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- 1. Vedic Mathematics Composed by A V Prakasam for the benefit of his daughters - from the book ‘Vedic Mathematics’ By Jagadguru Swami Sri Bharati Krsna Tirthaji MaharajaIt will be difficult to understand the concepts at the first reading, but if practiced with the examples given, the beauty of the vedic mathematics can be appreciated. 1
- 2. Multiplication• Nikhilam Sutra • Principle : (x+a)*(x+b) = x(x+a+b)+ab Thumb rule: -Make the number of digits in the Multiplier and Multiplicand equal, by adding zeros to the left, if necessary, to the number with less number of digits - Chose the Base – multiple of ten like 10|100|1000 etc– nearest to one of the numbers. The Base will have as many zeros to the right of 1 as the number of digits in the number(s) - Find the difference of each number from the base and write them after the corresponding number with an appropriate sign (+ or -) i. Add the Multiplier and Multiplicand and subtract the ‘Base’ from the total. ii. Multiply the differences and place the product by the side of the sum. iii. The number of digits in this product (of differences) should be the same as the number of zeros in the base. Any excess digit in the product, the left most digit should be added to the sum at (i). Any deficiency should be made up by adding zeros to the left of the product. iv. You got the product of the numbers To understand better, see the worked out examples in the succeeding slides. 2
- 3. Nikhilam Sutra (Contd.)ExamplesI Case: Multiplicand and Multiplier are below multiples of 10.a. Base 10 1. 8*5 2. 6*5 Note : In example 1, 5*2=10, but 8 -2 6 -4 because the Base is 10, the 5 -5 5 -5 number after The ‘|’ should be 8+5-10 5*2 6+5-10 4*5 one. Hence the 1 ( shown as a 3 10 1 20 subscript for clarity) is a overflow And so carried over to the left of = 40 = 30 the ‘|’ and is added to the sum on the left hand side.b. Base 100 Similar is the case with example 2 3. 91*91 4. 88*91 5. 99*97 91 -9 88 -12 99 -1 91 -9 91 -9 97 -3 91+91-100 9*9 179-100 108 99+97-100 1*3 82 81 79 108 96 03 = 8281 = 8008 = 9603 3
- 4. Nikhilam Sutra (Contd.)c. Base 1000 1. 888*998 2. 112*998 888 -112 112 -888 998 -002 998 -002 888+998-1000 112*002 112+998-1000 888*2 886 224 110 1776 = 886224 = 111776d. Base 100000 1. 99979*99999 99979 -00021 99999 -00001 99979+99999-100000 00021*1 99978 00021 = 9997821 4
- 5. Nikhilam Sutra (Contd.)II. Multiplicand or Multiplier slightly above or below Multiple of 10 1. 12*11 2. 18*12 3. 111*109 12 +2 18 +8 111 +11 11 +1 12 +2 109 +09 12+11-10 2*1 18+12-10 16 111+109-100 11*9 13 2 = 132 21 6 = 216 120 99 = 12099 4. 12*8 5. 108*97 12 +2 108 +8 97 -3 Note : In Example 4, 2*- 2=-4, (-4 is 8 -2 shown as 4 with a bar on top, 12+8-10 2*-2 108+97-100 8*-3 known as ‘vinkulum’, meaning that 10 4 = 96 105 24 the digit under the vinkulum only is 104 76 negative) The figure in the units place is -4 and hence 1, i.e. 10, is = 10476 borrowed from the left hand side 6. 10006*9999 number 10, leaving 9 behind, and 4 10006 +6 deducted from it giving 6. 9999 -1 Similar is the case with examples 5 and 6. 10006+9999-10000 6*-1 10005 0006 10004|9994 = 100049994 5
- 6. Nikhilam Sutra (Contd.)III. Multiples & sub-Multiples.Upa-Sutra : ‘Anurupena’Multiplication of two numbers neither of which is near a convenient BaseAnurupena means proportionately. If there is a rational ratio-wise relationship, theratio should be taken into account and should lead to a proportionatemultiplication or division, as the case may be.Example: 41*41 - 41 is nearer to 50 than to 100 and 50 is half of 100 - So workout with 50 as in the previous examples and Divide thesum of the numbers by 2. Base : 100|2 = 50, so the proportion is 2 41 -9 41 -9 (41+41-50) divided by 2 |81 16|81 = 1681 if the Base is 40, then 40 = 10*4 and the propotion is 4. The result is 41 +1 41 +1 = (41+41-40) Multiplied by 4 |01 = 1681 6
