You can find previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics

A Spectacular Illustration of Vedic Mathematics

10's Complements

Multiplication Part 1

Multiplication Part 2

Multiplication Part 3

The method we will use in this lesson is explained very cryptically and succintly in an Upasutra (sub-formula) of the main sutra. This upasutra simply reads Anurupyena, which literally means Proportionately. I will now explain how the method is applied by taking a simple example. Let us find the product 21 x 22.

We notice that both these numbers are not very close to 10, but they are very close to a multiple of 10, 20. So, we choose 10 as our base and 20 as our "working base" for this problem. Notice that our working base is derived from our base by multiplying by 2, so keep 2 in mind as our multiplier.

Following the method in this lesson, and using 20 as our working base, we come up with the figure below:

21 + 1

22 + 2

---------

23 | 2

Now, before we find the final answer, we just have to perform one additional computation: Take the left hand side of the last row of the figure and multiply it by our multiplier before combining it with the right hand side of the last row. This gives us 23 x 2 = 46, and the answer to our problem is simply 462. You may verify the answer using a calculator or computer.

The same method works when the numbers are just below a multiple of 10 also. Let us take 19 x 18 as an example. Let us choose 10 as our base and 20 as our working base. We get 2 as our multiplier. Use the method illustrated in this lesson to derive the figure below:

19 - 1

18 - 2

-----------

17 | 2

As in the previous problem, before we find the final answer, we just have to perform one additional computation: Take the left hand side of the last row of the figure and multiply it by our multiplier before combining it with the right hand side of the last row. This gives us 17 x 2 = 34, and the answer to our problem is simply 342, which is easily verifiable.

Let us examine the algebraic basis for why the method works. Let us assume that our base is b and our working base is w. Let us assume that our multiplier is m, so w = m x b. We need to multiply y with z, and their deficits (or excesses, it will work either way) with respect to w are d and e respectively. So, our problem is to find y x z, which can be written as (w - d) x (w - e). Expand this out to w x (w - d - e) + d x e. Expanding the outer w as m x b, we get m x b x (w - d - e) + d x e. Now, we derived (w - d - e) by cross subtracting the deficits across the rows. We then multiplied it by the base to make it the left hand side of the answer. The only odd piece left out is the multiplier itself, which can be brought inside the parentheses, giving us the following:

y x z = b x (m x (w - d - e)) + de

This is what was accomplished by the extra step of multiplying the original left hand side by the multiplier before combining it with the right hand side to get the final answer. It is a straight-forward explanation, but it also explains several things about the method that we might have had doubts about.

For instance, we knew how many digits the right hand side is supposed to have when we used a power of 10 as our base (the number of zeroes in the base), but what happens when we use a multiple of this power of 10 as our base. The algebraic explanation tells us that the number of digits does not change: We are still using the original base to multiply the left hand side by, so the number of digits of the right hand side has to be the number of zeroes in the original base (which, under most cases will be the same number of zeroes as in the working base, but need not be: there is absolutely nothing preventing us from choosing and using 2.5 or some other non-round number as the multiplier! Note that the number of zeroes in the working base would then not be the same as the number of zeroes in the base).

Using this knowledge, let us solve some problems using this method for practice, then we will conclude by solving the problem posed at the beginning of this post.

Let us try to solve 29 x 31 using this method. Let us use 10 as our base and 30 as our working base, giving us 3 as our multiplier. We get the figure below:

29 - 1

31 + 1

---------

30 | 1

Now we multiply 30 by 3 to 90 as our left hand side. Note that since there are both deficits and excesses (the numbers are on either side of the working base), we need to use the three-step process explained in this lesson to derive the final answer, which turns out to be 899. In the first step of that process, we see that the right hand side is not all zeroes. We then proceed to subtract 1 from the left hand side, to give us 89. Then we take the 10's complement of 1 to get 9 as the right hand side, giving us 899. Note that the multiplication by the multiplier happens before the three-step process is performed, not after. Once again, the order in which we do the operations is important for the method to succeed.

What about 22 x 25? Let us use 10 as the base and 20 as the working base, giving us 2 as our multiplier. We derive the figure below:

22 + 2

25 + 5

---------

27 | 10

Now, we multiply 27 by the multiplier to get 54. We then carry over the extra digit on the right hand side to the left hand side, giving us a left hand side of 55, a right hand side of 0, and a final answer of 550. Note again that the carryover happens after multiplication by the multiplier. A good rule to remember, therefore, is that multiplication of the left hand side by the multiplier precedes all other steps in finding the answer. This is obvious from the algebraic explanation of the method, but bears repeating nevertheless.

Let us compute 32 x 25 as a further illustration of the technique. We derive the figure below by using 10 as our base, 30 as our working base, and 3 as our multiplier:

32 + 2

25 - 5

---------

27 | 10

We first multiply the left hand side by our multiplier to get 81 on the left hand side. We then see that our right hand side contains 2 digits while our base (10) has only one zero. So, we carry over the extra 1 to the left hand side, but since this problem is one of numbers on either side of the working base, the carryover digit has to be subtracted from the left hand side rather than added to it. We get 80 on the left hand side. We then apply the three-step procedure in this lesson, and find that since the right hand side is all zeroes, nothing further needs to be done, giving us an answer of 800.

Let us illustrate one final example before we solve the problem we started out with. Let us compute 32 x 24 this time. Use 10 as the base and 30 as the working base. We then get the figure below:

32 + 2

24 - 6

----------

26 | 12

We multiply 26 by 3 (our multiplier) to get 78. Then we subtract the extra digit on the right hand side from this number to get 77. Now applying the three-step procedure, we get 768 as the final answer (subtract 1 from 77, take the 10's complement of 2 and then combine them).

Now, let us get to the problem we posed at the top of this post. How do we compute 5007 x 4950 using this method? We quickly conclude that we should use 1000 as our base, 5000 as our working base and therefore, 5 as our multiplier. That results in the figure below:

5007 + 7

4950 - 50

------------

4957 | 350

Now multiply 4957 by 5 to get 24,785 (this step may seem a little daunting, but it is actually easy to do since the multiplier is just a single-digit number. Just remember to keep track of the carryover digits as you perform the multiplication from right to left and you will find that it is not that difficult. But, it is one of the reasons why we will learn how to work with a submultiple of a power of 10 in the next lesson).

350 already has the same number of digits as the base, so there is no carryover from the right hand side. Applying steps 2 and 3 of the three-step procedure (the right hand side is not all zeroes), we get our final answer as 24,784,650 (subtract 1 from 24,785 to get the left hand side, and combine it with the 10's complement of 350).

All this may sound complicated at first, but I can not emphasize enough or enough times that practice is the key to success with Vedic Mathematics. The more you practice, the easier it gets. Practice also makes your brain sharper, making it easier to absorb new material. In the next lesson, we will take the problem above, but work at it using submultiples of a power of 10 rather than multiples. It will add one more weapon to your mental arithmetic arsenal. Happy practicing, and good luck!

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