax2+bx+c for x = 10, a ≠ 0, a, b, c • W., Now (ax2+bx+ c) 2 = a2 x4 + b2x2 + c2 + 2abx3 + 2bcx + 2cax2, = a2 x4+2ab. x3+ (b2 + 2ca)x2+2bc . x+ c2., This identity for x = 10, c = 5 becomes (a . 102 + b .10 + 5) 2, = a2.104 + 2.a.b.103 + (b2 + 2.5.a)102 + 2.b.5.10 + 52, = a2.104 + 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52, 3, , = a2.104 + 2ab.10 + b2.102 + a . 103 + b 102 + 52, = a2.104 + (2ab + a).103 + (b2+ b)102 +52, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (2 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , = [ a2.102 + 2ab.10 + a.10 + b2 + b] 102+ 52, = (10a + b) ( 10a+b+1).102 + 25, = P (P+1) 102 + 25, where P = 10a+b., Hence any three digit number whose last digit is 5 gives the same result as in (a), for P=10a + b, the ‘previous’ of 5., Example :, , 1652 = (1 . 102 + 6 . 10 + 5) 2., , It is of the form (ax2 +bx+c)2 for a = 1, b = 6, c = 5 and x = 10. It gives the, answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The, answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225., , Apply Ekadhikena purvena to find the squares of the numbers 95,, 225, 375, 635, 745, 915, 1105, 2545., , ii) Vulgar fractions whose denominators are numbers ending in NINE :, We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion of, such vulgar fractions into recurring decimals, Ekadhika process can be effectively, used both in division and multiplication., a) Division Method : Value of 1 / 19., The numbers of decimal places before repetition is the difference of numerator, and denominator, i.e.,, 19 -1=18 places., For the denominator 19, the purva (previous) is 1., Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2., The sutra is applied in a different context. Now the method of division is as, follows:, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (3 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Step. 1 : Divide numerator 1 by 20., i.e.,, 1 / 20 = 0.1 / 2 = .10 ( 0 times, 1 remainder), Step. 2 : Divide 10 by 2, i.e.,, 0.005( 5 times, 0 remainder ), Step. 3 : Divide 5 by 2, i.e.,, 0.0512 ( 2 times, 1 remainder ), , Step. 4 : Divide, , 12, , i.e.,, 12 by 2, , i.e.,, 0.0526 ( 6 times, No remainder ), Step. 5 :, i.e.,,, Step. 6 :, , Divide 6 by 2, 0.05263 ( 3 times, No remainder ), Divide 3 by 2, , i.e.,, 0.0526311(1 time, 1 remainder ), Step. 7 :, , Divide, , 11, , i.e.,, 11 by 2, , i.e.,, 0.05263115 (5 times, 1 remainder ), Step. 8 :, , Divide, , 15, , i.e.,, 15 by 2, , i.e.,, 0.052631517 ( 7 times, 1 remainder ), Step. 9 :, i.e.,,, , Divide 17, , i.e.,,, , 17 by 2, , 0.05263157 18 (8 times, 1 remainder ), , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (4 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Step. 10 : Divide, , 18, , i.e.,, 18 by 2, , i.e.,, 0.0526315789 (9 times, No remainder ), Step. 11 : Divide 9 by 2, i.e.,, 0.0526315789 14 (4 times, 1 remainder ), Step. 12 : Divide, , 14, , i.e.,, 14 by 2, , i.e.,, 0.052631578947 ( 7 times, No remainder ), Step. 13 : Divide 7 by 2, i.e.,, 0.05263157894713 ( 3 times, 1 remainder ), Step. 14 : Divide, , 13, , i.e.,,, , 13 by 2, , i.e.,, 0.052631578947316 ( 6 times, 1 remainder ), Step. 15 : Divide, i.e.,,, , 16, , i.e.,,, , 16 by 2, , 0.052631578947368 (8 times, No remainder ), , Step. 16 : Divide 8 by 2, i.e.,, 0.0526315789473684 ( 4 times, No remainder ), Step. 17 : Divide 4 by 2, i.e.,, 0.05263157894736842 ( 2 times, No remainder ), Step. 18 : Divide 2 by 2, i.e.,, 0.052631578947368421 ( 1 time, No remainder ), Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (5 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 0 __________________, ., ., 1 / 19 = 0.052631578947368421 or 0.052631578947368421, Note that we have completed the process of division only by using ‘2’. Nowhere, the division by 19 occurs., b) Multiplication Method: Value of 1 / 19, First we recognize the last digit of the denominator of the type 1 / a9. Here the, last digit is 9., For a fraction of the form in whose denominator 9 is the last digit, we take the, case of 1 / 19 as follows:, For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2., Therefore 2 is the multiplier for the conversion. We write the last digit in the, numerator as 1 and follow the steps leftwards., Step. 1 :, Step. 2 :, , 1, 21(multiply 1 by 2, put to left), , Step. 3 :, , 421(multiply 2 by 2, put to left), , Step. 4 :, , 8421(multiply 4 by 2, put to left), , Step. 5 :, , 168421, , (multiply 8 by 2 =16,, 1 carried over, 6 put to left), , Step. 6 :, , 1368421, , ( 6 X 2 =12,+1 [carry over], , = 13, 1 carried over, 3 put to left ), Step. 7 :, , 7368421 ( 3 X 2, = 6 +1 [Carryover], = 7, put to left), , Step. 8 :, , 147368421, , (as in the same process), , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (6 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Step. 9 :, , 947368421 ( Do – continue to step 18), , Step. 10 :, , 18947368421, , Step. 11 :, , 178947368421, , Step. 12 :, , 1578947368421, , Step. 13 :, , 11578947368421, , Step. 14 :, , 31578947368421, , Step. 15 :, , 631578947368421, , Step. 16 :, , 12631578947368421, , Step. 17 :, , 52631578947368421, , Step. 18 :, , 1052631578947368421, , Now from step 18 onwards the same numbers and order towards left continue., Thus 1 / 19 = 0.052631578947368421, It is interesting to note that we have, i) not at all used division process, ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued, to multiply the resultant successively by 2., Observations :, a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in, the units place and a is the set of remaining digits, the value of the fraction is in, recurring decimal form and the repeating block’s right most digit is 1., b) Whatever may be a9, and the numerator, it is enough to follow the said, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (7 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , process with (a+1) either in division or in multiplication., c) Starting from right most digit and counting from the right, we see ( in the, given example 1 / 19), Sum of 1st digit + 10th digit = 1 + 8 = 9, Sum of 2nd digit + 11th digit = 2 + 7 = 9, - - - - - - - - -- - - - - - - - - - - - - - - - - - Sum of 9th digit + 18th digit = 9+ 0 = 9, From the above observations, we conclude that if we find first 9 digits,, further digits can be derived as complements of 9., i) Thus at the step 8 in division process we have 0.052631517 and next step. 9, gives 0.052631578, Now the complements of the numbers, 0, 5, 2, 6, 3, 1, 5, 7, 8 from 9, 9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order, i.e.,, 0.052631578947368421, Now taking the multiplication process we have, Step. 8 :, , 147368421, , Step. 9 :, , 947368421, , Now the complements of 1, 2, 4, 8, 6, 3, 7, 4, 9 from 9, i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps,, giving the answer., 0.052631578947368421., d) When we get (Denominator – Numerator) as the product in the multiplicative, process, half the work is done. We stop the multiplication there and mechanically, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (8 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , write the remaining half of the answer by merely taking down complements from, 9., e) Either division or multiplication process of giving the answer can be put in a, single line form., Algebraic proof :, Any vulgar fraction of the form 1 / a9 can be written as, 1 / a9 = 1 / ( (a + 1 ) x - 1 ) where x = 10, =, , 1, ________________________, ( a + 1 ) x [1 - 1/(a+1)x, , ], , 1, =, , ___________, , [1 - 1/(a+1)x]-1, , (a+1)x, 1, 2, , = __________ [1 + 1/(a+1)x + 1/(a+1)x + ----------], (a+1)x, = 1/(a+1)x + 1/(a+1)2x2 +1/(a+1)3x3+ ---- ad infinitum, -1, , -2, , 2, , -3, , 3, , = 10 (1/(a+1))+10 (1/(a+1) )+10 (1/(a+1) ) + ---ad infinitum, This series explains the process of ekadhik., Now consider the problem of 1 / 19. From above we get, 1 / 19 = 10, , -1, , (1/(1+1)) + 10, , -1, , (1/2) + 10, , = 10, , -2, , -2, , 2, , -3, , (1/(1+1) ) + 10, , 2, , -3, , (1/2) + 10, , 3, , (1/(1+1) ) + ---( since a=1), , 3, , (1/3) + ----------, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (9 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , = 10, , -1, , (0.5) + 10, , -2, , -3, , (0.25) + 10, , (0.125)+ ----------, , = 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - = 0.052631 - - - - - - Example1 :, 1. Find 1 / 49 by ekadhikena process., Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5., Now by division right ward from the left by ‘5’., 1 / 49 = .10 - - - - - - - - - - - -(divide 1 by 50), = .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder ), = .0220 - - - - - - --(divide 20 by 5, 4 times), = .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder ), = .020440 -- - -- - ( divide 40 by 5, 8 times ), = .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder ), = .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder ), = .02040811 6 - - - - - - -, , continue, , = .0204081613322615306111222244448 - -- - - - On completing 21 digits, we get 48, i.e.,,Denominator - Numerator = 49 – 1 = 48 stands., i.e, half of the process stops here. The remaining half can be obtained as, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma.../Mathematics/MathematicalFormulae/Sutras/sutras.html (10 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , complements from 9., , ., , Thus 1 / 49 = 0.020408163265306122448, , ., , 979591836734693877551, Now finding 1 / 49 by process of multiplication left ward from right by 5, we get, 1 / 49 = ----------------------------------------------1, = ---------------------------------------------51, = -------------------------------------------2551, = ------------------------------------------27551, = ---- 483947294594118333617233446943383727551, i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3, ( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining, half is automatically obtained as complements of 9., Thus 1 / 49 = ---------------979591836734693877551, , ., , = 0.020408163265306122448, , ., , 979591836734693877551, Example 2: Find 1 / 39 by Ekadhika process., Now by multiplication method, Ekadhikena purva is 3 + 1 = 4, 1 / 39 = -------------------------------------1, = -------------------------------------41, = ----------------------------------1641, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma.../Mathematics/MathematicalFormulae/Sutras/sutras.html (11 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , = ---------------------------------25641, = --------------------------------225641, = -------------------------------1025641, Here the repeating block happens to be block of 6 digits. Now the rule predicting, the completion of half of the computation does not hold. The complete block has, to be computed by ekadhika process., Now continue and obtain the result. Find reasons for the non–applicability of, the said ‘rule’., , Find the recurring decimal form of the fractions 1 / 29, 1 / 59,, 1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether, the rule of completion of half the computation holds good in such cases., , Note : The Ekadhikena Purvena sutra can also be used for conversion of vulgar, fractions ending in 1, 3, 7 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - - -, 1 / 7, 1 / 17, - - - - - by writing them in the following way and solving them., , 3 4, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma.../Mathematics/MathematicalFormulae/Sutras/sutras.html (12 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Upa-Sutras anurupyena, ĀNURŨPYENA, The upa-Sutra 'anurupyena' means 'proportionality'. This Sutra is highly useful to, find products of two numbers when both of them are near the Common bases i.e, powers of base 10 . It is very clear that in such cases the expected 'Simplicity ' in, doing problems is absent., Example 1: 46 X 43, As per the previous methods, if we select 100 as base we get, 46 -54 This is much more difficult and of no use., 43 -57, ¯¯¯¯¯¯¯¯, Now by ‘anurupyena’ we consider a working base In three ways. We can solve the, problem., Method 1: Take the nearest higher multiple of 10. In this case it is 50., Treat it as 100 / 2 = 50. Now the steps are as follows:, i) Choose the working base near to the numbers under consideration., i.e., working base is 100 / 2 = 50, ii) Write the numbers one below the other, i.e., , 4 6, 4 3, ¯¯¯¯¯¯¯, , iii) Write the differences of the two numbers respectively from 50, against each number on right side, i.e., , 46 -04, 43 -07, ¯¯¯¯¯¯¯¯¯, , iv) Write cross-subtraction or cross- addition as the case may be under, the line drawn., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (1 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , v) Multiply the differences and write the product in the left side of the, answer., 46 -04, 43 -07, ____________, 39 / -4 x –7, = 28, vi) Since base is 100 / 2 = 50 , 39 in the answer represents 39X50., Hence divide 39 by 2 because 50 = 100 / 2, Thus 39 ÷ 2 gives 19½ where 19 is quotient and 1 is remainder . This 1 as, Reminder gives one 50 making the L.H.S of the answer 28 + 50 = 78(or, Remainder ½ x 100 + 28 ), i.e. R.H.S 19 and L.H.S 78 together give the answer 1978 We represent it as, 46 -04, 43 -07, ¯¯¯¯¯¯¯¯¯, 2) 39 / 28, ¯¯¯¯¯¯¯¯¯, 19½ / 28, = 19 / 78 = 1978, Example 2: 42 X 48., With 100 / 2 = 50 as working base, the problem is as follows:, 42, 48, , -08, -02, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (2 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , ¯¯¯¯¯¯¯¯¯, 2) 40 / 16, ¯¯¯¯¯¯¯¯¯, 20 / 16, 42 x 48 = 2016, Method 2: For the example 1: 46X43. We take the same working base 50. We, treat it as 50=5X10. i.e. we operate with 10 but not with 100 as in method, now, , (195 + 2) / 8 = 1978, [Since we operate with 10, the R.H.S portion shall have only unit place .Hence, out of the product 28, 2 is carried over to left side. The L.H.S portion of the, answer shall be multiplied by 5, since we have taken 50 = 5 X 10.], Now in the example 2: 42 x 48 we can carry as follows by treating 50 = 5 x 10, , Method 3: We take the nearest lower multiple of 10 since the numbers are 46, and 43 as in the first example, We consider 40 as working base and treat it as 4 X, 10., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (3 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Since 10 is in operation 1 is carried out digit in 18., Since 4 X 10 is working base we consider 49 X 4 on L.H.S of answer i.e. 196 and 1, carried over the left side, giving L.H.S. of answer as 1978. Hence the answer is, 1978., We proceed in the same method for 42 X 48, , Let us see the all the three methods for a problem at a glance, Example 3: 24 X 23, Method - 1:, , Working base = 100 / 5 = 20, , 24 04, 23 03, ¯¯¯¯¯¯¯¯, 5) 27 / 12, ¯¯¯¯¯¯¯¯, 5 2/5 / 12 = 5 / 52 = 552, [Since 2 / 5 of 100 is 2 / 5 x 100 = 40 and 40 + 12 = 52], Method - 2: Working base 2 X 10 = 20, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (4 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Now as 20 itself is nearest lower multiple of 10 for the problem under, consideration, the case of method – 3 shall not arise., Let us take another example and try all the three methods., Example 4: 492 X 404, Method - 1 : working base = 1000 / 2 = 500, 492 -008, 404 -096, ¯¯¯¯¯¯¯¯¯¯¯, 2) 396 / 768, ¯¯¯¯¯¯¯¯¯¯¯, 198 / 768, , since 1000 is in operation, = 198768, , Method 2: working base = 5 x 100 = 500, , Method - 3., Since 400 can also be taken as working base, treat 400 = 4 X 100 as, working base., Thus, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (5 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , No need to repeat that practice in these methods finally takes us to work out, all these mentally and getting the answers straight away in a single line., Example 5:, , 3998 X 4998, , Working base = 10000 / 2 = 5000, 3998 -1002, 4998 -0002, ¯¯¯¯¯¯¯¯¯¯¯¯, 2) 3996 / 2004, 1998 / 2004, , since 10,000 is in operation, = 19982004, , or taking working base = 5 x 1000 = 5,000 and, , What happens if we take 4000 i.e. 4 X 1000 as working base?, _____, 3998 0002, 4998 0998, Since 1000 is an operation, ¯¯¯¯¯¯¯¯¯¯¯¯, 4996 / 1996, ___, ___, As 1000 is in operation, 1996 has to be written as 1996 and 4000 as base, the, L.H.S portion 5000 has to be multiplied by 4. i. e. the answer is, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (6 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , A simpler example for better understanding., Example 6: 58 x 48, Working base 50 = 5 x 10 gives, , Since 10 is in operation., , Use anurupyena by selecting appropriate working base and method., Find the following product., 1., , 46 x 46, , 2. 