- 7. Nikhilam Sutra (Contd.) Examples:1. 49*49 Working Base 100|2 = 50 Working Base 10*5 = 50 49 -1 49 -1 49 -1 49 -1(49+49-50)|2 |01 (49+49-50)*5 |1 24|01 = 2401 240|1 = 2401 2. 46*44 Working Base 10*5=50 Working Base 100|2 = 50 46 -4 46 -4 44 -6 44 -6 (46+46-50)*5 | 24 (46+44-50)|2 |24 20|24 = 2024 200+2|4 = 2024 7
- 8. Nikhilam Sutra (Contd.)3. 54*46 Working Base 10*5 = 50 Working Base 100/2=50 54 +4 54 +4 46 -4 46 -4(54+46-50)*5 | 16 (54+46-50)/2 | 24 250-1 | 6 =249-1 | 4 = 2484 25 | 16 = 2484Note : 4*- 4=-16, which is written with a vinkulum as16, but because the Base is 10, the number after The‘|’ should be units. Hence the -1 is a overflow And socarried over to the left of the ‘|’ and because it is anegative number, is deducted from the sum on theleft hand side. Again, the remaining figure in the unitsplace is -6 and hence 1, i.e. 10, is borrowed from theleft hand side number and 6 deducted from it giving4. 4. 19*19 Working Base 10*2 = 20 19 -1 19 -1 (19+19-20)*2 | 1 = 36|1 = 361 8
- 9. Nikhilam Sutra (Contd.)5. 48*49 6. 249*246 Working Base 100/2 = 50 Working Base is 1000/4 = 250 48 -2 249 -1 49 -1 246 -4(48+49-50)/2 | 2 (249+246-250)/4 | 0004 47/2 | 2 = 23 2 | 2 = 23 / (50+2) = 2352 1/ 245/4 | 0004 =61 |250 + 4 = 61254Note : Here division of 47 by 2 gives us a fraction.The original base 100 is multiplied by this fraction Note : Here division of 245 by 4 gives us aand the result added to the number on the right fraction. The original base 1000 ishand side. multiplied by this fraction and the result added to the number on the right hand side 9
- 10. Nikhilam Sutra (Contd.)IV. Practice Session1. 87965*99998 8796240702. 48*97 46563. 889*9998 88882224. 77*9988 7690765. 532*472 2511046. 389*516 200724 10
- 11. Nikhilam Sutra (Contd.)V. Corollaries1. First Corollary: To find the Square of a number Principle: Find a base (power of 10) nearest to the number and find the differencebetween the number and the base. The difference may be less or more than the base.Whatever is the extent of its deficiency (surplus), lessen (add ) it stillfurther to that very extent; and also set up the square of the deficiency. Algebraical Expression: (a+or-b)2 = a2 +or- 2ab +b2 or a2 = (a+b)(a-b)+ b2 Thumb rule: to find the square of 9: i. Take the power of 10 nearest to the given number, 10 in this case ii. Find the difference between the nearest power and the given number, it is 1 in this case. ( called the deficiency. If the number is 12 instead of 9, the suplus will be 2 which should be added to 12 ) iii. Deduct this deficiency from the given number ( lessen it still further to that extent), in this case the result is 8. This is the left hand side portion of the answer. iv. Put down the square of the deficiency, i.e. square of 1 in this example, on the right hand side. v. you got the answer (92 = 8 | 1) 11
- 12. Nikhilam Sutra (Contd.)Examples:i. With Base multiples of 10 1. 72 = (7-3) | 32 = 4 | 9 = 49 2. 112 = (11+1) | 12 = 12 | 1 = 121 3. 1082 = (108+8) | 82 = 116 | 64 = 11664 4. 9892 = (989 – 11) | 112 = 978 | 121 = 978121ii. With base other than multiples of 10 1. 412 Base 40 = 10*4 ( 41+1)*4 | 12 168 | 1 = 1681 2. 592 Base 50 = 100/2 (59+9)/2 | 92 Note: In example 3, the right hand side figure comes to 625, but because the base is a 68/2 | 81 = 3481 Mutile of 100, the right hand side contains only 2 3. 7752 Digits. Consequently 6, shown as subscript, is carry forward and added to 6000. Base 800 = 100*8 (775-25)*8 | 252 750*8 | 625 = 6000 | 625 = 600625 12
