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1. Multiplication Urdhva-Tiryak Sutra "Vertically and Crosswise": That's all you need to remember to multiply two numbers! Line diagrams 2 digit numbers 32 x 54 = ? 32 54 CARRY 2 1728 3 digit numbers 749 x 322 = ? 749 322 CARRY 1 2 5 3 241178 4 digit numbers 6724 x 1823 = ? 6724 1283 CARRY 1 3 4 8 7 2 8626892 Multiplying a number by 11 To multiply any 2 digit number by 11, we just put their total between the 2 figures. 26 x 11 = 2 | 2+6 | 6 = 286 48 x 11 = 4 | 4+8 | 8 = 4 |1 2 | 8 = 528 89 x 11 = 8 | 8+9 | 9 = 8 |1 7 | 9 = 979 Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE) 2. Squaring & Cubing of 2 digit numbers Nikhilam Sutra for Squaring 2 digit numbers 1. For a given two digit number always take the base as 10 and the working base as the multiple of ten nearest to the given number. 2. Find the difference between the given number and the working base and add this to the given number. 3. The number now obtained is multiplied by the number of times the working base is a multiple of 10. This forms the LHS of the answer. 4. Square the difference of the number and the working base. This is the RHS of the answer. This method is much simpler than it may appear at first and is best understood through examples. 642: Base = 10, Working Base = 60(10 x 6). 64 – 60 = 4. Adding 4 to 64 we get 68. 68 x 6 = 408 and 42 = 16 ANS = 408 | 16 = 4096 2 78 : WB = 80 (10 x 8), 78 – 80 = –2, 78 + (–2) = 76, 76 x 8 = 608, (–2)2 = 4, ANS =6084 Anurupya Sutra for Cubing 2 digit numbers Remember cubes of the first 9 numbers 13 = 1 23 = 8 33 = 27 3 3 6 = 216 7 = 343 83 = 512 43 = 64 93 = 729 1. The formula for a 2 digit no. with digits a and b is 2. Below the a2b and ab2 terms write twice their values. 3. Add ‘vedically’ to get the cube of the number! 53 = 125 a3 a2b ab2 b3 Examples: 1 5 6 143 = 1 4 8 7 16 32 4 2 4 5 2 233 = 8 12 24 1 18 36 6 12 64 = 2744 = 12167 4 27 7 Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE) 3. Simultaneous Equations in 2 variables Paravartya Sutra This method is essentially similar to what we know today as “Cramer’s Rule”. This can be understood with the help of the following example Solve and 2x + 3y = 8 4x + 5y = 14 To find ‘x’: Consider the next column to the ‘x’ column, i.e. that containing y and cross multiply forwards (with – sign in between) to get numerator. Therefore, numerator = (3 x 14) – (8 x 5) = 2 Cross multiply backwards for the denominator. Denominator = (3 x 4) – (2 x 5) = 2 Hence, x = 2/2 = 1 To find ‘y’: Consider the next column to ‘y’ column, i.e. the constant column and cross multiply forwards with – sign in between. Therefore, numerator = (8 x 4) – (2 x 14) = 4 Denominator of ‘y’ term is same as that of ‘x’ term = 2 (note this point to avoid confusion). Hence, y = 4/2 = 2 ANS: x = 1, y = 2 Evidently this method can be done mentally after just a few practice examples! Examples: Solve x – y = 7 and 5x + 2y = 42 ANS: x = – 42 – 14 = – 56 = 8 ; –5 – 2 –7 y = 35 – 42 = 1 –7 Solve 11x + 6y = 28 and 7x – 4y = 10 ANS: x = 60 – (–112) = 172 = 2 ; 42 – (–44) 86 y = 196 – 110 = 1 86 Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE) 4. Factorisation of Simple Quadratics Aadyam-aadyena Aantam-antyena Sutra Instead of following the cumbersome method of splitting and taking common terms to factorise a quadratic expression, the Vedic method can be employed effectively. Procedure: For a quadratic of the form ax2 + bx + c 1. Split middle coefficient (i.e. b) into 2 parts such that: ratio of 1st coeff (a) to 1st part = ratio of 2nd part to constant term (c) This ratio gives first factor 2. To find second factor: divide first term of quadratic by first term of first factor and divide last term of quadratic by last term of first factor After going through just a few examples, this can be done mentally! Examples: 2x2 + 5x + 2 : Step 1 Split 5 = 4 + 1 since 2 : 4 :: 1 : 2 Ratio is 1:2 … therefore 1st factor is (x + 2) Step 2 2x2 ÷ x = 2x and 2 ÷ 2 = 1 … 2nd factor is (2x + 1) ANS: 2x2 + 5x + 2 = (x + 2)(2x + 1) 6x2 + 11x + 3 : Step 1 Split 11 = 9 + 2 since 6 : 9 :: 2 : 3 Ratio is 2:3 … therefore 1st factor is (2x + 3) Step 2 6 x2 ÷ 2 = 3x and 3 ÷ 3 = 1 … 2nd factor is (3x + 1) ANS : 6x2 + 11x + 3 = (2x + 3)(3x + 1) 8x2 – 22x + 5 : Step 1 Split -22 = -2 - 20 since 8 : -2 :: -20 : 5 =- 4:-1 Ratio is 4:-1 … therefore 1st factor is (4x - 1) Step 2 8x2 ÷ 4 = 2x and 5 ÷ -1 = -5 … 2nd factor (2x - 5) ANS: 8x2 – 22x + 5 = (4x - 1)(2x - 5) 12x2 – 23xy + 10y2: Step 1 Split -23 = -8 - 15 since 12 : -8 :: -15 : 10 =- 3:-2 Ratio is 3:-2 … therefore 1st factor is (3x – 2y) Step 2 12x2 ÷ 3 = 4x and 10y2 ÷ -2y = -5y … 2nd factor (4x – 5y) ANS: 12x2 – 23xy + 10y2 = (3x – 2y)(4x – 5y) Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE) 5. Quadratic Equations: A few types A general formula for solving any quadratic equation is First Differential of Quadratic = Square root of Discriminant Consider x2 + 4x - 3 = 0 Then its first differential is 2x + 4 Discriminant is 42 – [4 x 1 x (-3)] = 28 Therefore, the solution is given by 2x + 4 = (281/2 ) Hence x = (1/2) – 2 Numerous categories have been developed to further simplify the solution of quadratic equations. One such category is the “reciprocals type” Examples: 1. Solve x + 1 = 17 . x 4 => x + 1 = 4 + 1 x 4 2. Solve => x – 1 = 35– 25 x 2 3 3. Solve => x – 1 = 49 – 4 = 7 – 2 x 14 14 2 7 Hence, directly x = 4 or x = 1/4 x – 1 = 55 x 6 x – 1 = 45 x 14 Hence, directly x = 3/2 or x = -2/3 [14 = 7 x 2 and 45 = 72 – 22] Hence, directly x = 7/2 or x = -2/7 [Express numerator as a difference of squares of factors of denominator] Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE) 6. Method of Auxiliary Fractions Ekadhika Purva Sutra TYPE 1: Dividing by a number ending with 9 Consider 7 ., 149 To write the auxiliary fraction (AF), first note the number of 9s at the end of denominator. Then shift the decimal point of the numerator by this number and in the denominator increase the digit immediately preceding the 9s by 1. AF = .7 ., 15 0.7 ., 15 R Divide the numerator successively, each time prefixing the ‘remainder’ with the ‘quotient’ arrived at after dividing. This is continued till the required number of decimal places. 0.0 7 4 6 10 9 14 11 Another example: Consider 83 . AF = 0.83 399 4 0.83 ., 0 . 20 80 20 4 R 3 0 0 7 ANS = 0.04697 14 05 ANS = 0.20802005 0 Note: This method can also be applied for numbers ending in 3, by multiplying numerator and denominator by 3 to get denominator ending in 9. TYPE 2: Dividing by a number ending with 1 Drop the 1 from the denominator (eg. 31 becomes 30, 291 becomes 290), and subtract 1 from numerator. The denominator will always end with 0; get rid of this by dividing numerator and denominator by 10. This is the auxiliary fraction (AF). Successively divide in similar fashion to Type 1 but each time replace the ‘quotient’ by its nine’s complement (9 – the number, for eg. 9’s complement of 3 = 6, 9’s complement of 1 = 8) Consider 39 ., 171 3.8 ., 17 C R AF = 38 . = 3.8 170 17 7 0.2 4 7 2 13 1 8 1 9 0 11 2 9 7 0 1 0 [9’s COMPLEMENT] ANS = 0.2280701 2 12 Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE) 7. Miscellaneous Topics Difference of Squares Did you know that any number can be expressed as a difference of two squares? Procedure: 1. Express the number as a product of either two even or two odd factors, say ‘a’ and ‘b’. (if not possible then express as any two factors) 2. Then the number can be written as a + b 2 a–b 2 2 2 Examples: 45 = 9 x 5 => 45 = 72 - 22 316 = 158 x 2 => 316 = 802 - 782 ( )( ) Special type of Linear Equations If the denominators contain linear expressions in x and the numerical values of the numerators are same with RHS = 0, then we can add denominators directly Examples: 3 + 3 = 0 2x + 4 x – 3 => (2x + 4) + (x – 3) = 0 i.e. 3x + 1 = 0 x = –1/3 1 + 1 = 0 x–4 x+3 => (x – 4) + (x + 3) = 0 i.e. 2x – 1 = 0 x = 1/2 Equation of a Straight Line passing through two known points Say the points are (p,q) and (r,s) The equation will be of the form ax – by = c, where a, b and c have to be found Procedure: 1. To find a, i.e. coefficient of x, find the difference of y coordinates [q – s] 2. To find b, i.e. coefficient of y, find the difference of x coordinates [p – r] 3. To find c, i.e. constant term, find middle product – extreme product [qr – ps] Examples: (11,6) and (4,7) a = -1, b = 7, c = 24-77 = -53 ANS: –x – 7y = –53 (17,9) and (13,-8) a = 17, b = 4, c = 117 – (-136) = 253 ANS: 17x – 4y = 253 Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)