The base system of multiplication of numbers in Vedic mathematics produces one line result and one is not required to multiply in steps as we usually do in the available method in use where the steps become more tedious especially when the numbers are large. It involves multiplication and addition. The steps to multiply 995 and 1005 involve five lines as below:

• Step 1:           1005×5 = 0050025

• Step 2:           1005×9 = 09045

• Step 3:           1005×9 = 9045

• Step 4:           Addition = 999975

But by base system of Vedic mathematics, we write the result of multiplication as:

995×1005 = 1000/0 = 999975

(1)

In (Equation 1), 25 indicates that 25 is a negative number.

                We can see that by using the base system of multiplication one can save time and energy. Even use of calculator takes some time in feeding the numbers. This becomes more important where one is required to complete the given task in the given time, such as in competitive examinations. One has to choose judiciously that where to use which system as the base system is useful when the numbers to be multiplied are close to some base and the digits in the numbers are also large, such as 9999 and 9995.

Base System of Multiplication

In Vedic mathematics the numbers can be represented as having only positive, negative or both positive and negative digits. This system of positive and negative digits together in a number has been explained in Chandra [1]. The following formula suggested by the author is for the bases like 10, 100, 1000 and likewise.

                The result of multiplication of:

x and y = A/C

(2)

where, x = B"m, y = B"n, A = B+(algebraic sum of m and n), C = m×n (having number of digits equal to the number of zeros in B) and B ­= the base.

                The following illustrative examples will explain the use of the Equation 2.

Example 1

Multiplication of 6 and 8.

Solution

It is given that x = 6 and y = 8.

As the given numbers are close to 10, the base can be taken as 10. Therefore:

B = 10

Now writing the given numbers in terms of the base, we have:

x = B"m     or     6 = 10-4,      thus     m = -4

y = B"n     or     8 = 10-2,      thus     n = -2

The value of C = m×n = (-4)×(-2) = 8 having only one digit as the base 10 has only one zero.

                The value of A = B+(algebraic sum of m and n)  = 10+{(-4)+(-2)} = 4.

                Thus x×y = A/C

6×8 = 4/8 = 48

The above steps can be done mentally very easily and the result 4/8 = 48 is obtained directly.

Example 2

Multiplication of 7 and 12. 

Solution

It is given that x = 7 and y = 12.

                As the given numbers are close to 10, the base B will be 10. Therefore: m = -3 and      n = 2.

                C = (-3)×(-2) = -6 having only one digit as the base 10 has only one zero.

A = 10+{(-3)+(2)} = 9

Thus:

7×12 = 9/ = 84

The above answer 9 is also correct but it has positive and negative digits and looks awkward. Therefore, the digit which is negative is converted into positive digit and this process converts 9 into 84 which have only positive digits.

Example 3

Multiplication of 94 and 116. 

Solution

It is given that x = 94 and y = 116.

                As the given numbers are close to 100, the base B will be 100. Therefore:

m = -6 and n = 16

C = (-6)×(16) = -96

having only two digits as the base 100 has two zeros.

=

and:

A = 100+{(-6)+(16)} = 110

Thus:

94×116 = 110/ = 10904

Example 4

Multiplication of 9989 and 9997.

Solution

It is given that x =9989 and y = 9997.

                As the given numbers are close to 10000, the base B will be 10000. Therefore:

m = -11  and n = -3

Thus:

9989×9997 = [10000+{(-11)+(-3)}/00{(-11)×(-3)}

= 9986/0033 = 99860033

The normal method of multiplication of such numbers will consume lot of time and there are also chances of committing mistakes in multiplication.

General Method of Base System of Multiplication 

The general method of base system of multiplication developed by the author does not restrict to choosing only the bases like 10, 100, 1000, etc. but one can choose any number allowing that the numbers may not be close to 10, 100, 1000 or 10000 and thus depending on the given numbers the base can be selected close to the numbers to be multiplied. For example, if the numbers are 35 and 45, the base, to be called as sub base (SB), can be taken as 40. The selected sub base is between the two main bases 10 (lower) and 100 (higher). Now the value of A is given by the following expression:

A =

where:

S =

The main base is always the higher base.

The following illustrative examples will explain the general method of base system of multiplication.

Example 5

Multiplication of 35 and 45. 

Solution

It is given that x = 35 and y = 45.

                As the given numbers are close to 40, the sub base is taken as 40. Therefore:

B = 40

and:

S = = 2.5

Now writing the given numbers in terms of the sub base, we have:

x = B+m     or     35 = 40-5,      thus     m = -5

y = B+n     or     45 = 40+5,      thus     n = 5

The value of C = m×n = (-5)×(5) = having only two digits as the higher base 100 has two zeros.