57 x 57, , 3. 54 x 45, , 4., , 18 x 18, , 5. 62 x 48, , 6. 229 x 230, , 7., , 47 x 96, , 8. 87965 x 99996, , 9. 49x499, , 10. 389 x 512, , 3 4, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...hematicalFormulae/Upa-Satras/UpaSutrasanurupyena.html (7 of 7)12/22/2005 8:49:43 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Nikhilam Navatas'caramam Dasatah, NIKHILAM NAVATAS’CARAMAM DASATAH, , The formula simply means : “all from 9 and the last from 10”, The formula can be very effectively applied in multiplication of numbers, which are, nearer to bases like 10, 100, 1000 i.e., to the powers of 10 . The procedure of, multiplication using the Nikhilam involves minimum number of steps, space, time, saving and only mental calculation. The numbers taken can be either less or more, than the base considered., 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, 14, , Base, , Number – Base, , Deviation, , 10, , 14 - 10, , 4, , 8, , 10, , 8 - 10, , 97, , 100, , 97 - 100, , _, -2 or 2, __, -03 or 03, , 112, , 100, , 112 - 100, , 12, , 993, , 1000, , 993 - 1000, , ___, -007 or 007, , 1011, , 1000, , 1011 - 1000, , 011, , Some rules of the method (near to the base) in Multiplication, a) Since deviation is obtained by Nikhilam sutra we call the method as Nikhilam, multiplication., Eg : 94. Now deviation can be obtained by ‘all from 9 and the last from 10’ sutra i., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (1 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06., b) The two numbers under consideration are written one below the other. The, deviations are written on the right hand side., Eg : Multiply 7 by 8., Now the base is 10. Since it is near to both the numbers,, we write the numbers one below the other., Take the deviations of both the numbers from, the base and represent, Rekhank or the minus sign before the deviations, , 7, 8, ----_, 7 3, _, 8 2, -----------, , or 7 -3, 8 -2, ------------or remainders 3 and 2 implies that the numbers to be multiplied are both less, than 10, c) The product or answer will have two parts, one on the left side and the other on, the right. A vertical or a slant line i.e., a slash may be drawn for the demarcation, of the two parts i.e.,, , (or), d) The R.H.S. of the answer is the product of the deviations of the numbers. It, shall contain the number of digits equal to number of zeroes in the base., , i.e., 7, , _, 3, _, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (2 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 8 2, _____________, / (3x2) = 6, Since base is 10, 6 can be taken as it is., e) L.H.S of the answer is the sum of one number with the deviation of the other. It, can be arrived at in any one of the four ways., i) Cross-subtract deviation 2 on the second row from the original, number 7 in the first row i.e., 7-2 = 5., ii) Cross–subtract deviation 3 on the first row from the original, number8 in the second row (converse way of (i)), i.e., 8 - 3 = 5, iii) Subtract the base 10 from the sum of the given numbers., i.e., (7 + 8) – 10 = 5, iv) Subtract the sum of the two deviations from the base., i.e., 10 – ( 3 + 2) = 5, Hence 5 is left hand side of the answer., _, Thus, 7 3, _, 8 2, ¯¯¯¯¯¯¯¯¯¯¯¯, 5 /, Now (d) and (e) together give the solution, _, 7 3, 7, _, 8 2 i.e., X 8, ¯¯¯¯¯¯¯, ¯¯¯¯¯¯, 5/ 6, 56, f) If R.H.S. contains less number of digits than the number of zeros in the base,, the remaining digits are filled up by giving zero or zeroes on the left side of the R., H.S. If the number of digits are more than the number of zeroes in the base, the, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (3 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , excess digit or digits are to be added to L.H.S of the answer., The general form of the multiplication under Nikhilam can be shown as follows :, Let N1 and N2 be two numbers near to a given base in powers of 10, and D1 and, D2 are their respective deviations from the base. Then N1 X N2 can be represented, as, , Case (i) : Both the numbers are lower than the base. We have already considered, the example 7 x 8 , with base 10., Now let us solve some more examples by taking bases 100 and 1000 respectively., Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is as, follows:, , Ex. 2: 98 X 97 Base is 100., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (4 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Ex. 3: 75X95. Base is 100., , Ex. 4: 986 X 989. Base is 1000., , Ex. 5: 994X988. Base is 1000., , Ex. 6: 750X995., , Case ( ii) : Both the numbers are higher than the base., The method and rules follow as they are. The only difference is the positive, deviation. Instead of cross – subtract, we follow cross – add., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (5 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Ex. 7: 13X12. Base is 10, , Ex. 8: 18X14. Base is 10., , Ex. 9: 104X102. Base is 100., 104 04, 102 02, ¯¯¯¯¯¯¯¯¯¯¯¯, 106 / 4x2 =, ¯¯¯¯¯¯¯¯¯¯¯¯, , 10608, , ( rule - f ), , Ex. 10: 1275X1004. Base is 1000., 1275 275, 1004 004, ¯¯¯¯¯¯¯¯¯¯¯¯, 1279 / 275x4 =, ____________, , =, , 1279 / 1100, , ( rule - f ), , 1280100, , Case ( iii ): One number is more and the other is less than the base., In this situation one deviation is positive and the other is negative. So the product, of deviations becomes negative. So the right hand side of the answer obtained will, therefore have to be subtracted. To have a clear representation and understanding, a vinculum is used. It proceeds into normalization., Ex.11: 13X7. Base is 10, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (6 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Algebraic Proof:, Case ( i ):, Let the two numbers N1 and N2 be less than the selected base say x., N1 = (x-a), N2 = (x-b). Here a and b are the corresponding deviations of the, numbers N1 and N2 from the base x. Observe that x is a multiple of 10., Now N1 X N2, , = (x-a) (x-b) = x.x – x.b – a.x + ab, , = x (x – a – b ) + ab. [rule – e(iv), d ], = x [(x – a) – b] + ab = x (N1–b) + ab[rule–e(i),d], or, , = x [(x – b) – a] = x (N2 – a) + ab. [rule –e (ii),d], x (x – a – b) + ab can also be written as, , x[(x – a) + (x – b) – x] + ab = x[N1+N2 – x] + ab [rule – e(iii),d]., A difficult can be faced, if the vertical multiplication of the deficit digits or, deviations i.e., a.b yields a product consisting of more than the required digits., Then rule-f will enable us to surmount the difficulty., Case ( ii ) :, When both the numbers exceed the selected base, we have N1 = x + a, N2 = x +, b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds good,, of course with relevant details mentioned in case (i)., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (8 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Case ( iii ) :, When one number is less and another is more than the base, we can use (x-a)(x, +b) = x(x–a+ b)–ab. and the procedure is evident from the examples given., , Find the following products by Nikhilam formula., 1) 7 X 4, , 2) 93 X 85, , 3) 875 X 994, , 4) 1234 X 1002, , 5) 1003 X 997, , 6) 11112 X 9998, , 7) 1234 X 1002, , 8) 118 X 105, , Nikhilam in Division, Consider some two digit numbers (dividends) and same divisor 9. Observe the, following example., i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4., since 9 ) 13 ( 1, 9, ____, 4, ii), , 34 ÷ 9, Q is 3, R is 7., , iii), , 60 ÷ 9, Q is 6, R is 6., , iv), , 80 ÷ 9, Q is 8, R is 8., , Now we have another type of representation for the above examples as given, hereunder:, i) Split each dividend into a left hand part for the Quotient and right - hand part, for the remainder by a slant line or slash., Eg., , 13 as 1 / 3,, , 34 as 3 / 4 ,, , 80 as 8 / 0., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...alFormulae/Sutras/NikhilamnavatascaramamDasatah.html (9 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , ii) Leave some space below such representation, draw a horizontal line., Eg., , 1/3, , 3/4, , ______, , ,, , 8/0, , ______ ,, , ______, , iii) Put the first digit of the dividend as it is under the horizontal line. Put the same, digit under the right hand part for the remainder, add the two and place the sum i., e., sum of the digits of the numbers as the remainder., Eg., , 1/3, 1, ______ ,, 1/4, , 3/4, 8/0, 3, 8, ______ , ______, 3/7, 8/8, , Now the problem is over. i.e.,, 13 ÷ 9 gives Q = 1, R = 4, 34 ÷ 9 gives Q = 3, R = 7, 80 ÷ 9 gives Q = 8, R = 8, Proceeding for some more of the two digit number division by 9, we get, a) 21 ÷ 9 as, 9) 2 / 1, 2, ¯¯¯¯¯¯, 2 / 3, b) 43 ÷ 9 as, 9) 4 / 3, 4, ¯¯¯¯¯¯, 4 / 7, , i.e, , Q=2, R=3, , i.e, , Q = 4, R = 7., , The examples given so far convey that in the division of two digit numbers by 9,, we can mechanically take the first digit down for the quotient – column and that,, by adding the quotient to the second digit, we can get the remainder., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...lFormulae/Sutras/NikhilamnavatascaramamDasatah.html (10 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Now in the case of 3 digit numbers, let us proceed as follows., i), , ii), , iii), , 9 ) 104 ( 11, 99, ¯¯¯¯¯¯, 5, , 9 ) 212 ( 23, 207, ¯¯¯¯¯, 5, , 9 ) 401 ( 44, 396, ¯¯¯¯¯, 5, , as, , as, , as, , 9 ) 10 / 4, 1 / 1, ¯¯¯¯¯¯¯, 11 / 5, , 9 ) 21 / 2, 2 / 3, ¯¯¯¯¯¯¯, 23 / 5, , 9 ) 40 / 1, 4 / 4, ¯¯¯¯¯¯¯¯, 44 / 5, , Note that the remainder is the sum of the digits of the dividend. The first digit of, the dividend from left is added mechanically to the second digit of the dividend to, obtain the second digit of the quotient. This digit added to the third digit sets the, remainder. The first digit of the dividend remains as the first digit of the quotient., Consider, , 511 ÷ 9, , Add the first digit 5 to second digit 1 getting 5 + 1 = 6. Hence Quotient is 56. Now, second digit of 56 i.e., 6 is added to third digit 1 of dividend to get the remainder i., e., 1 + 6 = 7, Thus, , 9), , 51 / 1, 5/ 6, ¯¯¯¯¯¯¯, 56 / 7, , Q is 56, R is 7., Extending the same principle even to bigger numbers of still more digits, we can, get the results., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...lFormulae/Sutras/NikhilamnavatascaramamDasatah.html (11 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Eg : 1204 ÷ 9, i) Add first digit 1 to the second digit 2. 1 + 2 = 3, ii) Add the second digit of quotient 13. i.e., 3 to third digit ‘0’ and, obtain the Quotient. 3 + 0 = 3, 133, iii) Add the third digit of Quotient 133 i.e.,3 to last digit ‘4’ of the, dividend and write the final Quotient and Remainder. R = 3 + 4 = 7, Q, = 133, In symbolic form, , Another example., gives, , 9 ) 120 / 4, 13 / 3, ¯¯¯¯¯¯¯¯, 133 / 7, , 9 ) 13210, , /1, , 1467 / 7, ¯¯¯¯¯¯¯¯¯¯, 14677 / 8, , 132101 ÷ 9, Q = 14677, R = 8, , In all the cases mentioned above, the remainder is less than the divisor. What, about the case when the remainder is equal or greater than the divisor?, Eg., , 9) 3 / 6, 9) 24 / 6, 3, 2 / 6, ¯¯¯¯¯¯, or, ¯¯¯¯¯¯¯¯, 3 / 9 (equal), 26 / 12 (greater)., , We proceed by re-dividing the remainder by 9, carrying over this Quotient to the, quotient side and retaining the final remainder in the remainder side., 9) 3 / 6, / 3, ¯¯¯¯¯¯¯, 3 / 9, ¯¯¯¯¯¯¯, 4 / 0, , 9 ) 24 / 6, 2 / 6, ¯¯¯¯¯¯¯¯, 26 / 12, ¯¯¯¯¯¯¯¯, 27 / 3, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...lFormulae/Sutras/NikhilamnavatascaramamDasatah.html (12 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Q = 4,, , R=0, , Q = 27,, , R = 3., , When the remainder is greater than divisor, it can also be represented as, 9 ) 24 / 6, 2 / 6, ¯¯¯¯¯¯¯¯, 26 /1 / 2, /1, ¯¯¯¯¯¯¯¯, 1 /3, ¯¯¯¯¯¯¯¯, 27 / 3, Now consider the divisors of two or more digits whose last digit is 9,when divisor, is 89., We Know, , 113 = 1 X 89 + 24,, , Q =1,, , 10015 = 112 X 89 + 47,, , R = 24, , Q = 112, R = 47., , Representing in the previous form of procedure, we have, 89 ) 1 / 13, / 11, ¯¯¯¯¯¯¯, 1 / 24, , 89 ) 100 / 15, 12 / 32, ¯¯¯¯¯¯¯¯¯¯, 112 / 47, , But how to get these? What is the procedure?, Now Nikhilam rule comes to rescue us. The nikhilam states “all from 9 and the last, from 10”. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply, nikhilam formula on 89 and get the complement 11.Further while carrying the, added numbers to the place below the next digit, we have to multiply by this 11., 89 ) 1 / 13, / 11, ¯¯¯¯¯¯¯¯, 1 / 24, , 89 ) 100 / 15, ¯¯, 11, 11 /, , first digit 1 x 11, , 1/ 1, total second is 1x11, 22 total of 3rd digit is 2 x 11, ¯¯¯¯¯¯¯¯¯¯, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...lFormulae/Sutras/NikhilamnavatascaramamDasatah.html (13 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 112 / 47, What is 10015 ÷ 98 ? Apply Nikhilam and get 100 – 98 = 02. Set off the 2 digits, from the right as the remainder consists of 2 digits. While carrying the added, numbers to the place below the next digit, multiply by 02., Thus, , 98 ), ¯¯, 02, , In the same way, , gives, , 100, , / 15, , 02 /, 0 / 0, / 04, ¯¯¯¯¯¯¯¯¯¯, 102 / 19, , i.e., 10015 ÷ 98 gives, Q = 102,, R = 19, , 897 ) 11 / 422, ¯¯¯, 103, 1 / 03, / 206, ¯¯¯¯¯¯¯¯¯, 12 / 658, , 11,422 ÷ 897,, , Q = 12, R=658., , In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the case, of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the, divisors complement from 10. In case of more digited numbers we apply Nikhilam, and proceed. Any how, this method is highly useful and effective for division when, the numbers are near to bases of 10., , * Guess the logic in the process of division by 9., * Obtain the Quotient and Remainder for the following problems., 1) 311 ÷ 9, 4) 2342 ÷ 98, , 2) 120012 ÷ 9, , 3) 1135 ÷ 97, , 5) 113401 ÷ 997, , 6) 11199171 ÷ 99979, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...lFormulae/Sutras/NikhilamnavatascaramamDasatah.html (14 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Observe that by nikhilam process of division, even lengthier divisions involve no, division or no subtraction but only a few multiplications of single digits with small, numbers and a simple addition. But we know fairly well that only a special type of, cases are being dealt and hence many questions about various other types of, problems arise. The answer lies in Vedic Methods., , 3 4, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...lFormulae/Sutras/NikhilamnavatascaramamDasatah.html (15 of 15)12/22/2005 8:49:47 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Why Vedic Mathematics?, I. WHY VEDIC MATHEMATICS?, Many Indian Secondary School students consider Mathematics a very difficult, subject. Some students encounter difficulty with basic arithmetical operations., Some students feel it difficult to manipulate symbols and balance equations. In, other words, abstract and logical reasoning is their hurdle., Many such difficulties in learning Mathematics enter into a long list if prepared by, an experienced teacher of Mathematics. Volumes have been written on the, diagnosis of 'learning difficulties' related to Mathematics and remedial techniques., Learning Mathematics is an unpleasant experience to some students mainly, because it involves mental exercise., Of late, a few teachers and scholars have revived interest in Vedic Mathematics, which was developed, as a system derived from Vedic principles, by Swami, Bharati Krishna Tirthaji in the early decades of the 20th century., Dr. Narinder Puri of the Roorke University prepared teaching materials based on, Vedic Mathematics