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Vedic Mathematics Techniques

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Information about Vedic Mathematics Techniques
Education

Published on May 7, 2007

Author: Santoshmehta

Source: authorstream.com

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Vedic Mathematics:  Vedic Mathematics Session Topics Multiplication with common base Squares Cubes Two digit multiplication Three digit multiplication By Munish Kumar Slide2:  Multiplication with common base # Multiply 104 by 103 1 0 4 1 0 3 x 3 1 2 1 0 4 x x 0 0 0 x 1 0 7 1 2 Magic of Vedic 1 0 4 1 0 3 Here Common base is 100 4 3 1 2 1 0 7 (1 0 4 + 3 ) or (1 0 3 + 4) 4 x 3 Slide3:  Some problems to practice 105 x 107 102 x 108 107 x 109 102 x 103 104 x 112 Slide4:  Let us do some more multiplications 1 0 0 1 1 0 1 9 01 19 1 9 1 0 2 0 (1001 + 19 ) or (1 019 + 01) 01 x 19 Since the base is 1000, we take 3 digits on the right side 1,0 2 0,0 1 9 x 0 1 9 Slide5:  Some practice problems 1005 x 1007 1002 x 1008 1004 x 1012 1013 x 1003 Slide6:  Multiplication with less than base 9 7 8 9 - 3 - 11 3 3 8 6 (9 7 - 11 ) or (89 - 03) -11 x -3 8 6 3 3 x Slide7:  Let us do some practice 95 x 97 92 x 98 97 x 89 92 x 93 Slide8:  Squares Squares from 101 to 125 (101)2 (109)2 (112)2 (117)2 (123)2 Slide9:  Squares from 75 to 99 Squares (97)2 (93)2 (89)2 (85)2 (77)2 Slide10:  Squares of numbers ending with 5 Squares Formula used (a5)2 = a(a+1) Ι 25 Where, a = Digits in the number other than 5 (25)2 (45)2 (95)2 (995)2 (1035)2 (25)2 =2(2+1) Ι 25 = 6 2 5 Here a = 2 (95)2 = 9(9+1) Ι 25 = 9025 Here a= 9 (45)2 = 4(4+1) Ι 25 = 2 0 2 5 Here a= 4 (995)2 = 99(99 + 1) Ι 25 = 990025 Here a = 99 (1035)2 = 103(103 + 1) Ι 25 = 1071225 Here a = 103 Slide11:  Squares Squares from 25 to 49 (47)2 (43)2 (39)2 (35)2 (27)2 Formula used N2 = ( 25 – X ) Ι X2 Where, X = By how much a number less than 50 (47)2 = (25 – 3) Ι 32 = 2 2 0 9 Here X = 3 (39)2 = (25 – 11) Ι 112 = 14 Ι 121 = 1 5 2 1 Here X = 11 (43)2 = (25 – 7) Ι 72 = 1 8 4 9 Here X = 7 (35)2 = (25 – 15) Ι 152 = 10 Ι 225 = 1 2 2 5 Here X = 15 (27)2 = (25 – 23) Ι 232 = 02 Ι 529 = 7 2 9 Here X = 23 Slide12:  Squares Squares from 51 to 74 (52)2 (58)2 (63)2 (65)2 (72)2 Formula used N2 = ( 25 + X ) Ι X2 Where, X = By how much a number is more than 50 (52)2 = (25 + 2) Ι 22 = 2 7 0 4 Here X = 2 (63)2 = (25 + 13) Ι 132 = 3 8 Ι 169 = 3 9 6 9 Here X = 13 (58)2 = (25 + 8) Ι 82 = 3 3 6 4 Here X = 8 (65)2 = (25 + 15) Ι 152 = 40 Ι 225 = 4 2 2 5 Here X = 15 (72)2 = (25 + 22) Ι 222 = 47 Ι 484 = 5 1 8 4 Here X = 22 Slide13:  Ending with 5 and difference 10 25 x 35 = 45 x 55 = 115 x 125 = 195 x 205 = 505 x 495 = 8 7 5 ALWAYS (32 – 1) = 8 2,475 14,375 39,975 2,49,975 What is the similarity in the following Slide14:  Base same and unit digits sum 10 53 x 57 = 26 x 24 = 117 x 113 = 109 x 101 = 55 x 55 = 30 21 7 x 3 5 x 6 624 13,221 11,009 3,025 What is the similarity in the following Slide15:  Cubes Formula used (N)3 = b + 3x Ι 3x2 Ι x3 Where, x = Difference from the base & b = base (104)3 (12)3 (95)3 (995)3 (1007)3 (104)3 = (100 + 3 x 4) Ι 3 x 42 Ι 43 = (100 +12) Ι 3 x 16 Ι 64 = 1 1 2 4 8 6 4 Here b = 100 & x = 4 (12)3 = (10 + 3 x 2) Ι 3 x 22 Ι 23 = (10 +6 ) Ι 3 x 4 Ι 8 = 16 Ι 12 Ι 8 = 1 7 2 8 Here b = 10 & x = 2 (95)3 = (100 + 3 x -5) Ι 3 x (-5)2 Ι (-5)3 = (100 -15) Ι 75 Ι -125 = 85 Ι 73 Ι 200 -125 = 8 5 7 3 7 5 Here, b = 100 & x = -5 Since the base is 10 . We will take only on digit in each block and carry forward the extra to consecutive next block in the left. Since we have a negative term in the extreme right block. So have to make it positive by borrowing 2 carry from consecutive left block and add to -125. (995)3 = (1000 + 3 x -5) Ι 3 x (-5)2 Ι (-5)3 = (1000 -15) Ι 75 Ι -125 = 985 Ι 073 Ι 200 -125 = 9 8 5, 0 7 3, 0 7 5 Here, b = 1000 & x = -5 As the base here is 1000, so we will take three digits in each block after making negative term in the extreme right block positive (1007)3 = (1000 + 3 x 7) Ι 3 x (7)2 Ι (7)3 = (1000 +21) Ι 147 Ι 343 = 1021 Ι 147 Ι 343 = 1 0, 2 1 1, 4 7, 3 4 3 Here, b = 1000 & x = 7 Slide16:  Three step multiplication # Multiply 36 by 94 3 6 9 4 x 6 x 4 = 24 3 x 4 + 9 x 6 = 66 9 x 3 = 27 2 7 6 6 2 4 Slide17:  Let us do some more multiplications 1 1 1 1 1 3 11 13 1 4 3 1 2 4 (1 1 1 + 13 ) or (1 1 3 + 11) 11 x 13 This 1 is carried forward to the other side 1 2 5 4 3 x Slide18:  Multiplication in different mood 9 7 1 0 9 - 3 + 9 -2 7 1 0 6 (9 7 + 9 ) or (109 - 03) 9 x -3 x Since we have a negative term on the right side of it So we will make it positive . For that we will borrow carry from left side, which is equal to 100 and add -27 to it 100 - 27 = 73 106 – 1 = 105 7 3 105 1 0,5 7 3 Slide19:  Thank you Best of luck for your future

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