# Vedic Mathematics – 1

## Vedic Mathematics – 1

Firstly I bow down to Shri Bharati Krishna Tirtha Ji for doing a great work in discovering the mystic profound knowledge given in our Vedas. He did an extensive research on Atharvaveda and found out ‘Sixteen Mathematical Formulas’ which are known today as Vedic Mathematics. I also bow down to Aryabhatta, Bhaskaracharya, Baudhayana and other great scholars of ancient India who laid down a strong foundation of Mathematics and Science in India and worked hard on it much before Copernicus, Pythagoras and Newton.

Before we start, I would also like to add few common sources which I used during my initial and later study of Vedic Mathematics and in my Vedic notes also.

In this first part of the series, we will study three methods or formulas under Vedic Mathematics which are:

• Urdhva Tiryak Sutra (Vertically Crosswise Multiplication Method)
• Nikhilam Sutra
• Ekadhiken Poorven Sutra (Vulgar Fraction)

So let’s begin:

Warning : Follow me along on a separate sheet of paper for better understanding.

1.  Urdhva Tiryak Sutra – Vertically Cross-wise Multiplication Method

a) Two digit multiplication:

Two digit multiplication is done in three steps.

Consider two numbers AB & CD. To multiply both numbers, we will follow the below steps:

• Multiply the rightmost digits of two numbers B & D with each other.

•  Carry out a cross multiplication between A , D and B , C and add them.

• Lastly multiply the first digits of both numbers A & C with each other and solution is completed.

Note: After writing all three parts of the answer. Start from the last digit and carry the previous digit forward.

For Example – We have to multiply 38×52. We multiply the rightmost digit 8 & 2 with each other to get :16. Then we multiply 3, 2 & 8, 5 with each other cross-wise and then add them to get 46. So now we have 46:16.

Lastly we multiply 3 and 5 with each other to get 15.

So we have 15:46:16.

Now start from the last, write the last digit as an answer and carry the previous digit to add it with preceding digit.

b) Three digit multiplication:

Three digit multiplication is done in 5 steps.

Consider two 3 digit numbers ABC & DEF. To multiply these numbers, we follow these steps:

• Multiplying the rightmost digits C and F with each other, we write the last part of the answer.

• Now we multiply B, F and C , E cross-wise with each other and then add them to get next part of the answer.

• Then we multiply A, F ; B, E & C, D and we add them.

• Again similarly multiply A, E and B, D with each other and add them.

• Lastly multiply the leftmost digit A and D with each other in order to get the first part of the answer.

Hence, solution is completed.

For Example– We have to multiply 326 × 258. We multiply the rightmost digits 6 & 8 to get 48. So we get :48 for the rightmost part of answer. Then carry out a two digit cross multiplication between 2 & 8 i.e., 16 and 6 & 5 i.e., 30. Add the products 16 & 30 to get 46. Now we have :46:48 for the answer.

Again carry out a 3 digit cross multiplication between 3 & 8 (24), and 6 & 2 (12), and 2 & 5 (10). Add the products 24, 12 and 10 to get 46. So we have now :46:46:48 for the answer. Then we carry out a 2 digit cross multiplication between 3 & 5 (15) and 2 & 2 (4). Add 15 and 4 to get 19. We get :19:46:46:48 for the answer.

Lastly we multiply the left most digits 3 and 2 to get 6. So the final look for the answer is 6:19:46:46:48.

Now very important, we start from rightmost part. We write 8 for the final answer and carry 4 to add with preceding no. 46. We will continue similarly yo get the final product 84108.

2.  Nikhilam Sutra

a) For Integers less than the base:

Consider 93 × 98. Here base is 100.

• Both numbers are less than 100. Write two nos. one below other and write the difference of base no. (100) from each no. on the right side.
• Take algebraic sum of digits across any cross-wise pair, e.g., 93 + (-2) or 98 + (-7).
• Both give the same answer 91, which is first part of the answer.
• Now just multiply the two differences (-2) × (-7) to get 14, which is second part of the answer.

The final answer is then 9114, which can also be interpreted as (9100+14).

b) For integers greater than the base (100):

Consider 103 × 105. Here base no. is again 100.

• The first two steps are same as above given method.In third step, add crosswise any pair to get the same answer, e.g., 103 + 5 or 105 + 3.
• Now just multiply the two differences 3 × 5 to get 15, which is second part of the answer.

•   Final result is 10815.

Note: We have to ensure that we have exactly two digits in the second part of the answer as the base contains two zeroes. Suppose differences are 2 & 3, and their multiplication result in 6 then we must write 06 in the second part of the answer in order to get the correct answer.

c) For integers below & above the base (100):

Consider 97 × 104.

• The procedure remains same but the second part of the answer is negative (-3) × 4 = (-12). So, 101/(-12) can be interpreted as 10100 – 12 = 10088.

• So final answer is 10088.

Same procedure is followed as mentioned above for numbers having bases 1000 & 10000.

The conversion of vulgar fractions whose numerator is 1 and whose denominator ends with a 9 i.e. 1/19, 2/29, 1/39 etc. can be easily done using the ‘one more than previous’ formula. To demonstrate let us take 1/19.

• Step 1) Start by placing 1 as the last digit (i.e. the right-hand-most digit) of the answer.
• Step 2) Proceed to multiply leftward, continually multiplying by the ‘factor’. The factor = 1 + the penultimate (second-to-last) number of the denominator (19). In this case the penultimate number is 1. So the factor = 1 + 1 = 2.
• Step 3) So our first multiplication should be 1×2 which gives 2. The result should look as follows: 21
• Step 4) Next multiply the 2 by the factor (2) to get 4. The result should look as follows: 421
• Step 5) The 4 is multiplied by the factor to get 8. Then when 8 is multiplied by the factor the result is 16. Here the 6 is placed in the answer while the 1 is carried. The result should be as follows: 168421.
• Step 6) Then, when 6 is multiplied by the factor the result is 12, but the 1 which was carried in the previous step has to be added, giving 13. The 3 is placed in the answer and again the 1 is carried. At this point your answer should look as follows: 1368421.
• Step 7) This process of multiplying and carrying should be continued until the multiplication results in a figure equivalent to the difference between the denominator (19) and the numerator (1) which is 18 (i.e. 19-1 = 18). So in this case we stop at 947368421. Since 9 x 2 (the factor) = 18 and at this point, for this example, our answer should look as follows: 947368421
• Step 8) Now take the 9’s compliment of this figure to arrive at the first half of the answer as follows:
999999999-
947368421
052631578
• Step 9) Now put the two halves together to get the complete answer. You should arrive at the following answer:
.052631578947368421

To arrive at the answer from left to right we use the division process:

• Step 1) Dividing 1 (the first digit of the dividend) by 2, (the factor) we see the quotient is zero and the remainder is 1. We therefore set 0 down as the first digit of the quotient and prefix the remainder (1) to that very digit of the quotient and thus obtain 10 as our next dividend.
10
• Step 2) Dividing this 10 by 2, we get 5 as the second digit of the quotient; and as there is no remainder (to be prefixed thereto), we take up that digit 5 itself as our next dividend.
105
• Step 3) So, the next quotient digit is 2; and the remainder is 1. We therefore, put 2 down as the third digit of the quotient and prefix the remainder (1) to that quotient digit (2) and thus have 12 as our next dividend:
10512
• Step 4) This is continued until we reach the figure equivalent to the difference between numerator and denominator. And then take the 9’s compliment as in the first method. Combining the two figures we get the answer.