during 1986 - 89. A few of his opinions are stated hereunder:, i) Mathematics, derived from the Veda, provides one line, mental and super- fast, methods along with quick cross checking systems., ii) Vedic Mathematics converts a tedious subject into a playful and blissful one, which students learn with smiles., iii) Vedic Mathematics offers a new and entirely different approach to the study of, Mathematics based on pattern recognition. It allows for constant expression of a, student's creativity, and is found to be easier to learn., iv) In this system, for any problem, there is always one general technique, applicable to all cases and also a number of special pattern problems. The element, of choice and flexibility at each stage keeps the mind lively and alert to develop, clarity of thought and intuition, and thereby a holistic development of the human, brain automatically takes place., v) Vedic Mathematics with its special features has the inbuilt potential to solve the, psychological problem of Mathematics - anxiety., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...matics/WhyVedicMathematics/whyvedicmathematics.html (1 of 2)12/22/2005 8:49:52 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , J.T.Glover (London, 1995) says that the experience of teaching Vedic, Mathematics' methods to children has shown that a high degree of mathematical, ability can be attained from an early stage while the subject is enjoyed for its own, merits., A.P. Nicholas (1984) puts the Vedic Mathematics system as 'one of the most, delightful chapters of the 20th century mathematical history'., Prof. R.C. Gupta (1994) says 'the system has great educational value because the, Sutras contain techniques for performing some elementary mathematical, operations in simple ways, and results are obtained quickly'., Prof. J.N. Kapur says 'Vedic Mathematics can be used to remove math-phobia, and, can be taught to (school) children as enrichment material along with other high, speed methods'., Dr. Michael Weinless, Chairman of the Department of Mathematics at the M.I.U,, Iowa says thus: 'Vedic Mathematics is easier to learn, faster to use and less prone, to error than conventional methods. Furthermore, the techniques of Vedic, Mathematics not only enable the students to solve specific mathematical, problems; they also develop creativity, logical thinking and intuition.', Keeping the above observations in view, let us enter Vedic Mathematics as given, by Sri Bharati Krishna Tirthaji (1884 - 1960), Sankaracharya of Govardhana Math,, Puri. Entering into the methods and procedures, one can realize the importance, and applicability of the different formulae (Sutras) and methods., , 3 4, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20m...matics/WhyVedicMathematics/whyvedicmathematics.html (2 of 2)12/22/2005 8:49:52 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , In the text, the words Sutra, aphorism, formula are used synonymously. So are, also the words Upa-sutra, Sub-sutra, Sub-formula, corollary used., Now we shall have the literal meaning, contextual meaning, process, different, methods of application along with examples for the Sutras. Explanation, methods,, further short-cuts, algebraic proof, etc follow. What follows relates to a single, formula or a group of formulae related to the methods of Vedic Mathematics., , 3 4, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...matics/MathematicalFormulae/mathematicalformulae.html (2 of 2)12/22/2005 8:49:59 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Antyayor Dasakepi, ANTYAYOR DA•AKE′PI, The Sutra signifies numbers of which the last digits added up give 10. i.e. the, Sutra works in multiplication of numbers for example: 25 and 25, 47 and 43, 62, and 68, 116 and 114. Note that in each case the sum of the last digit of first, number to the last digit of second number is 10. Further the portion of digits or, numbers left wards to the last digits remain the same. At that instant use, Ekadhikena on left hand side digits. Multiplication of the last digits gives the right, hand part of the answer., Example 1 : 47 X 43, See the end digits sum 7 + 3 = 10 ; then by the sutras antyayor dasakepi and, ekadhikena we have the answer., 47 x 43 = ( 4 + 1 ) x 4 / 7 x 3, = 20 / 21, = 2021., Example 2: 62 x 68, 2 + 8 = 10, L.H.S. portion remains the same i.e.,, 6., Ekadhikena of 6 gives 7, 62 x 68 = ( 6 x 7 ) / ( 2 x 8 ), = 42 / 16, = 4216., Example 3:, , 127 x 123, , As antyayor dasakepi works, we apply ekadhikena, 127 x 123 = 12 x 13 / 7 x 3, = 156 / 21, = 15621., Example 4: 65 x 65, We have already worked on this type. As the present sutra is applicable., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...athematicalFormulae/Upa-Satras/AntyayorDasakepi.html (1 of 3)12/22/2005 8:52:17 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , We have 65 x 65 = 6 x 7 / 5 x 5, = 4225., Example 5: 3952, 3952 =, =, =, =, , 395 x 395, 39 x 40 / 5 x 5, 1560 / 25, 156025., , Use Vedic sutras to find the products, 1. 125 x 125, , 2. 34 x 36, , 3. 98 x 92, , 4. 401 x 409, , 5. 693 x 697, , 6. 1404 x 1406, , It is further interesting to note that the same rule works when the sum of the last, 2, last 3, last 4 - - - digits added respectively equal to 100, 1000, 10000 -- - - ., The simple point to remember is to multiply each product by 10, 100, 1000, - - as, the case may be . Your can observe that this is more convenient while working, with the product of 3 digit numbers., Eg. 1: 292 x 208, Here 92 + 08 = 100, L.H.S portion is same i.e. 2, 292 x 208 = ( 2 x 3 ) / 92 x 8, 60 / =736 ( for 100 raise the L.H.S. product by 0 ), = 60736., Eg. 2:, , 848 X 852, , Here 48 + 52 = 100, L.H.S portion is 8 and its ‘ekhadhikena’ is 9., Now R.H.S product 48 X 52 can be obtained by ‘anurupyena’ mentally., _, 48 2, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...athematicalFormulae/Upa-Satras/AntyayorDasakepi.html (2 of 3)12/22/2005 8:52:17 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 52 2, _______, _, 2) 50 4, ¯¯, 25, , =, , 24 / ( 100 – 4 ), = 96, , = 2496, and write 848 x 852 = 8 x 9 / 48 x 52, 720 = 2496, = 722496., [Since L.H.S product is to be multiplied by 10 and 2 to be carried over as the, base is 100]., Eg. 3:, , 693 x 607, 693 x 607 = 6 x 7 / 93 x 7, = 420 / 651, = 420651., , Find the following products using ‘antyayordasakepi’, 1. 318 x 312, , 2. 425 x 475, , 3. 796 x 744, , 4. 902 x 998, , 5. 397 x 393, , 6. 551 x 549, , 34, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...athematicalFormulae/Upa-Satras/AntyayorDasakepi.html (3 of 3)12/22/2005 8:52:17 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Ekanyunenapurvena, EKANYŨ•ENA PŨRVENA, The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives the, meaning 'One less than the previous' or 'One less than the one before'., 1) The use of this sutra in case of multiplication by 9,99,999.. is as follows ., Method :, a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena, purvena i.e. by deduction 1 from the left side digit (digits) ., e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit ), b) The right hand side digit is the complement or difference between the multiplier, and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3., c) The two numbers give the answer; i.e. 7 X 9 = 63., Example 1:, , 8 x 9 Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit ), Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit ), Step ( c ) gives the answer 72, , Example 2: 15 x 99, , Step ( a ) : 15 – 1 = 14, Step ( b ) : 99 – 14 = 85 ( or 100 – 15 ), Step ( c ) : 15 x 99 = 1485, , Example 3: 24 x 99, Answer :, , Example 4:, 356 x 999, Answer :, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (1 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Example 5:, 878 x 9999, Answer :, , Note the process : The multiplicand has to be reduced by 1 to obtain the LHS and, the rightside is mechanically obtained by the subtraction of the L.H.S from the, multiplier which is practically a direct application of Nikhilam Sutra., Now by Nikhilam, 24 – 1 = 23 L.H.S., x 99 – 23 = 76 R.H.S. (100–24), _____________________________, 23 / 76, = 2376, Reconsider the Example 4:, 356 – 1 = 355 L.H.S., x 999 – 355 = 644 R.H.S., ________________________, 355 / 644, =, , 355644, , and in Example 5: 878 x 9999 we write, 0878 – 1 = 877, L.H.S., x 9999 – 877 = 9122 R.H.S., __________________________, 877 / 9122, = 8779122, Algebraic proof :, As any two digit number is of the form ( 10x + y ), we proceed, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (2 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , ( 10x + y ) x 99, = ( 10x + y ) x ( 100 – 1 ), = 10x . 102 – 10x + 102 .y – y, = x . 103 + y . 102 – ( 10x + y ), = x . 103 + ( y – 1 ) . 102 + [ 102 – ( 10x + y )], Thus the answer is a four digit number whose 1000th place is x, 100th place is, ( y - 1 ) and the two digit number which makes up the 10th and unit place is the, number obtained by subtracting the multiplicand from 100.(or apply nikhilam)., Thus in 37 X 99. The 1000th place is x i.e. 3, 100th place is ( y - 1 ) i.e. (7 - 1 ) = 6, Number in the last two places 100-37=63., Hence answer is 3663., Apply Ekanyunena purvena to find out the products, 1. 64 x 99, , 2. 723 x 999, , 3. 3251 x 9999, , 4. 43 x 999, , 5. 256 x 9999, , 6. 1857 x 99999, , We have dealt the cases, i) When the multiplicand and multiplier both have the same number of digits, ii) When the multiplier has more number of digits than the multiplicand., In both the cases the same rule applies. But what happens when the multiplier has, lesser digits?, i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc.,, For this let us have a re-look in to the process for proper understanding., Multiplication table of 9., a, 2x9 =, 3x9 =, , b, 1, 2, , 8, 7, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (3 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 4x9, ---8x9, 9x9, 10 x 9, , = 3 6, -----= 7 2, = 8 1, = 9 0, , Observe the left hand side of the answer is always one less than the multiplicand, (here multiplier is 9) as read from Column (a) and the right hand side of the, answer is the complement of the left hand side digit from 9 as read from Column, (b), Multiplication table when both multiplicand and multiplier are of 2 digits., a, b, 11 x 99 = 10 89 = (11–1) / 99 – (11–1) =, 12 x 99 = 11 88 = (12–1) / 99 – (12–1) =, 13 x 99 = 12 87 = (13–1) / 99 – (13–1) =, ------------------------------------------------18 x 99 = 17 82 ---------------------------19 x 99 = 18 81, 20 x 99 = 19 80 = (20–1) / 99 – (20–1) =, , 1089, 1188, 1287, , 1980, , The rule mentioned in the case of above table also holds good here, Further we can state that the rule applies to all cases, where the multiplicand and, the multiplier have the same number of digits., Consider the following Tables., (i), , (ii), , a b, 11 x 9 = 9 9, 12 x 9 = 10 8, 13 x 9 = 11 7, ---------------------18 x 9 = 16 2, 19 x 9 = 17 1, 20 x 9 = 18 0, , 21 x 9 = 18, 22 x 9 = 19, , 9, 8, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (4 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 23 x 9 = 20 7, ----------------------28 x 9 = 25 2, 29 x 9 = 26 1, 30 x 9 = 27 0, (iii), , 35 x 9 = 31 5, 46 x 9 = 41 4, 53 x 9 = 47 7, 67 x 9 = 60 3, -------------------------so on., , From the above tables the following points can be observed:, 1) Table (i) has the multiplicands with 1 as first digit except the last one. Here L., H.S of products are uniformly 2 less than the multiplicands. So also with 20 x 9, 2) Table (ii) has the same pattern. Here L.H.S of products are uniformly 3 less, than the multiplicands., 3) Table (iii) is of mixed example and yet the same result i.e. if 3 is first digit of, the multiplicand then L.H.S of product is 4 less than the multiplicand; if 4 is first, digit of the multiplicand then, L.H.S of the product is 5 less than the multiplicand, and so on., 4) The right hand side of the product in all the tables and cases is obtained by, subtracting the R.H.S. part of the multiplicand by Nikhilam., Keeping these points in view we solve the problems:, Example1 : 42 X 9, i) Divide the multiplicand (42) of by a Vertical line or by the Sign : into a right, hand portion consisting of as many digits as the multiplier., i.e. 42 has to be written as 4/2 or 4:2, ii) Subtract from the multiplicand one more than the whole excess portion on the, left. i.e. left portion of multiplicand is 4., one more than it 4 + 1 = 5., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (5 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , We have to subtract this from multiplicand, i.e. write it as, 4 : 2, :-5, --------------3 : 7, This gives the L.H.S part of the product., This step can be interpreted as "take the ekanyunena and sub tract from the, previous" i.e. the excess portion on the left., iii) Subtract the R.H.S. part of the multiplicand by nikhilam process., i.e. R.H.S of multiplicand is 2, its nikhilam is 8, It gives the R.H.S of the product, i.e. answer is 3 : 7 : 8 = 378., Thus 42 X 9 can be represented as, 4:2, :-5 : 8, -----------------3 : 7 : 8 = 378., Example 2 :, , 124 X 9, , Here Multiplier has one digit only ., We write 12 : 4, Now step (ii), 12 + 1 = 13, i.e., , 12 : 4, -1 : 3, -----------Step ( iii ) R.H.S. of multiplicand is 4. Its Nikhilam is 6, 124 x 9 is, , 12 : 4, -1 : 3 : 6, -----------------, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (6 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 11 : 1 : 6, , =, , 1116, , The process can also be represented as, 124 x 9 = [ 124 – ( 12 + 1 ) ] : ( 10 – 4 ) = ( 124 – 13 ) : 6 = 1116, Example 3:, , 15639 x 99, , Since the multiplier has 2 digits, the answer is, [15639 – (156 + 1)] : (100 – 39) = (15639 – 157) : 61 = 1548261, , Find the products in the following cases., 1., , 58 x 9, , 2., , 62 x 9, , 3., , 4., , 832 x 9, , 5., , 24821 x 999, , 427 x 99, , 6. 