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Monday, July 6, 2009

Vedic Mathematics Lesson 7: Multiplication Part 5

In this lesson, we are going to deal with products of numbers that are close to a sub-multiple (or factor) of a power of 10, rather than being close to a multiple of a power of 10 or a power of 10 itself. This will come in handy when we want to work out products like 5007 x 4950 mentally (this is the same problem we dealt with in the previous lesson. We will calculate the answer again for this problem so that you can compare the two methods and choose which one is more convenient for you to use based on your personal preferences), without the use of a calculator or computer.

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
Multiplication Part 4

The method derives from the same upasutra that we used in the previous lesson. As you may recall, this upasutra reads simply Anurupyena, which literally means Proportionately. I will now explain how the method is applied by taking a simple example. Let us find the product 53 x 55.

We notice that these numbers are not very close to 100, but they are very close to a sub-multiple (or factor) of 100, 50. So, let us choose 100 as our base and 50 as our working base and because 100/2 = 50, we will will designate 2 as our divider. Using 50 as our working base, we now derive the figure below:

53 + 3
55 + 5
------------
58 | 15

If you have questions about the derivation of the figure above, please review the previous lessons, in particular, Lesson 4.

As in the previous lesson, the first step after we get the figure above is to apply the upasutra (Proportionately). In this case, our divider is 2, so divide the left hand side of the last row by 2 to get 29. Combine it with the right hand side to get 2915, which can be verified to be our answer.

Let us examine the algebraic basis of the method, which is almost identical to the algebraic basis of the method illustrated in the previous lesson. That will once again answer some questions we might have about special cases related to this method.

Let us assume that our base is b and our working base is w. Let us assume that our divider is m, so w = b/m. We need to multiply y with z, and their excesses (or deficits, 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 b/m, we get (b/m) x (w + d + e) + d x e. Now, we derived (w + d + e) by cross adding the excesses 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 divider itself, which can be brought inside the parentheses, giving us the following:

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

This is what was accomplished by the extra step of dividing the original left hand side by the divider before combining it with the right hand side to get the final answer.

Now, as long as the division does not result in a fractional result, things are fine. Let us examine a case when the division actually does result in a fractional result. Let us take the example of finding 23 x 24 by using 100 as our base and 25 as our working base. Our divider becomes 4. We first draw the figure below:

23 - 2
24 - 1
-----------
22 | 2

Now, when we divide 22 by 4 (our divider), we get a quotient of 5 and a remainder of 2. So, the answer to the division is actually 5 and a half. Since we are multiplying the left hand side by the base, 100, the result of the division is actually 550, not 5 and a half. Thus, the rule for dealing with fractional results in division becomes clear: express the remainder as a fraction using the base as the denominator and then carry over the numerator to the right hand side. Our remainder here is 2 out of 4 or 2/4 and expressing it as a fraction with 100 as the denominator gives us a numerator of 50 (50/100 = 2/4). Carrying over the 50 to the right hand side gives us the answer 52 for the right hand side. Our left hand side was 5 (the quotient of dividing 22 by 4), giving us the final answer of 552.

Let us now illustrate this method with one more example: 48 x 51. Let us take 100 as our base and 50 as our working base. We get the figure below:

48 - 2
51 + 1
-----------
49 | 2

Dividing 49 by 2 gives us 24 with a remainder of 1. Since 1/2 is the same as 50/100, we carry 50 over to the right hand side. But notice that this is a problem of numbers on either side of the working base, so the right hand side has to be subtracted from the left hand side. So, to get the right hand side of the answer, we actually have to compute 50 (our carryover from the left hand side) - 2 (the number originally on the right hand side, which needs to be subtracted). This gives us a right hand side of 48, and a final answer of 2448.

Hopefully, this, in combination with the algebraic basis of the method, and all the examples we have worked out in previous lessons is enough for us to solve any problems we encounter using this method. Therefore, as a final illustration, let us solve the problem we started this lesson with: 5007 x 4950.

Let us use 10000 as our base and 5000 as our working base. Our divider is 2. Let us draw the figure below:

5007 + 7
4950 - 50
---------------
4957 | 350

When we divide 4957 by 2, we get a quotient of 2478 and a remainder of 1. Notice that 1/2 is the same as 5000/10000, so we need to carry 5000 over to the right hand side. The computation on the right hand side is therefore 5000 - 350 = 4650 (this is a case of numbers on either side of the working base, so the right hand side has to be subtracted from the left hand side). The final answer is 24784650, which is what we derived in the previous lesson using 1000 as the base.

Hope you are practicing these techniques so that you can do them mentally without the need for even a paper or pen. Practice is the key to speed and mental agility. Happy computing and good luck!

7 comments:

onkar said...

hi I have been reading your blog about this vedic maths.It is reallly cool.I am really thankful.It will makes maths fun for anyone.In this chapter in the example of 53*55 it will be 15 not 12 so please correct it.

Blogannath said...

You are right! I must have changed the problem for some reason without changing the derivation of the solution. I have corrected the error now. Thank you.

selpahi said...

How do I calculate problems in which both factors have a different number of digits? For example 365*21.
Thank you. And great website, I learned a lot!

Blogannath said...

Please take a look at later posts in the series such as the following two: http://blogannath.blogspot.com/2009/07/vedic-mathematics-lesson-11-vertically.html, and http://blogannath.blogspot.com/2009/07/vedic-mathematics-lesson-12-vertically.html.

ajay said...

in 53*55 = when we apply the first part that is multiple of power 10. 58*5 = 290 but here it is 58/2 = 29 so answer differs that is 29015 and 2915 please clear my doubt thank you.

ajay said...

in 53*55 = when we apply the first part that is multiple of power 10. 58*5 = 290 but here it is 58/2 = 29 so answer differs that is 29015 and 2915 please clear my doubt thank you.

Blogannath said...

The answer is not 29015 when you use 10 as the base because 15 is a two-digit number, and so you need to carry over the 1 to the left (remember that the number of digits on the right has to be the same as the number of zeroes in the base). Once you do the carryover, you will once again get 2915, not 29015.

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