- 13. Nikhilam Sutra (Contd.)2.1 Second corollary: Squaring of numbes ending in 5This is a special case of the previous corollary, applicable to numbersending in 5. The principle says ‘ By one more than the previous one’.Thumb rule: to find the square of 15: i. Here the last digit is 5 and the ‘previous one’ is 1 ii. One more than the ‘previous one’ is 1+1=2 iii. Multiply the previous one i.e. 1 by one more than the previous one i.e. 2. This is the left hand side digit which is 1*2 = 2 iv. Put down the square of 5 on the right hand side i.e. square of = 25 v. you got the answer (152 = 2 | 25 = 225)Examples: 1. 252 = (2*3) | 25 = 625 2. 1052 = (10*11) | 25 = 11025 3. 1852 = (18*19) | 25 = 34225 13
- 14. Nikhilam Sutra (Contd.) 2.2 Second Corollary (Contd.): Multiplication of numbers whose first digits are same and whose last digits add up to 10. The above rule is applicable not only to the squaring of a number ending In 5, but also to the multiplication of two numbers whose last digits together total 10 and whose previous part is exactly the same. Examples: 1. 27*23 2*(2+1) | 7*3 = 621 2. 96*94 9*(9+1) | 6*4 = 9024 The above rule is capable of further application for the multiplication of numbers whose last digits in sets of 2,3 and so on together total 100, 1000 Etc. Examples: 1. 191* 109 = 1*(1+1)*10 | 91*9 = 20 819 2. 884*816 = 8*(8+1)*10 | 84*16 = 720 | 1344 ( 1 is a carry) = 721344 14
- 15. Nikhilam Sutra (Contd.)2.3. Third corollary: Multiplier digits are only 9s. Principle: This relates to a very special case where the multiplier digits consists entirely of 9sCase 1 Multiplicand and the multiplier contain the same number of digits i.e. case like 9*1, 99*11 etc. In these cases, the left hand side of the result is one less than the multiplicand and the right hand side is the 9’s complement of the left hand side. Examples: 1. 2*9 = 18 Explanation – left hand is one less than 2 ( the multiplicand) and the right hand side 8 is 9’s complement of1 2. 3*9 = 27 3. 7*9 = 63 4. 17*99 = 1683 5. 19*99 = 1881 6. 777*999 = 776223 7. 9879*9999 = 98780121 15
- 16. Nikhilam Sutra (Contd.) Case 2: The multiplicand consists of a smaller number of digits than the multiplier In this case, add zeros to the left hand side of the multiplier to make the number of digits equal to those in the mutiplier. Example: 1. 7*99 = 07*99 = 0693 2. 79*999 = 079*999 = 078921 3. 79*9999999 = 0000079*9999999 = 00000789999921 Case 3: The multiplier contains a smaller number of digits than the multiplicand. In this case, the following procedure applies: i. Divide the mutiplicand off by a vertical line into a right hand portion consisting of as many digits as the multiplier ii. Subtract from the multiplicand one more than the whole excess portion on the left iii This gives the right hand side portion of the product iv. The right hand portion of the product would be 10’s complement of the right hand side portion of the multiplicand 16
- 17. Nikhilam Sutra (Contd.)Examples: Note: - Refer to Example 11. 43*9 There is one digit in the multiplier i.e. 9. ( 43-5) | (10 – 3) - Therefore draw a vertical line after one digit 38 | 7 = 387 from the right in the multiplicand 43 -To the right of the line is 3 and to the left of the line is 42. 63*9 -Add 1 to 4 giving 5 (63-7) | (10 – 3) -Subtract 5 from 43 giving 38. This is the left 56 | 3 = 567 hand portion of the product - The 10’s complement of 3 is 7 which is the3. 122*9 -Right hand side portion of the product. (122-13) | (10 – 2) -The product is therefore 387 109 | 8 = 10984. 112*99 (112-2) | (100 – 12) = 110885. 11119*99 (11119-112) | (100-19) = 11007816. 1000001*99999 (1000001-11) | 100000 – 00001) = 9999099999 End of Chapter 1A – Nikhilam Sutra 17

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