111011 x 99, , Ekanyunena Sutra is also useful in Recurring Decimals. We can take up this under, a separate treatment., Thus we have a glimpse of majority of the Sutras. At some places some Sutras are, mentioned as Sub-Sutras. Any how we now proceed into the use of Sub-Sutras., As already mentioned the book on Vedic Mathematics enlisted 13 Upa-Sutras., But some approaches in the Vedic Mathematics book prompted some serious, research workers in this field to mention some other Upa-Sutras. We can observe, those approaches and developments also., , 3 4, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/ekanyunenapurvena.html (7 of 7)12/22/2005 8:52:32 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | AnurupyeSunyamanyat, ĀNURŨPYE •ŨNYAMANYAT, The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is zero'., We use this Sutra in solving a special type of simultaneous simple equations in, which the coefficients of 'one' variable are in the same ratio to each other as the, independent terms are to each other. In such a context the Sutra says the 'other', variable is zero from which we get two simple equations in the first variable, (already considered) and of course give the same value for the variable., Example 1:, , 3x + 7y = 2, 4x + 21y = 6, , Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is same as, the ratio of independent terms i.e., 2 : 6 i.e., 1 : 3. Hence the other variable x = 0, and 7y = 2 or 21y = 6 gives y = 2 / 7, Example 2:, , 323x + 147y = 1615, 969x + 321y = 4845, , The very appearance of the problem is frightening. But just an observation and, anurupye sunyamanyat give the solution x = 5, because coefficient of x ratio is, 323 : 969 = 1 : 3 and constant terms ratio is 1615 : 4845 = 1 : 3., y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5., , Solve the following by anurupye sunyamanyat., 1., 3., , 12x + 78y = 12, 16x + 96y = 16, 4x – 6y = 24, 7x – 9y = 36, , 2., 4., , 3x + 7y = 24, 12x + 5y = 96, ax + by = bm, cx + dy = dm, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...MathematicalFormulae/Sutras/AnurupyeSunyamanyat.html (1 of 2)12/22/2005 8:52:36 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Urdhava Tiryagbhyam, ŨRDHVA TIRYAGBHYĀM, Urdhva – tiryagbhyam is the general formula applicable to all cases of, multiplication and also in the division of a large number by another large number., It means “Vertically and cross wise.”, (a) Multiplication of two 2 digit numbers., Ex.1: Find the product 14 X 12, i) The right hand most digit of the multiplicand, the first number (14) i.e., 4 is, multiplied by the right hand most digit of the multiplier, the second number (12) i., e., 2. The product 4 X 2 = 8 forms the right hand most part of the answer., ii) Now, diagonally multiply the first digit of the multiplicand (14) i.e., 4 and, second digit of the multiplier (12) i.e., 1 (answer 4 X 1=4); then multiply the, second digit of the multiplicand i.e., 1 and first digit of the multiplier i.e., 2, (answer 1 X 2 = 2); add these two i.e., 4 + 2 = 6. It gives the next, i.e., second, digit of the answer. Hence second digit of the answer is 6., iii) Now, multiply the second digit of the multiplicand i.e., 1 and second digit of the, multiplier i.e., 1 vertically, i.e., 1 X 1 = 1. It gives the left hand most part of the, answer., Thus the answer is 16 8., Symbolically we can represent the process as follows :, , The symbols are operated from right to left ., Step i) :, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (1 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Step ii) :, , Step iii) :, , Now in the same process, answer can be written as, 23, , 13, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, 2 : 6 + 3 : 9 = 299, Ex.3, , (Recall the 3 steps), , 41, X 41, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, 16 : 4 + 4 : 1 = 1681., , What happens when one of the results i.e., either in the last digit or in the middle, digit of the result, contains more than 1 digit ? Answer is simple. The right – hand, – most digit there of is to be put down there and the preceding, i.e., left –hand –, side digit or digits should be carried over to the left and placed under the previous, digit or digits of the upper row. The digits carried over may be written as in Ex. 4., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (2 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Ex.4:, , 32 X 24, , Step (i) :, , 2X4=8, , Step (ii) :, , 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16., , Here 6 is to be retained. 1 is to be carried out to left side., Step (iii) :, , 3 X 2 = 6. Now the carried over digit 1 of 16 is to be added., i.e., 6 + 1 = 7., , Thus 32 X 24 = 768, We can write it as follows, , 32, 24, ¯¯¯¯, 668, 1, ¯¯¯¯, 768., , Note that the carried over digit from the result (3X4) + (2X2) = 12+4 = 16 i.e., 1, is placed under the previous digit 3 X 2 = 6 and added., After sufficient practice, you feel no necessity of writing in this way and simply, operate or perform mentally., Ex.5, , 28 X 35., Step (i) :, 8 X 5 = 40. 0 is retained as the first digit of the answer, and 4 is carried over., Step (ii) :, 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried, over 4 to 34. Now the result is 34 + 4 = 38. Now 8 is retained as the, second digit of the answer and 3 is carried over., Step (iii) : 2 X 3 = 6; add the carried over 3 to 6. The result 6 + 3, = 9 is the third or final digit from right to left of the answer., Thus 28 X 35 = 980., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (3 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Ex.6, , 48, 47, ¯¯¯¯¯¯, 1606, 65, ¯¯¯¯¯¯¯, 2256, Step (i):, 8 X 7 = 56; 5, the carried over digit is placed below the, second digit., Step (ii):, ( 4 X 7) + (8 X 4) = 28 + 32 = 60; 6, the carried over, digit is placed below the third digit., Step (iii):, , Respective digits are added., , Algebraic proof :, a) Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now, consider the product, (ax + b) (cx + d) = ac.x2 + adx + bcx + b.d, = ac.x2 + (ad + bc)x + b.d, Observe that, i) The first term i.e., the coefficient of x2 (i.e., 100, hence the digit in, the 100th place) is obtained by vertical multiplication of a and c i.e.,, the digits in 10th place (coefficient of x) of both the numbers;, ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th, place) is obtained by cross wise multiplication of a and d; and of b and, c; and the addition of the two products;, iii) The last (independent of x) term is obtained by vertical, multiplication of the independent terms b and d., b) Consider the multiplication of two 3 digit numbers., Let the two numbers be (ax2 + bx + c) and (dx2 + ex + f). Note that x=10, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (4 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Now the product is, ax2 + bx + c, x dx2 + ex + f, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, ad.x4+bd.x3+cd.x2+ae.x3+be.x2+ce.x+af.x2+bf.x+cf, = ad.x4 + (bd + ae). x3 + (cd + be + af).x2 + (ce + bf)x + cf, Note the following points :, i) The coefficient of x4 , i.e., ad is obtained by the vertical, multiplication of the first coefficient from the left side :, , ii)The coefficient of x3 , i.e., (ae + bd) is obtained by the cross –wise, multiplication of the first two coefficients and by the addition of the two, products;, , iii) The coefficient of x2 is obtained by the multiplication of the first, coefficient of the multiplicand (ax2+bx +c) i.e., a; by the last, coefficient of the multiplier (dx2 +ex +f) i.e.,f ; of the middle one i.e.,, b of the multiplicand by the middle one i.e., e of the multiplier and of, the last one i.e., c of the multiplicand by the first one i.e., d of the, multiplier and by the addition of all the three products i.e., af + be, +cd :, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (5 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , iv) The coefficient of x is obtained by the cross wise multiplication of, the second coefficient i.e., b of the multiplicand by the third one i.e., f, of the multiplier, and conversely the third coefficient i.e., c of the, multiplicand by the second coefficient i.e., e of the multiplier and by, addition of the two products, i.e., bf + ce ;, , v) And finally the last (independent of x) term is obtained by the, vertical multiplication of the last coefficients c and f i.e., cf, , Thus the process can be put symbolically as (from left to right), , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (6 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Consider the following example, 124 X 132., Proceeding from right to left, i) 4 X 2 = 8. First digit = 8, ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is, carried over to left side. Second digit = 6., iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of, above step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is, carried over to left side. Thus third digit = 3., iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is, added i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6, v) ( 1 X 1 ) = 1. As there is no carried over number from the previous, step it is retained. Thus fifth digit = 1, 124 X 132 = 16368., Let us work another problem by placing the carried over digits under, the first row and proceed., 234, x 316, ¯¯¯¯¯¯¯, 61724, 1222, ¯¯¯¯¯¯¯, 73944, i) 4 X 6 = 24 : 2, the carried over digit is placed below the second, digit., ii) (3 X 6) + (4 x 1) = 18 + 4 = 22 ; 2, the carried over digit is placed, below third digit., iii) (2 X 6) + (3 X 1) + (4 X 3) = 12 + 3 + 12 = 27 ; 2, the carried, over digit is placed below fourth digit., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (7 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , iv) (2 X 1) + ( 3 X 3) = 2 + 9 = 11; 1, the carried over digit is placed, below fifth digit., v) ( 2 X 3 ) = 6., vi) Respective digits are added., Note :, 1. We can carry out the multiplication in urdhva - tiryak process from, left to right or right to left., 2. The same process can be applied even for numbers having more, digits., 3. urdhva –tiryak process of multiplication can be effectively used in, multiplication regarding algebraic expressions., Example 1 : Find the product of (a+2b) and (3a+b)., , Example 2 :, 3a2 + 2a + 4, x 2a2 + 5a + 3, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯, i), ii), , 4X3, , = 12, , (2 X 3) + ( 4 X 5 ) = 6 + 20 = 26, , i.e., 26a, , iii) (3 X 3) + ( 2 X 5 ) + ( 4 X 2 ) = 9 + 10 + 8 = 27 i.e., 27a2, iv) (3 X 5) + ( 2 X 2 ) = 15 + 4 = 19 i.e., 19 a3, v), , 3X2, , =, , 6, , i.e., 6a4, , Hence the product is 6a4 + 19a3 + 27a2 + 26a + 12, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Urdhvatiryagbhyam.html (8 of 11)12/22/2005 8:52:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Paravartya-yojayet, PARĀVARTYA – YOJAYET, ‘Paravartya – Yojayet’ means 'transpose and apply', (i) Consider the division by divisors of more than one digit, and when the divisors, are slightly greater than powers of 10., Example 1 : Divide 1225 by 12., Step 1 : (From left to right ) write the Divisor leaving the first digit,, write the other digit or digits using negative (-) sign and place them, below the divisor as shown., 12, -2, ¯¯¯¯, Step 2 : Write down the dividend to the right. Set apart the last digit, for the remainder., i.e.,,, , 12, -2, , 122, , 5, , Step 3 : Write the 1st digit below the horizontal line drawn under the, dividend. Multiply the digit by –2, write the product below the 2nd digit, and add., i.e.,,, , 12, 122, -2, -2, ¯¯¯¯¯ ¯¯¯¯, 10, , 5, , Since 1 x –2 = -2 and 2 + (-2) = 0, Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’, sum thus obtained by –2 and writes the product under 3rd digit and, add., 12, -2, ¯¯¯¯, , 122, 5, -20, ¯¯¯¯¯¯¯¯¯¯, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/Paravartyayogayet.html (1 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 102, , 5, , Step 5 : Continue the process to the last digit., i.e.,, , 12, -2, ¯¯¯¯¯, , 122, 5, -20, -4, ¯¯¯¯¯¯¯¯¯¯, 102, 1, , Step 6: The sum of the last digit is the Remainder and the result to its, left is Quotient., Thus Q = 102 and R = 1, Example 2 :, , Divide, , 1697 by 14., , 14, -4, ¯¯¯¯, , 169 7, -4–8–4, ¯¯¯¯¯¯¯, 121 3, , Q = 121, R = 3., Example 3 :, , Divide, , 2598 by 123., , Note that the divisor has 3 digits. So we have to set up the last two digits of, the dividend for the remainder., 123, 25, 98, Step ( 1 ) & Step ( 2 ), -2-3, ¯¯¯¯¯, ¯¯¯¯¯¯¯¯, Now proceed the sequence of steps write –2 and –3 as follows :, 123, -2-3, ¯¯¯¯¯, , 25, -4, , 98, -6, -2–3, ¯¯¯¯¯¯¯¯¯¯, 21, 15, , Since, , 2 X (-2, -3)= -4 , -6; 5 – 4 = 1, and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5., Hence Q = 21 and R = 15., Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last 4 digits, of the dividend are to be set up for Remainder., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/Paravartyayogayet.html (2 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 11213, -1-2-1-3, ¯¯¯¯¯¯¯¯, , 23, 9 479, -2 -4-2-6, -1-2-1-3, ¯¯¯¯¯¯¯¯¯¯¯¯¯, 21, 4006, , with 2, with 1, , Hence Q = 21, R = 4006., Example 5 : Divide, , 13456 by 1123, , 112 3, -1–2–3, ¯¯¯¯¯¯¯, , 134 5, 6, -1-2-3, -2-4, –6, ¯¯¯¯¯¯¯¯¯¯¯¯¯, 1 2 0–2, 0, , Note that the remainder portion contains –20, i.e.,, a negative quantity. To over, come this situation, take 1 over from the quotient column, i.e.,, 1123 over to the, right side, subtract the remainder portion 20 to get the actual remainder., Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103., Find the Quotient and Remainder for the problems using paravartya –, yojayet method., 1) 1234 ÷ 112, 2) 11329 ÷ 1132, 3) 12349 ÷ 133, 4) 239479 ÷ 1203, Now let us consider the application of paravartya – yojayet in algebra., Example 1 :, , Divide, , 6x2 + 5x + 4 by x – 1, , X-1, ¯¯¯¯¯¯, 1, , Example 2 :, , Divide, X-5, , 6x2 + 5x + 4, 6 + 11, ¯¯¯¯¯¯¯¯¯¯¯¯, 6x + 11 + 15, , Thus Q = 6x+11, R=15., , x3 – 3x2 + 10x – 4 by x - 5, x3 – 3x2 + 10x –, , 4, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/Paravartyayogayet.html (3 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , ¯¯¯¯¯, 5, , 5, + 10, 100, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, x2 + 2x + 20, + 96, , Thus Q= x2 + 2x + 20, R = 96., The procedure as a mental exercise comes as follows :, i) x3 / x gives x2 i.e.,, 1 the first coefficient in the Quotient., ii) Multiply 1 by + 5,(obtained after reversing the sign of second term, in the Quotient) and add to the next coefficient in the dividend. It gives, 1 X( +5) = +5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is, next coefficient in Quotient., iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to, the next coefficient 10 + 10 = 20. This is next coefficient in Quotient., Thus Quotient is x2 + 2x + 20, iv) Now multiply 20 by + 5 i.e.,, 20 x 5 = 100. Add to the next (last), term,, 100 + (-4) = 96, which becomes R, i.e.,, R =9., Example 3:, x4 – 3x3 + 7x2 + 5x + 7, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, x+4, Now thinking the method as in example ( 1 ), we proceed as follows., x+4, ¯¯¯¯¯, -4, , x4 - 3x3 + 7x2 + 5x +, , 7, , - 4 + 28 - 140 + 540, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, x3 - 7x2 + 35x - 135 547, , Thus Q = x3 – 7x2 + 35x – 135 and R = 547., or we proceed orally as follows:, x4 / x gives 1 as first coefficient., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/Paravartyayogayet.html (4 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7 which gives, next coefficient in Q., ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q., iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q., iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R., Thus Q = x3 – 7x2 + 35x – 135 , R = 547., Note :, 1. We can follow the same procedure even the number of terms is, more., 2. If any term is missing, we have to take the coefficient of the term as, zero and proceed., Now consider the divisors of second degree or more as in the following example., Example :4, , 2x4 – 3x3 – 3x + 2 by x2 + 1., , Here x2 term is missing in the dividend. Hence treat it as 0 . x2 or 0 . And, the x term in divisor is also absent we treat it as 0 . x. Now, x2, +1, x2 + 0 . x + 1, ¯¯¯¯¯¯¯¯¯¯¯¯, 0, -1, , 2x4 - 3x3 + 0 . x2 - 3x + 2, 0, -2, 0, , +3, 0 +2, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, 2 -3, -2, 0, 4, , Thus Q = 2x2 - 3x - 2 and R = 0 . x + 4 = 4., Example 5 :, , 2x5 – 5x4 + 3x2 – 4x + 7 by x3 – 2x2 + 3., , We treat the dividend as 2x5 – 5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as x3, – 2x2 + 0 . x + 3 and proceed as follows :, x3 – 2x2 + 0 . x + 3, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, , 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/Paravartyayogayet.html (5 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 2, , 0, , -3, , 4, , 0, -2, , -6, 0 + 3, -4, 0 + 6, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, 2 - 1, - 2, - 7 - 1 +13, , Thus Q = 2x2 – x – 2, R = - 7 x2 – x + 13., You may observe a very close relation of the method paravartya in this aspect, with regard to REMAINDER THEOREM and HORNER PROCESS of Synthetic division., And yet paravartya goes much farther and is capable of numerous applications in, other directions also., Apply paravartya – yojayet to find out the Quotient and Remainder in, each of the following, problems., 1), 2), 3), 4), 5), , (4x2 + 3x + 5) ÷ (x+1), (x3 – 4x2 + 7x + 6) ÷ (x – 2), (x4 – x3 + x2 + 2x + 4) ÷ (x2 - x – 1), (2x5 + x3 – 3x + 7) ÷ (x3 + 2x – 3), (7x6 + 6x5 – 5x4 + 4x3 – 3x2 + 2x – 1) ÷ (x-1), , Paravartya in solving simple equations :, Recall that 'paravartya yojayet' means 'transpose and apply'. The rule relating to, transposition enjoins invariable change of sign with every change of side. i.e., +, becomes - and conversely ; and X becomes ÷ and conversely. Further it can be, extended to the transposition of terms from left to right and conversely and from, numerator to denominator and conversely in the concerned problems., Type ( i ) :, , Consider the problem 7x – 5 = 5x + 1, 7x – 5x = 1 + 5, i.e.,, 2x = 6, , x = 6 ÷ 2 = 3., , Observe that the problem is of the type