                The value of:

A = = 16

Thus:

35×45 = 16/= 15/75 = 1575

Example 6

Multiplication of 63 and 69. 

Solution

It is given that x = 63 and y = 69.

                As the given numbers are close to 70, the sub base is taken as 70. Therefore:

B = 70

and:

S = = 1.43

Now writing the given numbers in terms of the sub base, we have:

63 = 70-7,     thus    m = -7

69 = 70-1,     thus    n = -1

The value of C = m×n = (-7)×(-1) = 07 having only two digits as the higher base 100 has two zeros.

                The value of:

A = = 43.356 = 43.4

Thus:

63×69 = 43.4/07 = 43/47 = 4347

The right side of slash (/) is found by multiplying 0.4 by 100 and then adding to 07, thus 40+07 = 47.

                In this example the sub base can also be taken as 60 and then:

63×69 = [() /(3×9) = 43.2/27 = 43/(20+27)

= 43/47 = 4347

For this given example, the sub base can also be taken as 50 and then:

63×69 = [() /(13×19) = 41/(247)

As the right side of the slash can have only two digits because the main base 100 has two zeros, the 2 of 247 has to go to the left side of the slash, hence it is added to 41 making it 43. Thus 41/(247) becomes (41+2)/47 which is:

= 43/47 = 4347

Now we see that the sub base system of multiplication provides various options of choosing a sub base. 

Apply the Base System Judiciously 

It has been said above that the base system of multiplication is to be applied judiciously where it is found that the results can be obtained immediately. Of course, the method of base system can be applied to any case of multiplication but some cases may put you in long unending situations of multiplications and complicated steps to get the results. The example given below will illustrates such situations.

Example 7

Multiplication of 593 and 64. 

Solution

It is given that x = 593 and y = 64.

                The numbers 593 and 64 are such that no one base will be close to both the numbers, so let us take a base which is close to one of the numbers and let it be 500. Thus:

B = 500

and:

S = = 2

Now writing the given numbers in terms of the sub base, we have:

593 = 500+93,     thus    m = m = 93

64 = 500-436,     thus    n = -36

The value of C = m×n = 93×(-436) = . It should have only three digits as the main base 1000 is having only three zeros and therefore, it would be /.

                The value of:

A = = 78.5

Thus:

63×69 = (78.5 + )/= 3/8.5/= 38/(500-548)

= 38/= 37/(1000-48) = 37/952 = 37952

We find in this example that the multiplication of m and n to get the value of C is itself another problem of multiplication similar to the one which we started solving. Also there are many complicated steps afterwards to get the result. Therefore, such situations do not warrant the use of base system of multiplication.

Checking the Results 

There is a very simple method of checking the result of the multiplication. Let the two numbers be x (abc), y (def) and the r (pqrst) be the result of the multiplication, where a, b, c, d, f, p, q, r, s, t, etc., are the digits in the numbers. Now the digits of a number are added till it becomes a single digit. For Example, if:

x = 537, then 5+3+7 = 15!1+5 = 6 = X

x = 6928, then 6+9+2+8 = 25!2+5 = 7 = Y

and:

r = 537×6928 = 3720336

then:

3+7+2+0+3+3+6 = 24!2+4 = 6 = R

The check provides that the digits of the result of multiplication of X and Y when added as above must be equal to R. Thus:

X×Y = R

6×7 = 42!4+2 = 6 = R

It may be noted that while adding the digits 9 can be ignored or may be taken as zero.

                Let us check the result 99860033 of multiplication of 9989 and 9997 in Example 4.

X = 9+9+8+9 = 8

Y = 9+9+9+7 = 7

R = 9+9+8+6+0+0+3+3 = 20!2+0 = 2

X×Y = R

8×7 = 56!5+6 = 11!1+1 = 2 = R

The base system of multiplication is an easier and faster method of multiplication of numbers if a selected base is close to the numbers. It can be said that when there are digital calculators these methods are not required but still the author feels that if one does mastery on the suggested method he can do the calculations faster without committing mistakes and picking up the calculator again and again. The base system has to be employed judiciously only under those conditions where they can be applied advantageously compared to the existing method of multiplication. The sub-base system developed by the author is extension of the base system and provides more flexibility in choosing an appropriate base though it requires more clear understanding of converting the obtained result into the desired value.

1. Chandra, A.M. "Method of Squaring Any Number Retaining All the Digits in the Result and Use of Positive and Negative Digits in a Number." IOSR Journal of Mathematics, vol. X, no. X, September-October 2012.