Career Help & Jobs in IndiaInformation on careers & higher education in India

Vedic Multiplication : Quick Math for competitive exams

Nikhilam

1. Base 10 9 -1 13 +3 12 +2

X 7 -3 X 14 +4 8 -2

——————— ———————– ————

9-3 | 3 17 | ₁2 10 -4 = 96.

Or 7-1 | 3 18 | 2 = 182.

Or 9+7-10 | 3

Or 10-1-3 | 3

———————

6 | 3 = 63.

2. Base 100 93 -7 88 -12 107 +7

97 -3 94 -6 112 +12

———— ————– ————-

90 21 = 9021 82 72 = 8272 119 84 = 11984

112 +12 63 +13 89 +79 887 -13

84 -16 54 +4 13 +3 907 +7

————– —————- ————– ————-

96 -₁92 = 9408 2/67 52 92 ₂₃7 = 1157 9*894 -91

33 ½ 52 = 3402 8046 -91

804509

3. Base 1000 887 -113 1047 +47 888 -12

997 –3 991 -9 888 -12

————— ————- —————

884 339 = 884339 1038 -423 = 1037577 9*876 ₁44

7884 ₁44 = 788544

9997 -0003 988 -012 99979 -00021

9997 -0003 998 -002 99999 -00001

—————- ————– ——————

9994 0009 = 99940009 986 024 = 986024 99978 00021

249 -001 532 +32 635 +135

246 -004 472 -28 502 +002

————- ————– ————————

4/245 004 5*504 -₈96 2/637 270

61 ¼ 004 = 61254 2520 -₈96 = 251104. 318 ½ 270

= 318770

389 –111

516 +16

—————

2/405 -₁776

202 ½ -₁776

=202 -₁276

=201 -276 = 200724

Squaring:

1. Add the positive difference from the base to the quantity or Subtract the negative difference from the base from the quantity.

2. Append the square of difference to the result obtained from 1 above.

3. Take carry over as per rules followed earlier, that is carry over tens in case of base 10, and hundreds in case of base 100.

4. Remember to multiply or divide the result obtained from 1 above, if base is other than any power of 10.

Ex.

19 +9 28/₈1 = 361.

19 -1 18*2/1 = 361.

91 -9 82/81 = 8281.

108 +8 116/64 = 11664

989 -11 978/121 = 978121

59 +9 68¸2/81 = 3481

43 +3 46*4/9 = 1849

Square of Number ending in 5

1. Multiply number of tens with next higher integer and annex 25 to the product.

Multiplication of Cognate numbers

1. This is applicable to numbers with number of tens identical and units totaling together 10. e.g. 23 & 27, 73 & 77, 84 & 86, 62 & 68, 154 & 156, 173 & 177, etc.

2. Multiply number of tens with next higher integer and annex the product of units to the roduct so arrived.

Ex.

23 * 27 = 2*3/21 = 621.

153 * 157 = 15*16/21 = 24021

Multiplication with a number of 9’s

Case:1 When the number of digits are same in multiplicand and in multiplier.

1. Arithmetically 1 less than the multiplicand forms left part of the answer.

2. Complement of left part forms right part of the answer.

Ex.

7*9 = 6/3

15*99 = 14/85

673*999 = 672/327

 

9876*9999 = 9875/0124

Case: 2 When the digits in multiplicand are less than the digits in multiplier.

1. Prefixes as many zeros as required to make both multiplicand and multiplier of same number of digits.

2. Proceed as in case 1 above.

Ex.

6*99 = 06*99 = 05/94

73*999 = 073*999 = 072/927

798*9999999 = 0000798*9999999 = 0000797/9999202 = 7979999202.

Case:3 When the number of digits in multiplicand are more than the number of digits in the multiplier.

1. Divide the multiplicand vertically so as to have same number of digits as in multiplier on the right side of the division.

2. Subtract 1 more than the number on left of division from the multiplicand, which gives left part of the answer.

3. Right hand part should be subtracted from the multiplier+1 to get right side part of the answer.

Ex.

43 * 9 => 4 | 3 122 * 9 => 12 | 2 111011*99 => 1110 | 11

-5 | 9+1-3 -13 | 9+1-2 -1111

————— —————- ————–

38/7 = 387 109/8 = 1098 109900/89

Urdhva-Tiryak

This is a general method applicable to any kind of multiplication. In this method starting from first digit on left, we go on multiplying crossways all digits and add them to get individual digits of the answer.

Each position is allowed only one digit, so digits in excess of one ar carried over to next multiplication-addition on the left side.

Ex.

a b c

f g h

————-

af | ag+bf | ah+bg+cf | bh+cg | ch

1. First just first digit on left is multiplied.

2. Then First two digits from the left are multiplied and added.

3. Similarly we go on getting digits till we reach upto first and the last digits from the left.

4. Then we start moving towards right and go on reducing digits to be multiplied and get

successive digits of the answer.

1 1 1 1 0 8 5 8 2

1 1 1 1 0 8 2 3 1

———– ———– —————-

1 | 2 | 3 | 2 |1 1 | 0 | 6 | 0 | 4 10| 1 | 3 | 4 | 2

= 12321 1 6 3 3 1

—————- —————–

1 | 1 | 6 | 6 | 4 = 11664 13| 4 | 4 | 4 | 2 =134442

7 8 5 3 2 1

3 6 2 0 5 2

—————- —————-

21| 6 | 7 | 6 | 0 0 | 5 | 6 | 9 | 2

6| 7 | 4 | 1 1 1

—————– —————-

28| 4 | 1 | 7 | 0 = 284170 1 | 6 | 6 | 9 | 2 = 16692

8 7 2 6 5

3 2 1 1 7

————————

24| 7 | 8 | 7 | 2 | 7 | 5 | 7 | 5

3 2 3 9 6 2 4 3

———————————

28| 0 | 2 | 6 | 9 | 0 | 0 | 0 | 5

Ex. ( a+b) ( a+9b) a + b x⁵+3x⁴+5x³+3x²+x+1

a + 9b 7x⁵+5x⁴+3x³+x²+3x+5

——– ——— ——– ———————————————————-

a²+10ab+9b² 7x¹⁰+26x⁹+53x⁸+56x⁷+33x⁶+40x⁵+41x⁴+38x³+19x²+ 8x+5

When a certain power is not present in any of the expression, we put zero in its place and proceed same as above.

Student should use his/her prudence in selecting the method to multiply different numbers either with the help of Nikhilam or with the help of Urdhva-Tiryak.

When, all the digits are more than 5, viniculam method is also useful.

Ex. 889*898 1 1 1 1

1 1 0 2

—————–

1 | 2 | 0 | 2 | 3 | 2 | 2

7 9 8 3 2 2

---