ax + b = cx + d from which we get by, ‘transpose’ (d – b), (a – c) and, x=, , d - b., ¯¯¯¯¯¯¯¯, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...cs/MathematicalFormulae/Sutras/Paravartyayogayet.html (6 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 12, 18, _____ + _____, 6x + 3, 6x + 4, , 30, = _____, 6x + 2, , Now proceed., , Simultaneous simple equations:, By applying Paravartya sutra we can derive the values of x and y which are given, by two simultaneous equations. The values of x and y are given by ration form., The method to find out the numerator and denominator of the ratio is given below., Example 1:, , 2x + 3y = 13, 4x + 5y = 23., , i) To get x, start with y coefficients and the independent terms and, cross-multiply forward, i.e.,, right ward. Start from the upper row and, multiply across by the lower one, and conversely, the connecting link, between the two cross-products being a minus. This becomes, numerator., i.e.,, 2x + 3y = 13, 4x + 5y = 23, Numerator of the x – value is 3 x 23 – 5 x 13 = 69 – 65 = 4, ii) Go from the upper row across to the lower one, i.e.,, the xcoefficient but backward, i.e.,, leftward., Denominator of the x – value is 3 x 4 – 2 x 5 = 12 – 10 = 2, Hence value of x = 4 ÷ 2 = 2., iii) To get y, follow the cyclic system, i.e.,, start with the independent, term on the upper row towards the x–coefficient on the lower row. So, numerator of the y–value is, 13 x 4 – 23 x 2 = 52 – 46 = 6., iv) The denominator is the same as obtained in Step(ii) i.e.,, 2. Hence, value of y is 6÷2=3., Thus the solution to the given equation is x = 2 and y = 3., Example 2:, , Now, , 5x – 3y = 11, 6x – 5y = 09, , Nr. of x is (-3) (9) – (5) (11) = - 27 + 55 = 28, Dr. of x is (-3) (6) – (5) (-5) = - 18 + 25 = 07, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/Paravartyayogayet.html (15 of 17)12/22/2005 8:52:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Sankalana-Vyavkalanabhyam, SA•KALANA – VYAVAKALANĀBHYAM, This Sutra means 'by addition and by subtraction'. It can be applied in solving a, special type of simultaneous equations where the x - coefficients and the y coefficients are found interchanged., Example 1:, , 45x – 23y = 113, 23x – 45y = 91, , In the conventional method we have to make equal either the coefficient of x or, coefficient of y in both the equations. For that we have to multiply equation ( 1 ), by 45 and equation ( 2 ) by 23 and subtract to get the value of x and then, substitute the value of x in one of the equations to get the value of y or we have, to multiply equation ( 1 ) by 23 and equation ( 2 ) by 45 and then subtract to get, value of y and then substitute the value of y in one of the equations, to get the, value of x. It is difficult process to think of., From Sankalana – vyavakalanabhyam, add them,, i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91, i.e., 68x – 68y = 204, x–y=3, subtract one from other,, i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91, x+y=1, i.e., 22x + 22y = 22, and repeat the same sutra, we get x = 2 and y = - 1, Very simple addition and subtraction are enough, however big the coefficients, may be., Example 2:, , 1955x – 476y = 2482, 476x – 1955y = -4913, , Oh ! what a problem ! And still, just add, 2431( x – y ) = - 2431, , x – y = -1, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...ematicalFormulae/Sutras/SankalanaVyavkalanabyam.html (1 of 2)12/22/2005 8:53:23 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Calana-Kalanabhyam, CALANA - KALANĀBHYĀM, In the book on Vedic Mathematics Sri Bharati Krishna Tirthaji mentioned the, Sutra 'Calana - Kalanabhyam' at only two places. The Sutra means 'Sequential, motion'., i) In the first instance it is used to find the roots of a quadratic equation 7x2 – 11x, – 7 = 0. Swamiji called the sutra as calculus formula. Its application at that point, is as follows. Now by calculus formula we say: 14x–11 = ±√317, A Note follows saying every Quadratic can thus be broken down into two binomial, factors. An explanation in terms of first differential, discriminant with sufficient, number of examples are given under the chapter ‘Quadratic Equations’., ii) At the Second instance under the chapter ‘Factorization and Differential, Calculus’ for factorizing expressions of 3rd, 4th and 5th degree, the procedure is, mentioned as 'Vedic Sutras relating to Calana – Kalana – Differential Calculus'., Further other Sutras 10 to 16 mentioned below are also used to get the required, results. Hence the sutra and its various applications will be taken up at a later, stage for discussion., But sutra – 14 is discussed immediately after this item., Now the remaining sutras :, 10., 11., 12., 13., 15., 16., , YĀVADŨNAM ( The deficiency ), VYA••ISAMA••IH ( Whole as one and one as whole ), •E•ĀNYA• KENA CARAME•A ( Remainder by the last digit ), SOPĀNTYADVAYAMANTYAM ( Ultimate and twice the penultimate ), GU•ITASAMUCCAYAH ( The whole product is the same ), GU•AKA SAMUCCAYAH ( Collectivity of multipliers ), , The Sutras have their applications in solving different problems in different, contexts. Further they are used along with other Sutras. So it is a bit of, inconvenience to deal each Sutra under a separate heading exclusively and also, independently. Of course they will be mentioned and also be applied in solving the, problems in the forth coming chapter wherever necessary. This decision has been, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...s/MathematicalFormulae/Sutras/CalanaKalanabhyam.html (1 of 2)12/22/2005 8:53:34 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Adyamadyenantya-mantyena, ĀDYAMĀDYENĀNTYA - MANTYENA, The Sutra ' adyamadyenantya-mantyena' means 'the first by the first and the last, by the last'., Suppose we are asked to find out the area of a rectangular card board whose, length and breadth are respectively 6ft . 4 inches and 5 ft. 8 inches. Generally we, continue the problem like this., Area = Length X Breath, = 6’ 4" X 5’ 8" Since 1’ = 12", conversion, = ( 6 X 12 + 4) ( 5 X 12 + 8) in to single unit, = 76" 68" = 5168 Sq. inches., Since 1 sq. ft. =12 X 12 = 144sq.inches we have area, 5168, ¯¯¯¯, 144, , =, , 144) 5168 (35, 432, ¯¯¯¯, 848, 720 i.e., 35 Sq. ft 128 Sq. inches, ¯¯¯¯¯, 128, , By Vedic principles we proceed in the way "the first by first and the last by last", i.e. 6’ 4" can be treated as 6x + 4 and 5’ 8" as 5x + 8,, Where x= 1ft. = 12 in; x2 is sq. ft., Now ( 6x + 4 )(5x + 8 ), = 30x2 + 6.8.x + 4.5.x + 32, = 30x2 + 48x + 20x + 32, = 30x2 + 68. x + 32, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...icalFormulae/Upa-Satras/Adyamadyenantyamamtyena.html (1 of 5)12/22/2005 8:54:03 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , =, =, =, =, =, , 30x2 + ( 5x + 8 ). x + 32 Writing 68 = 5 x 12 + 8, 35x2 + 8. x + 32, 35 Sq. ft. + 8 x 12 Sq. in + 32 Sq. in, 35 Sq. ft. + 96 Sq. in + 32 Sq. in, 35 Sq. ft. + 128 Sq. in, , It is interesting to know that a mathematically untrained and even uneducated, carpenter simply works in this way by mental argumentation. It goes in his mind, like this, 6’, , 4", , 5’, , 8", , First by first i.e. 6’ X 5’ = 30 sq. ft., Last by last i.e. 4" X 8" = 32 sq. in., Now cross wise 6 X 8 + 5 x 4 = 48 +20 = 68., Adjust as many '12' s as possible towards left as 'units' i.e. 68 = 5 X 12 +8 , 5, twelve's as 5 square feet make the first 30+5 = 35 sq. ft ; 8 left becomes 8 x 12, square inches and go towards right i.e. 8 x 12 = 96 sq. in. towards right ives 96, +32 = 128sq.in., Thus he got area in some sort of 35 squints and another sort of 128 sq. units. i.e., 35 sq. ft 128 sq. in, Another Example:, , Now 12 + 2 = 14, 10 x 12 + 24 = 120 + 24 = 144, Thus 4′ 6″ x 3′ 4″ = 14 Sq. ft. 144 Sq. inches., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...icalFormulae/Upa-Satras/Adyamadyenantyamamtyena.html (2 of 5)12/22/2005 8:54:03 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Since 144 sq. in = 12 X 12 = 1 sq. ft The answer is 15 sq. ft., We can extend the same principle to find volumes of parallelepiped also., , I. Find the area of the rectangles in each of the following situations., 1)., , l = 3’ 8" , b = 2’ 4 ", , 2)., , l = 12’ 5" , b = 5’ 7", , 3)., , l = 4 yard 3 ft. b = 2 yards 5 ft.(1yard =3ft), , 4)., , l = 6 yard 6 ft. b = 5 yards 2 ft., , II. Find the area of the trapezium in each of the following cases. Recall, area = ½ h (a + b) where a, b are parallel sides and h is the distance, between them., 1)., , a = 3’ 7", b = 2’ 4", h = 1’ 5", , 2)., , a = 5’ 6", b = 4’ 4", h = 3’ 2", , 3)., , a = 8’ 4", b = 4’ 6", h = 5’ 1"., , Factorization of quadratics:, The usual procedure of factorizing a quadratic is as follows:, 3x2 + 8x + 4, = 3x2 + 6x + 2x + 4, = 3x ( x + 2 ) + 2 ( x + 2 ), = ( x + 2 ) ( 3x + 2 ), But by mental process, we can get the result immediately. The steps are as, follows., i). Split the middle coefficient in to two such parts that the ratio of the, first coefficient to the first part is the same as the ratio of the second, part to the last coefficient. Thus we split the coefficient of middle term, of 3x2 + 8x + 4 i.e. 8 in to two such parts 6 and 2 such that the ratio, of the first coefficient to the first part of the middle coefficient i.e. 3:6, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...icalFormulae/Upa-Satras/Adyamadyenantyamamtyena.html (3 of 5)12/22/2005 8:54:03 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Yavadunam Tavadunikritya Varganca Yojayet, YĀVADŨNAM TĀVADŨNĪK•TYA, VARGAÑCA YOJAYET, The meaning of the Sutra is 'what ever the deficiency subtract that deficit from the, number and write along side the square of that deficit'., This Sutra can be applicable to obtain squares of numbers close to bases of, powers of 10., Method-1 : Numbers near and less than the bases of powers of 10., Eg 1: 92 Here base is 10., The answer is separated in to two parts by a’/’, Note that deficit is 10 - 9 = 1, Multiply the deficit by itself or square it, 12 = 1. As the deficiency is 1, subtract it from the number i.e., 9–1 = 8., 8/1., , Now put 8 on the left and 1 on the right side of the vertical line or slash i.e.,, Hence 81 is answer., , Eg. 2: 962 Here base is 100., Since deficit is 100-96=4 and square of it is 16 and the deficiency subtracted, from the number 96 gives 96-4 = 92, we get the answer 92 / 16 Thus 962 =, 9216., Eg. 3: 9942 Base is 1000, Deficit is 1000 - 994 = 6. Square of it is 36., Deficiency subtracted from 994 gives 994 - 6 = 988, Answer is 988 / 036, , [since base is 1000], , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (1 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Eg. 4:, , 99882, , Base is 10,000., , Deficit = 10000 - 9988 = 12., Square of deficit = 122 = 144., Deficiency subtracted from number = 9988 - 12 = 9976., Answer is 9976 / 0144, Eg. 5:, , 882, , [since base is 10,000]., , Base is 100., , Deficit = 100 - 88 = 12., Square of deficit = 122 = 144., Deficiency subtracted from number = 88 - 12 = 76., Now answer is 76 / 144 = 7744, , [since base is 100], , Algebraic proof:, The numbers near and less than the bases of power of 10 can be treated as (x-y),, where x is the base and y, the deficit., Thus, (1) 9 = (10 -1) (2) 96 = ( 100-4) (3) 994 = (1000-6), (4) 9988 = (10000-12 ) (v) 88 = (100-12), ( x – y )2 = x2 – 2xy + y2, = x ( x – 2y ) + y2, = x ( x – y – y ) + y2, = Base ( number – deficiency ) + ( deficit )2, Thus, 9852 = ( 1000 – 15 )2, = 1000 ( 985 – 15 ) + (15)2, = 1000 ( 970 ) + 225, = 970000 + 225, = 970225., or we can take the identity a2 - b2 = (a + b) ( a - b) and proceed as, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (2 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , a2 - b2 = (a + b) ( a - b)., a2 = (a + b) ( a - b) + b2, , gives, , Thus for a = 985 and b = 15;, a2= (a + b) ( a - b) + b2, 9852 = ( 985 + 15 ) ( 985 - 15 ) + (15)2, = 1000 ( 970 ) + 225, = 970225., Method. 2 : Numbers near and greater than the bases of powers of 10., Eg.(1): 132 ., Instead of subtracting the deficiency from the number we add and proceed, as in Method-1., for 132 , base is 10, surplus is 3., Surplus added to the number = 13 + 3 = 16., Square of surplus = 32 = 9, Answer is 16 / 9 = 169., Eg.(2):, , 1122, , Base = 100, Surplus = 12,, Square of surplus = 122 = 144, add surplus to number = 112 + 12 = 124., Answer is 124 / 144 = 12544, Or think of identity a2 = (a + b) (a – b) + b2, , for a = 112, b = 12:, , 1122 = (112 + 12) (112 – 12) + 122, = 124 (100) + 144, = 12400 + 144, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (3 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , = 12544., (x + y)2 = x2 + 2xy + y2, = x ( x + 2y ) + y2, = x ( x + y + y ) + y2, = Base ( Number + surplus ) + ( surplus )2, gives, 1122 =, =, =, =, Eg. 3:, , 100 ( 112 + 12 ) + 122, 100 ( 124 ) + 144, 12400 + 144, 12544., , 100252, = ( 10025 + 25 ) / 252, = 10050 / 0625, , [ since base is 10,000 ], , = 100500625., Method - 3: This is applicable to numbers which are near to multiples of 10, 100,, 1000 .... etc. For this we combine the upa-Sutra 'anurupyena' and 'yavadunam, tavadunikritya varganca yojayet' together., Example 1:, , 3882 Nearest base = 400., , We treat 400 as 4 x 100. As the number is less than the base we proceed as, follows, Number 388, deficit = 400 - 388 = 12, Since it is less than base, deduct the deficit, i.e. 388 - 12 = 376., multiply this result by 4 since base is 4 X 100 = 400., 376 x 4 = 1504, Square of deficit = 122 = 144., Hence answer is 1504 / 144 = 150544, , [since we have taken multiples of, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (4 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 100]., Example 2: 4852 Nearest base = 500., Treat 500 as 5 x 100 and proceed, , Example 3:, , 672 Nearest base = 70, , Example 4:, , 4162 Nearest ( lower ) base = 400, , Here surplus = 16 and 400 = 4 x 100, , Example 5:, , 50122 Nearest lower base is 5000 = 5 x 1000, , Surplus = 12, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (5 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Apply yavadunam to find the following squares., 1. 72, , 2. 982, , 3. 9872, , 4. 142, , 5. 1162, , 6. 10122, , 7. 192, , 8. 4752, , 9. 7962, , 10. 1082, , 11. 99882, , 12. 60142., , So far we have observed the application of yavadunam in finding the squares of, number. Now with a slight modification yavadunam can also be applied for finding, the cubes of numbers., Cubing of Numbers:, Example : Find the cube of the number 106., We proceed as follows:, i), , For 106, Base is 100. The surplus is 6., Here we add double of the surplus i.e. 106+12 = 118., (Recall in squaring, we directly add the surplus), This makes the left-hand -most part of the answer., i.e. answer proceeds like 118 / - - - - -, , ii) Put down the new surplus i.e. 118-100=18 multiplied by the initial, surplus, i.e. 6=108., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (6 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Since base is 100, we write 108 in carried over form 108 i.e. ., As this is middle portion of the answer, the answer proceeds like, 118 / 108 /...., iii) Write down the cube of initial surplus i.e. 63 = 216 as the last, portion, i.e. right hand side last portion of the answer., Since base is 100, write 216 as 216 as 2 is to be carried over., Answer is 118 / 108 / 216, Now proceeding from right to left and adjusting the carried over, we get the, answer, 119 / 10 / 16 = 1191016., Eg.(1):, , 1023 = (102 + 4) / 6 X 2 / 23, =, , 106, , =, , 12 = 08, , = 1061208., Observe initial surplus = 2, next surplus =6 and base = 100., Eg.(2):, , 943, , Observe that the nearest base = 100. Here it is deficit contrary to the above, examples., i) Deficit = -6. Twice of it -6 X 2 = -12, add it to the number = 94 -12 =82., ii) New deficit is -18., Product of new deficit x initial deficit = -18 x -6 = 108, iii) deficit3 = (-6)3 = -216., __, Hence the answer is 82 / 108 / -216, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (7 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Since 100 is base 1 and -2 are the carried over. Adjusting the carried over in, order, we get the answer, ( 82 + 1 ) / ( 08 – 03 ) / ( 100 – 16 ), = 83, , /, , = 05, , /, , = 84, , = 830584, , __, 16 becomes 84 after taking1 from middle most portion i.e. 100. (100-16=84)., _, Now 08 - 01 = 07 remains in the middle portion, and 2 or 2 carried to it, makes the middle as 07 - 02 = 05. Thus we get the above result., Eg.(3):, 9983 Base = 1000; initial deficit = - 2., 9983 = (998 – 2 x 2) / (- 6 x – 2) / (- 2)3, =, , 994, , /, , = 012, , =, , 994 / 011 / 1000 - 008, , =, , 994 / 011 / 992, , =, , 994011992., , / = -008, , Find the cubes of the following numbers using yavadunam sutra., 1. 105, , 2. 114, , 3. 1003, , 4. 10007, , 5. 92, , 6. 96, , 7. 993, , 8. 9991, , 9. 1000008, , 10. 999992., , 34, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...pa-Satras/YavadunamTavadunikrityaVargancayojayet.html (8 of 8)12/22/2005 8:54:07 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , ÷ x2 gives, , x2 + 2x + 3 ------ (i), , Subtract after multiplying the first by x and the second by 2., Thus, , (2x4 + x3 – 9x) - (2x4 + 4x2 + 18), = x3 - 4x2 – 9x – 18 ------ ( ii ), , Multiply (i) by x and subtract from (ii), x3 – 4x2 – 9x – 18 – (x3 + 2x2 + 3x), = - 6x2 – 12x – 18, ÷ - 6 gives, , x2 + 2x + 3., , Thus ( x2 + 2x + 3 ) is the H.C.F. of the given expressions., Algebraic Proof:, Let P and Q be two expressions and H is their H.C.F. Let A and B the Quotients, after their division by H.C.F., P, Q, i.e., __ = A and __ = B, H, H, , which gives P = A.H and Q = B.H, , P + Q = AH + BH and P – Q = AH – BH, = (A+B).H, = (A–B).H, Thus we can write P ± Q = (A ± B).H, Similarly MP = M.AH and NQ = N.BH gives MP ± NQ = H (MA ± NB), This states that the H.C.F. of P and Q is also the H.C.F. of P±Q or MA±NB., i.e. we have to select M and N in such a way that highest powers and lowest, powers (or independent terms) are removed and H.C.F appears as we have seen, in the examples., , Find the H.C.F. in each of the following cases using Vedic sutras:, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...ematicalFormulae/Upa-Satras/LopanaSthapanabhyam.html (5 of 6)12/22/2005 8:54:34 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , x – y = √ 25 = ± 5, , x + y = 9 gives 7 + y = 9, y = 9 – 7 = 2., Thus the solution is x = 7, y = 2 or x = 2, y = 7., But by Vilokanam, xy = 14 gives x = 2, y = 7 or x = 7, y = 2 and these two sets, satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9. Hence the solution., Example 2:, , 5x – y = 7 and xy = 6., , xy = 6 gives x = 6, y = 1; x = 1, y = 6;, x = 2, y = 3; x = 3, y = 2 and of course negatives of all these., Observe that x = 6, y = 1; x = 1, y = 6: are not solutions because they do not, satisfy the equation 5x – y = 7., But for x = 2, y = 3; 5x – y = 5 (2) – 3 = 10 – 3 = 7 we have 5(3)–2≠7., Hence x = 2, y = 3 is a solution., For x = 3, y = 2 we get 5 (3) – 2 = 15 – 2 ≠ 7., Hence it is not a solution., Negative values of the above are also not the solutions. Thus one set of the, solutions i.e., x = 2, y = 3 can be found. Of course the other will be obtained from, solving 5x – y = 7 and 5x + y = -13., i.e., x = -3 / 5, y = -10., Partial Fractions:, Example 1:, , Resolve, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...matics/MathematicalFormulae/Upa-Satras/Vilokanam.html (4 of 7)12/22/2005 8:54:39 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 2x + 7, ___________, (x + 3) (x + 4), We write, , into partial fractions., , 2x + 7, ____________, (x + 3)(x + 4), , A, ______, (x + 3), , =, , +, , B, ______, (x + 4), , A (x + 4) + B (x + 3), = __________________, (x + 3) (x + 4), 2x + 7, , ≡, , A (x + 4) + B (x + 3)., , We proceed by comparing coefficients on either side, coefficient of x : A + B = 2 ..........(i) X 3, Independent of x : 4A + 3B = 7 .............(ii), Solving (ii) – (i) x 3, , 4A + 3B = 7, 3A + 3B = 6, ___________, A=1, , A = 1 in (i) gives, 1 + B = 2 i.e., B = 1, Or we proceed as, , ≡ A (x + 4) + B (x + 3)., 2 (-3) + 7 ≡ A (-3 + 4) + B (-3 + 3), 2x + 7, , Put, , x = -3,, , ... A = 1., , 1 = A (1), x = -4,, , 2 (- 4) + 7 = A (-4 + 4) + B (-4 + 3), -1 = B(-1), , Thus, , 2x + 7, ____________, (x + 3) (x + 4), , =, , ... B = 1., , 1, 1, _____ + _____, (x + 3), (x + 4), , 2x + 7, But by Vilokanam ____________, (x + 3) (x + 4), , can be resolved as, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...matics/MathematicalFormulae/Upa-Satras/Vilokanam.html (5 of 7)12/22/2005 8:54:39 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Gunita samuccaya - Samuccaya Gunitah, GUNÌTA SAMUCCAYAH - SAMUCCAYA GU•ÌTAH, In connection with factorization of quadratic expressions a sub-Sutra, viz. 'Gunita, samuccayah-Samuccaya Gunitah' is useful. It is intended for the purpose of, verifying the correctness of obtained answers in multiplications, divisions and, factorizations. It means in this context:, 'The product of the sum of the coefficients sc in the factors is equal to the sum of, the coefficients sc in the product', Symbolically we represent as sc of the product = product of the sc (in the factors), Example 1:, , (x + 3) (x + 2) = x2 + 5x + 6, , Now ( x + 3 ) ( x + 2 ) = 4 x 3 = 12 : Thus verified., Example 2:, , (x – 4) (2x + 5) = 2x2 – 3x – 20, , Sc of the product 2 – 3 – 20 = - 21, Product of the Sc = (1 – 4) (2 + 5) = (-3) (7) = - 21. Hence verified., In case of cubics, biquadratics also the same rule applies., We have (x + 2) (x + 3) (x + 4) = x3 + 9x2 + 26x + 24, Sc of the product = 1 + 9 + 26 + 24 = 60, Product of the Sc = (1 + 2) (1 + 3) (1 + 4), = 3 x 4 x 5 = 60. Verified., Example 3:, , (x + 5) (x + 7) (x – 2) = x3 + 10x2 + 11x – 70, (1 + 5) (1 + 7) (1 – 2) = 1 + 10 + 11 – 70, , i.e.,, i.e.,, , 6 x 8 x –1 = 22 – 70, -48 = -48 Verified., , We apply and interpret So and Sc as sum of the coefficients of the odd powers, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...MathematicalFormulae/Upa-Satras/Gunitasamuccaya.html (1 of 3)12/22/2005 8:54:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , We now proceed to deal the Sutras with reference to their variety, applicability,, speed, generality etc. Further we think how 'the element of choice in the Vedic, system, even of innovation, together with mental approach, brings a new, dimension to the study and practice of Mathematics. The variety and simplicity of, the methods brings fun and amusement, the mental practice leads to a more, agile, alert and intelligent mind and innovation naturally follow' (Prof. K.R., Williams, London)., , 34, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...MathematicalFormulae/Upa-Satras/Gunitasamuccaya.html (3 of 3)12/22/2005 8:54:44 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | VedicMathematics A briefing, III. VEDIC MATHEMATICS – A BRIEFING, In the previous chapters we have gone through the Vedic Mathematics Sutras and, upa - Sutras: their application in solving problems. In this approach we have, missed to note some points and merits of one method over the other methods at, some instances., Now we take a few steps in this direction. You may question why this book first, gives examples and methods and then once again try to proceed as if an, introduction to the Vedic Mathematics has been just started. This is because in, this approach the reader first feels that it is easy to solve problems using Vedic, Mathematics. This is clear from the examples given. But the reader may get doubt, why we are doing this way or that way some times very close and almost akin to, the conventional textual way; and some times very different from these, procedures? why new representations and different meanings for the same Sutra, (!) in different contexts? But observe that it is not uncommon to Mathematics., Question some body showing the symbol Π., Majority may say it is 22 / 7 (is it right?) some may say it is a radian measure., Very few may state it is a function or so., What does the representation A X B mean?, A boy thinking about numbers may answer that is A multi plied by B and gives the, product provided A and B are known. A girl thinking of set notation simply says, that it is Cartesian product of the sets A and B. No sort of multiplication at all., Another may conclude that it is a product of two matrices A and B . No doubt a, multiplication but altogether different from above., Some other may go deep in to elementary number theory and may take ' X ' to be, the symbol ' X ' (does not divide) and conclude 'A does not divide B', Now the question arises does a student fail to understand and apply contextual, meaning and representation of symbols and such forms in mathematical writings?, certainly not. In the same way the contextual meanings of the Sutras also can not, bring any problem to the practitioners of Vedic Mathematics., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...u.org/Mathematics/VedicBreifing/VedicMathematics.html (1 of 2)12/22/2005 8:54:49 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Again a careful observation brings all of us to a conclusion that even though the, Sutras are not like mathematical formulae so as to fit in any context under, consideration but they are intended to recognize the pattern in the problems and, suggest procedures to solve. Now recall the terms, rules and methods once again, to fill in some gaps that occur in the previous attempt., , 34, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...u.org/Mathematics/VedicBreifing/VedicMathematics.html (2 of 2)12/22/2005 8:54:49 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Terms and operations, 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, is, , 1 2 3 ..... 8., 9 from 10, 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 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., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (1 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , e) Addition and subtraction using Rekhank., Adding a bar-digit i.e. Rekhank to a digit means the digit is subtracted., _, _, _, e.g: 3 + 1 = 2, 5 + 2 = 3, 4 + 4 = 0, Subtracting a bar - digit i.e. Rekhank to a digit means the digit is added., _, _, _, e.g: 4 - 1 = 5, 6 - 2 = 8, 3 - 3 = 6, 1. Some more examples, e.g: 3 + 4 = 7, , _ _ _, (-2) + (-5) = 2 + 5 = 7 or -7, , f) Multiplication and Division using rekhank., 1. Product of two positive digits or two negative digits ( Rekhanks ), _ _, e.g: 2 X 4 = 8; 4 X 3 = 12 i.e. always positive, 2. Product of one positive digit and one Rekhank, _ _, _, __, e.g: 3 x 2 = 6 or -6; 5 X 3 = 15 or -15 i.e. always Rekhank or negative., 3. Division of one positive by another or division of one Rekhank by another, Rekhank., _ _, e.g: 8 ÷ 2 = 4, 6 ÷ 3 = 2 i.e. always positive, 4. Division of a positive by a Rekhank or vice versa., __, _, _ _, e.g: 10 ÷ 5 = 2, 6 ÷ 2 = 3 i.e. always negative or Rekhank., g) 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., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (2 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Beejank of 348 is 3 + 4 + 8 = 15, Further 1 + 5 = 6. i.e. 6 is Beejank., Beejank of 1567, , 1+5+6+7, , 19, , 1+9, , 1, , i.e. Beejank of 1567 is 1., ii) 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., eg 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., eg 2:, Beejank of 1256847, , 1+2+5+6+8+4+7, , 33, , 3+3, , 6., , But we can cancel 1& 8, 2& 7, 5 & 4 because in each case the sum is 9. Hence, remaining 6 is the Beejank., h) Check by Beejank method:, The Vedic sutra - Gunita Samuccayah gives ‘the whole product is same’. We apply, this sutra in this context as follows. It means that the operations carried out with, the numbers have same effect when the same operations are carried out with, their Beejanks., Observe the following examples., i) 42 + 39, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (3 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Beejanks of 42 and 39 are respectively 4 + 2 = 6 and 3 + 9 = 12 and 1+2=3, Now 6 + 3 = 9 is the Beejank of the sum of the two numbers, Further 42 + 39 = 81. Its Beejank is 8+ 1 = 9., we have checked the correctness., ii), , 64 + 125., 64, 125, , 6+4, , 10, , 1+0, , 1+2+5, , 1, , 8, , Sum of these Beejanks 8 + 1 = 9, Note that, 64 + 125 = 189, , 1+8+9, , 18, , 1+8, , 9, , we have checked the correctness., iii) 134 - 49, 134, 49, , 1+3+4, 4+9, , 8, , 13, , 1+3, , Difference of Beejanks 8 -4, Beejanks of 85 is 8 + 5, , 4, , 4, note that 134 – 49 = 85, 85, , 8+5, , 13, , 1+3, , 4 verified., , iv) 376 - 284, 376, , 7(, , 6+3, , 284, , 2+8+4, , 9), 14, , 1+4, , 5, , Difference of Beejanks = 7 – 5 = 2, 376 – 284 = 92, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (4 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Beejank of 92, , 9+2, , 11, , 1+1, , 2, , Hence verified., v) 24 X 16 = 384, Multiplication of Beejanks of, 24 and 16 is 6 X 7 = 42, Beejank of 384, , 4+2, , 3+8+4, , 6, , 15, , 1+5, , 6, , 12, , 1+2, , 3, , Hence verified., vi) 237 X 18 = 4266, Beejank of 237, Beejank of 18, , 2+3+7, 1+8, , 9, , Product of the Beejanks = 3 X 9, Beejank of 4266, , 27, , 4+2+6+6, , 2+7, 18, , 9, , 1+8, , 9, , Hence verified., vii) 242 = 576, Beejank of 24, , 2+4, , square of it 62, , 36, , Beejank of result = 576, , 6, 9, 5+7+6, , 18, , 1+8, , 9, , Hence verified., viii) 3562 = 126736, Beejank of 356, , 3+5+6, , 5, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (5 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Square of it = 52 = 25, , 2+5, , Beejank of result 126736, (, , 7, , 1+2+6+7+3+6, , 1+6, , 7, , 7 + 2 = 9; 6 + 3 = 9) hence verified., , ix) Beejank in Division:, Let P, D, Q and R be respectively the dividend, the divisor, the quotient and the, remainder., Further the relationship between them is P = ( Q X D ) + R, eg 1: 187 ÷ 5, we know that 187 = ( 37 X 5 ) + 2 now the Beejank check., We know that 187 = (37 X 5) +2 now the Beejank check., 187, , 1+8+7, , (37 X 5) + 2, , 7(, , 1 + 8 = 9), , Beejank [(3 + 7) X 5] + 2, 5+2, , 7, , Hence verified., eg 2: 7986 ÷ 143, 7896 = (143 X 55) + 121, Beejank of 7986, , 7+9+8+6, , (, , 2+1, , 9 is omitted), , Beejank of 143 X 55, 8 X 10, , 80, , 21, , 3, , (1 + 4 + 3) (5 + 5), (8 + 0), , Beejank of (143 X 55) + 121, , 8, 8 + (1 + 2 + 1), , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (6 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 8+4, , 12, , 1+2, , 3, , hence verified., , Check the following results by Beejank method, 1. 67 + 34 + 82 = 183, , 2. 4381 - 3216 = 1165, , 3. 632 = 3969, , 4. (1234)2 = 1522756, , 5. (86X17) + 34 = 1496, , 6. 2556 ÷ 127 gives Q =20, R = 16, , i) Vinculum : The numbers which by presentation contains both, positive and negative digits are called vinculum numbers., ii) 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 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, _, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (7 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , eg 4: 315., sankalanam (addition) = 315+315 = 630., , _, Vyvakalanam (subtraction) = 630 - 315 = 325, Working steps :, , _, 0-5=5, 3-1=2, 6-3=3, , Let’s apply this sutra in the already taken example 47768., Samkalanam = 47768 + 47768 = 95536, Vyavakalanam = 95536 - 47768., , Consider the convertion by sankalanavyavakalanabhyam and check it by Ekadhika, and Nikhilam., eg 5: 12637, 1. Sankalana ....... gives, 12637 + 12637 = 25274, , __, 25274 – 12637 = (2 – 1) / (5 – 2) / (2 – 6) / (7 – 3) / (4 – 7) = 13443, , 2. Ekadhika and Nikhilam gives the following., As in the number 1 2 6 3 7, the smaller and bigger digits (i.e. less than 5 and; 5,, greater than 5) are mixed up, we split up in to groups and conversion is made up, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (9 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , as given below., Split 1 2 6 and 3 7, , _, _, Now the sutra gives 1 2 6 as 134 and 37 as 43, _ _, Thus 12637 = 13443, , _, Now for the number 315 we have already obtained vinculum as 325 by, "sankalana ... " Now by ‘Ekadhika and Nikhilam ...’ we also get the same answer., 315 Since digits of 31 are less than 5,, We apply the sutras on 15 only as, Ekadhika of 1 is 2 and Charman of 5 is 5 ., Consider another number which comes under the split process., eg 6:, , 24173, , As both bigger and smaller numbers are mixed up we split the number 24173 as, 24 and 173 and write their vinculums by Ekadhika and Nikhilam sutras as, _, __, 24 = 36 and 173 = 227, _ __, Thus 24173 = 36227, , Convert the following numbers into viniculum number by, i. Ekadhika and Nikhilam sutras ii. Sankalana vyavakalana sutra., Observe whether in any case they give the same answer or not., 1. 64, , 2. 289, , 3. 791, , 4. 2879, , 5. 19182, , 6. 823672, , 7. 123456799, , 8. 65384738, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (10 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , ii) Conversion of vinculum number into general numbers., The process of conversion is exactly reverse to the already done., Rekhanks are converted by Nikhilam where as other digits by, ‘Ekanyunena’ sutra. thus:, _, i) 12 = (1 – 1) / (10 – 2) Ekanyunena 1 – 1, _, = 08 = 8, Nikhilam. 2 = 10 – 2, __, ii) 326 = (3 – 1) / (9 – 2) / (10 – 6), = 274, _ _, iii) 3344 = (3 – 1) / (10 – 3) / (4 – 1) / (10 – 4), = 2736, (note the split), __ __, iv) 20340121 = 2/(0–1)/(9–3)/(10–4)/(0–1)/(9–1)/(10–2)/1, _ _, = 21661881, _, _, = 21 / 6 / 61 / 881. once again split, = (2 – 1) / (10 –1) / 6 / (6 –1) / (10 –1) / 881, = 19659881, , ___, v) 303212 = 3 / 0321 / 2, = 3 / (0-1) / (9-3) / (9-2) / (10-1) / 2, _, 3 / 1 / 6792, (3 –1) / (10 –1) / 6792, = 296792., iii) Single to many conversions., It is interesting to observe that the conversions can be shown in many, ways., eg 1: 86 can be expressed in following many ways, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (11 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Beejank of 216, , 2 + ( -1 ) + 6, , 7., , Thus verified., , But there are instances at which, if beejank of vinculum number is rekhank i.e., negative. Then it is to be converted to +ve number by adding 9 to Rekhank, ( already we have practised) and hence 9 is taken as zero, or vice versa in finding, Beejank., eg:, , __, 213 = 200 - 13 = 187, _, Now Beejank of 213 = 2 + ( -1 ) + (-3 ) = -2, Beejank of 187 = 1 + 8 + 7 = 16, , 1+6=7, , The variation in answers can be easily understood as, _ _, 2=2+9, - 2 + 9 = 7 Hence verified., , Use Vedic check method of the verification of the following result., _, _ _, 1. 24 = 36, 2. 2736 = 3344., __, _ _, 3. 326 = 274, 4. 23213 = 17187, , Addition and subtraction using vinculum numbers., eg 1: Add 7 and 6 i.e., 7+6., i) Change the numbers as vinculum numbers as per rules already discussed., _, _, { 7 = 13 and 6 = 14 }, ii) Carry out the addition column by column in the normal process, moving from, top to bottom or vice versa., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (13 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , iii) add the digits of the next higher level i.e.,, 1 + 1 = 2, _, 13, _, 14, ____, _, 27, iv) the obtained answer is to be normalized as per rules already explained. rules, already explained., _, i.e., 27 = (2 - 1) (10- 7) = 13 Thus we get 7 + 6 = 13., eg 2 : Add 973 and 866., _, 973 = 1 0 3 3, ___, 866 = 1 1 3 4, , ___, But 2161 = 2000 - 161 = 1839., , _, 1033, ___, 1134, ______, ___, 2161, , Thus 973+866 by vinculum method gives 1839 which is correct., Observe that in this representation the need to carry over from the previous digit, to the next higher level is almost not required., eg 3 : Subtract 1828 from 4247., i.e.,, 4247, -1828, ______, , ____, Step (i) : write –1828 in Bar form i.e.,, 1828, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...u.org/Mathematics/VedicBreifing/Termsoperations.html (14 of 15)12/22/2005 8:55:08 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Addition and subtraction, ADDITION AND SUBTRACTION, ADDITION:, In the convention process we perform the process as follows., 234 + 403 + 564 + 721, write as 234, 403, 564, 721, Step (i): 4 + 3 + 4 + 1 = 12, , 2 retained and 1 is carried over to left., , Step (ii): 3 + 0 + 6 + 2 = 11, , the carried ‘1’ is added, , i.e., Now 2 retained as digit in the second place (from right to left) of the answer, and 1 is carried over to left., step (iii): 2 + 4 + 5 + 7 = 18, , carried over ‘1’ is added, , i.e., 18 + 1 = 19. As the addition process ends, the same 19 is retained in the left, had most part of the answer., thus, , 234, 403, 564, +721, _____, 1922, , is the answer, , we follow sudhikaran process Recall ‘sudha’ i.e., dot (.) is taken as an upa-sutra, (No: 15), consider the same example, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (1 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , i) Carry out the addition column by column in the usual fashion, moving from, bottom to top., (a) 1 + 4 = 5, 5 + 3 = 8, 8 + 4 = 12 The final result is more than 9., The tenth place ‘1’ is dropped once number in the unit place i.e., 2, retained. We say at this stage sudha and a dot is above the top 4. Thus, column (1) of addition (right to left), , ., , 4, 3, 4, 1, __, 2, b) Before coming to column (2) addition, the number of dots are to be, counted, This shall be added to the bottom number of column (2) and, we proceed as above., Thus second column becomes, , ., , 3, 0, 6, 2, __, 2, , dot=1,, , 1+2=3, 3+6=9, 9+0=9, 9 + 3 = 12, , 2 retained and ‘.’ is placed on top number 3, c) proceed as above for column (3), 2, 4, , i) dot = 1, iii) 8 + 5 = 13, , 5, , A dot is placed on 5 and proceed, , ., , ii) 1 + 7 = 8, iv) Sudha is said., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (2 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 7, __, 9, , with retained unit place 3., v) 3+4=7,7+2=9 Retain 9 in 3rd digit i.e.,in 100th place., , d) Now the number of dots is counted. Here it is 1 only and the, number is carried out left side ie. 1000th place, , .., , Thus, , 234, 403, , ., , 564, +721, _____, 1922, , is the answer., , Though it appears to follow the conventional procedure, a careful, observation and practice gives its special use., eg (1):, , ., 437, , . ., , 624, , ., , 586, +162, ______, 1809, Steps 1:, i) 2 + 6 = 8, 8 + 4 = 12 so a dot on 4 and 2 + 7 = 9 the answer, retained under column (i), ii) One dot from column (i) treated as 1, is carried over to column (ii),, thus 1 + 6 = 7, 7 + 8 = 15 A' dot’; is placed on 8 for the 1 in 15 and, the 5 in 15 is added to 2 above., 5 + 2 = 7, 7 + 3 = 10 i.e. 0 is written under column (ii) and a dot for, the carried over 1 of 10 is placed on the top of 3., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (3 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , (iii) The number of dots counted in column (iii) are 2., Hence the number 2 is carried over to column (ii) Now in column (iii), 2 + 1 = 3, 3 + 5 = 8, 8 + 6 = 14 A dot for 1 on the number 6 and 4 is, retained to be added 4 above to give 8. Thus 8 is placed under column, (iii)., iv) Finally the number of dots in column (iii) are counted. It is ‘1’ only., So it carried over to 1000th place. As there is no fourth column 1 is the, answer for 4th column. Thus the answer is 1809., Example 3:, , Check the result verify these steps with the procedure mentioned above., The process of addition can also be done in the down-ward direction i.e., addition, of numbers column wise from top to bottom, Example 1:, , Step 1:, , 6 + 4 = 10, 1 dot ; 0 + 8 = 8; 8 + 4 = 12;, 1 dot and 2 answer under first column - total 2 dots., , Step 2: 2+2 (, , 2 dots) = 4; 4+9 = 13: 1 dot and 3+0 = 3; 3+8 = 11;, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (4 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 1 dot and 1 answer under second column - total 2 dots., Step 3: 3+2 (, , 2 dots ) = 5; 5+6 = 11:1 dot and 1+7 = 8; 8+7 = 15;, , 1 dot and 5 under third column as answer - total 2 dots., Step 4: 4 + 2 (, , 2 dots ) = 6; 6 + 5 =11:, , 1 dot and 1+3 = 4; 4+2 = 6. - total 1 dot in the fourth 6 column as answer., Step 5: 1 dot in the fourth column carried over to 5th column (No digits in it) as 1, Thus answer is from Step5 to Step1; 16512, Example 2:, , Steps, (i): 8 + 9 = 17; 7 + 4 = 11; 1 + 1 = (2), , (2dots), , (ii): 7 + 2 = 9; 9 + 1 = 10; 0 + 8 = 8, 8 + 9 = 17, (7), , (2dots), , (iii): 2 + 2 = 4; 4 + 6 = 10; 0 + 0 = 0; 0 + 7 = (7), , (1 dot), , (iv): 3 + 1 = 4; 4 + 4 = 8; 8 + 3 = 11; 1 + 1 = (2), , (1 dot), , (v): 1, Thus answer is 12772., , Add the following numbers use ‘Sudhikaran’ whereever applicable., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (5 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 1., , 486, 395, 721, +609, ¯¯¯¯¯, ¯¯¯¯¯, , 2., , 5432, 3691, 4808, +6787, ¯¯¯¯¯¯, ¯¯¯¯¯¯, , 3., , 968763, 476509, +584376, ¯¯¯¯¯¯¯¯, ¯¯¯¯¯¯¯¯, , Check up whether ‘Sudhkaran’ is done correctly. If not write the correct, process. In either case find the sums., , SUBTRACTION:, The ‘Sudha’ Sutra is applicable where the larger digit is to be subtracted from the, smaller digit. Let us go to the process through the examples., Procedure:, i) If the digit to be subtracted is larger, a dot ( sudha ) is given to its left., ii) The purak of this lower digit is added to the upper digit or purak-rekhank of this, lower digit is subtracted., file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (6 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Example (i): 34 - 18, 34, , ., , -18, _____, , ., , Steps: (i): Since 8>4, a dot is put on its left i.e. 1, (ii) Purak of 8 i.e. 2 is added to the upper digit i.e. 4, _, 2 + 4 = 6. or Purak-rekhank of 8 i.e. 2 is, _, Subtracted from i.e. 4 - 2 =6., Now at the tens place a dot (means1) makes the ‘1’ in the number into 1+1=2., This has to be subtracted from above digit. i.e. 3 - 2 = 1. Thus, 34, , ., , -18, _____, 16, Example 2:, 63, , ., , -37, _____, , ., , Steps: (i) 7>3. Hence a dot on left of 7 i.e., 3, (ii) Purak of 7 i.e. 3 is added to upper digit 3 i.e. 3+3 = 6., This is unit place of the answer., Thus answer is 26., Example (3) :, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (7 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 3274, , .., , -1892, _______, Steps:, (i) 2 < 4. No sudha . 4-2 = 2 first digit (form right to left), , ., , (ii) 9 > 7 sudha required. Hence a dot on left of 9 i.e. 8, (iii) purak of 9 i.e. 1, added to upper 7 gives 1 + 7 = 8 second digit, , ., , (iv) Now means 8 + 1 = 9., , ., , (v) As 9 > 2, once again the same process: dot on left of i.e., 1, (vi) purak of 9 i.e. 1, added to upper 2 gives 1 + 2 = 3, the third digit., , ., , (vii) Now 1 means 1+1 = 2, (viii) As 2 < 3, we have 3-2 = 1, the fourth digit, Thus answer is 1382, Vedic Check :, Eg (i) in addition : 437 + 624 + 586 + 162 = 1809., By beejank method, the Beejanks are, 437, , 4+3+7, , 14, , 1+4, , 5, , 624, , 6+2+4, , 12, , 1+2, , 3, , 586, , 5+8+6, , 19, , 1+9, , 10, , 162, , 1+6+2, , 9, , 1+0, , 1, , Now, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (8 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 437 + 624 + 586 + 162, Beejank of 1809, , 5+3+1+9, , 1+8+0+9, , 18, , 18, , 1+8, , 1+8, , 9, , 9 verified, , Eg.(3) in subtraction :, 3274 – 1892 = 1382, now beejanks, 3274, , 3+2+7+4, , 3+4, , 1892, , 1+8+9+2, , 2, , 3292-1892, , 7-2, , 5, , 1+3+8+2, , 1382, , 7, , 5 Hence verified., , Mixed addition and subtraction using Rekhanks:, Example 1 : 423 - 654 + 847 - 126 + 204., In the conventional method we first add all the +ve terms, 423 + 847 + 204 = 1474, Next we add all negative terms, - 654 - 126 = -780, At the end their difference is taken, 1474 - 780 = 694, Thus in 3 steps we complete the problem, But in Vedic method using Rekhank we write and directly find the answer., 423, ___, 654, 847, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...rg/Mathematics/VedicBreifing/Additionsubtraction.html (9 of 10)12/22/2005 8:55:13 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Multiplication, MULTIPLICATION, We have already observed the application of Vedic sutras in multiplication. Let us, recall them., It enables us to have a comparative study of the applicability of these methods, to, assess advantage of one method over the other method and so-on., Example (i) : Find the square of 195., The Conventional method :, 1952, , =, , 195, x 195, ______, 975, 1755, 195, _______, 38025, ¯¯¯¯¯¯¯, , (ii) By Ekadhikena purvena, since the number ends up in 5 we write the answer, split up into two parts., The right side part is 52 where as the left side part 19 X (19+1) (Ekhadhikena), Thus 1952 = 19 X 20/52 = 380/25 = 38025, (iii) By Nikhilam Navatascaramam Dasatah; as the number is far from base 100,, we combine the sutra with the upa-sutra ‘anurupyena’ and proceed by taking, working base 200., a) Working Base = 200 = 2 X 100., Now 1952 = 195 X 195, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...amu.org/Mathematics/VedicBreifing/Multiplication.html (1 of 6)12/22/2005 8:55:33 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , _, 101, _, x 101, ______, _, 10201, , = 9801, , 3) By Nikhilam method, 99 -1, 99 -1, _________, 98 / 01 = 9801., 4) ‘Yadunam’ sutra : 992 Base = 100, Deficiency is 1 : It indicates 992 = (99 – 1) / 12 = 98 / 01 = 9801., In the above examples we have observed how in more than one way problems can, be solved and also the variety. You can have your own choice for doing, multiplication. Not only that which method suits well for easier and quicker, calculations. Thus the element of choice, divergent thinking, insight into properties, and patterns in numbers, natural way of developing an idea, resourcefulness play, major role in Vedic Mathematics methods., , 34, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...amu.org/Mathematics/VedicBreifing/Multiplication.html (6 of 6)12/22/2005 8:55:33 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Division, DIVISION, In the conventional procedure for division, the process is of the following form., Quotient, _______, Divisor ) Dividend, ------------------_________, Remainder, , or, , Divisor ) Dividend ( Quotient, ------------------_________, Remainder, , But in the Vedic process, the format is, Divisor ) Dividend, --------------__________________, Quotient / Remainder, The conventional method is always the same irrespective of the divisor. But Vedic, methods are different depending on the nature of the divisor., Example 1: Consider the division 1235 ÷ 89., i) Conventional method:, 89 ) 1235 ( 13, 89, _____, 345, 267, Thus Q = 13 and R = 78., _____, 78, ii) Nikhilam method:, This method is useful when the divisor is nearer and less than the base. Since for, 89, the base is 100 we can apply the method. Let us recall the nikhilam division, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (1 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , already dealt., Step (i):, Write the dividend and divisor as in the conventional method. Obtain the, modified divisor (M.D.) applying the Nikhilam formula. Write M.D. just below the, actual divisor., Thus for the divisor 89, the M.D. obtained by using Nikhilam is 11 in the last from, 10 and the rest from 9. Now Step 1 gives, 89 ) 1235, __, 11, Step (ii):, Bifurcate the dividend by by a slash so that R.H.S of dividend contains the, number of digits equal to that of M.D. Here M.D. contains 2 digits hence, 89 ) 12 / 35, __, 11, Step (iii): Multiply the M.D. with first column digit of the dividend. Here it is 1. i., e. 11 x 1 = 11. Write this product place wise under the 2nd and 3rd columns of, the dividend., 89 ) 12 / 35, __, 11, 1 1, Step (iv):, Add the digits in the 2nd column and multiply the M.D. with that result i.e. 2, +1=3 and 11x3=33. Write the digits of this result column wise as shown below,, under 3rd and 4th columns. i.e., 89 ) 12 / 35, __, 11, 1 1, 33, _______, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (2 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 13 / 78, Now the division process is complete, giving Q = 13 and R = 78., Example 2: Find Q and R for 121134 ÷ 8988., Steps (1+2):, 8988 ) 12 / 1134, ____, 1012, Step (3):, 8988 ) 12 / 1134, ____, 1012, 1 012, Step(4):, 8988 ) 12 / 1134, ____, 1012, 1 012, [ 2 + 1 = 3 and 3x1012 = 3036 ], 3036, Now final Step, 8988 ) 12 / 1134, ____, 1012 1, 012, 3036(Column wise addition), _________, 13 / 4290, Thus 121134¸ 8988 gives Q = 13 and R = 4290., iii) Paravartya method: Recall that this method is suitable when the divisor is, nearer but more than the base., Example 3: 32894 ÷ 1028., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (3 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , The divisor has 4 digits. So the last 3 digits of the dividend are set apart for the, remainder and the procedure follows., , Now the remainder contains -19, -12 i.e. negative quantities. Observe that 32 is, quotient. Take 1 over from the quotient column i.e. 1x1028 = 1028 over to the, right side and proceed thus: 32 - 1 = 31 becomes the Q and R = 1028+200 - 190, - 12 =1028-2 =1026., Thus 3289 ÷ 1028 gives Q = 31 and R = 1026., The same problem can be presented or thought of in any one of the following, forms., , _, *Converting the divisor 1028 into vinculum number we get 1028 = 1032 Now, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (4 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , get, , __, *Converting dividend into vinculum number 32894 = 33114 and proceeding we, , Now we take another process of division based on the combination of Vedic sutras, urdhva-tiryak and Dhvjanka. The word Dhvjanka means " on the top of the flag", Example 4: 43852 ÷ 54., Step1: Put down the first digit (5) of the divisor (54) in the divisor column as, operator and the other digit (4) as flag digit. Separate the dividend into two parts, where the right part has one digit. This is because the falg digit is single digit. The, representation is as follows., 5, , 4:4385:2, , Step2: i) Divide 43 by the operator 5. Now Q= 8 and R = 3. Write this Q=8 as, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (5 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , the 1st Quotient - digit and prefix R=3, before the next digit i.e. 8 of the dividend,, as shown below. Now 38 becomes the gross-dividend ( G.D. ) for the next step., 4:43 85:2, 5 :, 3, ________________, : 8, ii) Subtract the product of falg digit (4) and first quotient digit (8) from the G.D., (38) i.e. 38-(4X8)=38-32=6. This is the net - dividend (N.D) for the next step., Step3: Now N.D Operator gives Q and R as follows. 6 ÷ 5, Q = 1, R = 1. So Q =, 1, the second quotient-digit and R - 1, the prefix for the next digit (5) of the, dividend., 4:43 8 5:2, 5 :, 3 1, ________________, : 8 1, Step4: Now G.D = 15; product of flag-digit (4) and 2nd quotient - digit (1) is, 4X1=4 Hence N.D=15-4=11 divide N.D by 5 to get 11 ÷ 5, Q = 2, R= 1. The, representation is, 4:43 8 5:2, 5 :, 3 1 :1, ________________, : 8 1 2:, Step5: Now the R.H.S part has to be considered. The final remainder is obtained, by subtracting the product of falg-digit (4)and third quotient digit (2) form 12 i.e.,, 12:, Final remainder = 12 - (4 X 2) = 12 - 8 = 4. Thus the division ends into, 4:43 8 5:2, 5 :, 3 1 :1, ________________, : 8 1 2:4, Thus 43852 ÷ 54 gives Q = 812 and R = 4., Consider the algebraic proof for the above problem. The divisor 54 can be, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (6 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Step1: We take the divisor 524 as 5, the operator and 24, the flag-digit and, proceed as in the above example. We now seperate the dividend into two parts, where the RHS part contains two digits for Remainder., Thus, , 24 : 2 3 7 9 : 63, 5, , Step2:, i) 23÷5 gives Q = 4 and R = 3, G.D = 37., ii) N.D is obtained as, , = 37 – ( 8 + 0), = 29., Representation, , 24 : 2 3 7 9 : 63, 5, 3, _________________, :4, , Step3:, i) N.D ÷ Operator = 29 ÷ 5 gives Q = 5, R = 4 and G.D = 49., ii) N.D is obtained as, , = 49 – (10 + 16), = 49 – 26, = 23., i.e.,, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (8 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 24 : 2 3 7 9 : 63, 5 :, 3 4 :, _________________, : 4 5, :, Step 4:, i) N.D ÷ Operator = 23 ÷ 5 gives Q = 4, R = 3 and G.D = 363., Note that we have reached the remainder part thus 363 is total sub–remainder., 24 : 2 3 7 9 : 63, 5 :, 3 4 :3, _________________, : 4 5 4 :, Step 5: We find the final remainder as follows. Subtract the cross-product of the, two, falg-digits and two last quotient-digits and then vertical product of last flagdigit with last quotient-digit from the total sub-remainder., i.e.,,, , Note that 2, 4 are two falg digits: 5, 4 are two last quotient digits:, , represents the last flag - digit and last quotient digit., Thus the division 237963 ÷ 524 gives Q = 454 and R = 67., Thus the Vedic process of division which is also called as Straight division is a, simple application of urdhva-tiryak together with dhvajanka. This process has, many uses along with the one-line presentation of the answer., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20ma...w.vedamu.org/Mathematics/VedicBreifing/Division.html (9 of 10)12/22/2005 8:55:50 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Miscellaneous Items, MISCELLANEOUS ITEMS, 1. Straight Squaring:, We have already noticed methods useful to find out squares of numbers. But the, methods are useful under some situations and conditions only. Now we go to a, more general formula., The sutra Dwandwa-yoga (Duplex combination process) is used in two different, meanings. They are i) by squaring ii) by cross-multiplying., We use both the meanings of Dwandwa-yoga in the context of finding squares of, numbers as follows:, We denote the Duplex of a number by the symbol D. We define for a single digit, ‘a’, D =a2. and for a two digit number of the form ‘ab’, D =2( a x b ). If it is a 3, digit number like ‘abc’, D =2( a x c ) + b2., For a 4 digit number ‘abcd’, D = 2( a x d ) + 2( b x c ) and so on. i.e. if the digit is, single central digit, D represents ‘square’: and for the case of an even number of, digits equidistant from the two ends D represent the double of the cross- product., Consider the examples:, Number, , DuplexD, , 3, , 32 = 9, , 6, , 62 = 36, 2 (2 x 3) = 12, 2 (6 x 4) = 48, , 23, 64, 128, , 2 (1 x 8) + 22 = 16 + 4 = 20, , 305, , 2 (3 x 5) + 02 = 30 + 0 = 30, 2 (4 x 1) + 2 (2 x 3) = 8 + 12 = 20, 2 (7 x 6) + 2 (3 x 4) = 84 + 24 = 108, , 4231, 7346, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (1 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Further observe that for a n- digit number, the square of the number contains 2n, or 2n-1 digits. Thus in this process, we take extra dots to the left one less than, the number of digits in the given numbers., Examples:1 622 Since number of digits = 2, we take one extra dot to the left., Thus, .62, ____, 644, 32, _____, , 3844, , for 2, D = 22 = 4, for 62, D = 2 x 6 x 2 = 24, for 62, D = 2(0 x 2) + 62, = 36, , 622 = 3844., Examples:2 2342 Number of digits = 3. extradots =2 Thus, ..234, _____, 42546, 1221, _____, 54756, , for 4, D = 42 = 16, for 34, D = 2 x 3 x 4 = 24, for 234, D = 2 x 2 x 4 + 32 = 25, for .234, D = 2.0.4 + 2.2.3 = 12, for ..234, D = 2.0.4 + 2.0.3 + 22 = 4, , Examples:3 14262. Number of digits = 4, extra dots = 3, i.e, , ...1426, ________, 1808246, 22523, _________, 2033476, , 6, D = 36, 26, D = 2.2.6 = 24, 426, D = 2.4.6 + 22 = 52, 1426,, .1426,, ..1426,, ...1426,, , D = 2.1.6 + 2.4.2 = 28, D = 2.0.6 + 2.1.2 + 42 = 20, D = 2.0.6 + 2.0.2 + 2.1.4 = 8, D = 12 = 1, , Thus 14262 = 2033476., With a little bit of practice the results can be obtained mentally as a single line, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (2 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , answer., Algebraic Proof:, Consider the first example 622, Now 622 = (6 x 10 + 2)2 = (10a + b)2 where a = 6, b = 2, = 100a2 + 2.10a.b + b2, = a2 (100) + 2ab (10) + b2, i.e. b2 in the unit place, 2ab in the 10th place and a2 in the 100th place i.e. 22 = 4, in units place, 2.6.2 = 24 in the 10th place (4 in the 10th place and with carried, over to 100th place). 62=36 in the 100th place and with carried over 2 the 100th, place becomes 36+2=38., Thus the answer 3844., , Find the squares of the numbers 54, 123, 2051, 3146., Applying the Vedic sutra Dwanda yoga., , 2.CUBING, Take a two digit number say 14., i) Find the ratio of the two digits i.e. 1:4, ii) Now write the cube of the first digit of the number i.e. 13, iii) Now write numbers in a row of 4 terms in such a way that the first, one is the cube of the first digit and remaining three are obtained in a, geometric progression with common ratio as the ratio of the original, two digits (i.e. 1:4) i.e. the row is, 1, , 4, , 16, , 64., , iv) Write twice the values of 2nd and 3rd terms under the terms, respectively in second row., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (3 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , i.e.,, , 1, , 4, 8, , 16, 32, , 64, , (, , 2 x 4 = 8, 2 x 16 = 32), , v) Add the numbers column wise and follow carry over process., 1, , 4 16 64 Since 16 + 32 + 6 (carryover) = 54, 8 32, 4 written and 5 (carryover) + 4 + 8 = 17, ______________, 2 7, 4, 4, 7 written and 1 (carryover) + 1 = 2., This 2744 is nothing but the cube of the number 14, Example 1: Find 183, , Example 2: Find 333, , Algebraic Proof:, Let a and b be two digits., Consider the row a3, , a2b, , ab2, , b3, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (4 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , the first is a3 and the numbers are in the ratio a:b, since a3:a2b=a2b:b3=a:b, Now twice of a2b, ab2 are 2a2b, 2ab2, a3 +, , a2b + ab2 + b3, 2a2b + 2ab2, , ________________________________, a3 + 3a2b + 3ab2 + b3 = (a + b)3., , Thus cubes of two digit numbers can be obtained very easily by using the vedic, sutra ‘anurupyena’. Now cubing can be done by using the vedic sutra ‘Yavadunam’., Example 3:, , Consider 1063., , i) The base is 100 and excess is 6. In this context we double the excess and then, add., i.e. 106 + 12 = 118. (, , 2 X 6 =12 ), , This becomes the left - hand - most portion of the cube., i.e. 1063 = 118 / - - - ii) Multiply the new excess by the initial excess, i.e. 18 x 6 = 108 (excess of 118 is 18), Now this forms the middle portion of the product of course 1 is carried over, 08, in the middle., i.e. 1063 = 118 / 08 / - - - - 1, iii) The last portion of the product is cube of the initial excess., i.e. 63 = 216., 16 in the last portion and 2 carried over., i.e. 1063 = 118 / 081 / 16 = 1191016, , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (5 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 1, , 2, , Example 4: Find 10023., i) Base = 1000. Excess = 2. Left-hand-most portion of the cube becomes 1002, +(2x2)=1006., ii) New excess x initial excess = 6 x 2 = 12., Thus 012 forms the middle portion of the cube., iii) Cube of initial excess = 23 = 8., So the last portion is 008., Thus 10023 = 1006 / 012 / 008 = 1006012008., Example 5: Find 943., i) Base = 100, deficit = -6. Left-hand-most portion of the cube becomes 94+(2x-6), =94-12=82., ii) New deficit x initial deficit = -(100-82)x(-6)=-18x-6=108, Thus middle potion of the cube = 08 and 1 is carried over., iii) Cube of initial deficit = (-6)3 = -216, __, __, 3, Now 94 = 82 / 08 / 16 = 83 / 06 / 16, _, 1, 2, = 83 / 05 / (100 – 16), = 830584., , Find the cubes of the following numbers using Vedic sutras., 103, 112, 91, 89, 998, 9992, 1014., , file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (6 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , 3. Equation of Straight line passing through two given points:, To find the equation of straight line passing through the points (x1, y1) and (x2,, y2) , we generally consider one of the following methods., 1. General equation y = mx + c., It is passing through (x1, y1) then y1 = mx1 + c., It is passing through (x2, y2) also, then y2 = mx2 + c., Solving these two simultaneous equations, we get ‘m’ and ‘c’ and so the, equation., 2. The formula, , (y2 - y1), y – y1 = ________ (x – x1), , and substitution., , (x2 - x1), Some sequence of steps gives the equation. But the paravartya sutra enables us, to arrive at the conclusion in a more easy way and convenient to work mentally., Example1: Find the equation of the line passing through the points (9,7) and, (5,2)., Step1: Put the difference of the y - coordinates as the x - coefficient and vice versa., i.e. x coefficient = 7 - 2 = 5, y coefficient = 9 - 5 = 4., Thus L.H.S of equation is 5x - 4y., Step 2: The constant term (R.H.S) is obtained by substituting the co-ordinates of, either of the given points in, L.H.S (obtained through step-1), file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (7 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , i.e. R.H.S of the equation is, 5(9) - 4(7) = 45 - 28 = 17, or 5(5) - 4(2) = 25 - 8 = 17., Thus the equation is 5x - 4y = 17., Example 2: Find the equation of the line passing through (2, -3) and (4,-7)., Step 1 : x[-3-(-7)] –y[2-4] = 4x + 2y., Step 2 : 4(2) + 2(-3) = 8 –6 = 2., Step 3 : Equation is 4x + 2y =2 or 2x +y = 1., Example 3 : Equation of the line passing through the points (7,9) and (3,-7)., Step 1 : x[9 - (-7)] – y(7 - 3) = 16x - 4y., Step 2 : 16(7) - 4(9) = 112 – 36 = 76, Step 3 : 16x- 4y = 76 or 4x – y = 19, , Find the equation of the line passing through the points using Vedic, methods., 1., , (1, 2), (4,-3), , 3. (-5, -7), (13,2), , 2. (5,-2), (5,-4), 4., , (a, o) , (o,b), , 34, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...org/Mathematics/VedicBreifing/Miscellaneousitems.html (8 of 8)12/22/2005 8:55:56 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics, , Vedic Mathematics | Conclusion, IV.CONCLUSION, After going through the content presented in this book, you may, perhaps, have, noted a number of applications of methods of Vedic Mathematics. We are aware, that this attempt is only to make you familiar with a few special methods. The, methods discussed, and organization of the content here are intended for any, reader with some basic mathematical background. That is why the serious, mathematical issues, higher level mathematical problems are not taken up in this, volume, even though many aspects like four fundamental operations, squaring,, cubing, linear equations, simultaneous equations. factorization, H.C.F, recurring, decimals, etc are dealt with. Many more concepts and aspects are omitted, unavoidably, keeping in view the scope and limitations of the present volume., Thus the present volume serves as only an 'introduction'. More has to be, presented to cover all the issues in Swamiji's 'Vedic Mathematics'. Still more steps, are needed to touch the latest developments in Vedic Mathematics. As a result,, serious and sincere work by scholars and research workers continues in this field, both in our country and abroad. Sri Sathya Sai Veda Pratisthan intends to bring, about more volumes covering the aspects now left over, and also elaborating the, content of Vedic Mathematics., The present volume, even though introductory, has touched almost all the Sutras, and sub-Sutras as mentioned in Swamiji's 'Vedic Mathematics'. Further it has, given rationale and proof for the methods. As there is a general opinion that the, 'so called Vedic Mathematics is only rude, rote, non mathematical and none other, than some sort of tricks', the logic, proof and Mathematics behind the 'the so, called tricks' has been explained. An impartial reader can easily experience the, beauty, charm and resourcefulness in Vedic Mathematics systems. We feel that, the reader can enjoy the diversity and simplicity in Vedic Mathematics while, applying the methods against the conventional textbook methods. The reader can, also compare and contrast both the methods., The Vedic Methods enable the practitioner improve mental abilities to solve, difficult problems with high speed and accuracy., , 3, file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20maths/www.vedamu.org/Mathematics/VedicBreifing/Conclusion.html2/15/2006 7:42:40 PM