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Thursday, July 2, 2009

Vedic Mathematics Lesson 3: Multiplication Part 1

You can read about my interest in Vedic Mathematics in this earlier post, and read Lesson 1 and Lesson 2 to appreciate the power of Vedic Mathematics, and to learn the basics that we are going to use in this lesson.

In this lesson, our main goal will be to learn to multiply large numbers that are just under a power of 10 without doing any long multiplication. Imagine trying to compute 9986 x 9997 without going through 4 sets of laborious multiplications and then a laborious addition. When we are done with this lesson, we will be able to do the above problem in our minds and find the answer in about 3 to 5 seconds.

As always, it is best to start from the basics and build up to what we want to accomplish. Let us go through a simple example to illustrate the method, then I will explain the algebraic logic behind why the method works, then we will apply it to bigger problems and also explore some additional considerations.

Assume that we have to find the answer to 9x8. This is quite simple to do for most people who know their multiplication tables and should not require any Vedic Mathematics to accomplish. However, we will use this problem to illustrate the method and extend it to products that would not normally be covered by multiplication tables.

Follow these steps to find 9x8:

  • Find the appropriate base for our calculations. The base will be the power of 10 that is closest to the numbers to be multiplied. In this particular case, our base will be 10 itself.
  • Put the numbers in two rows on the left hand side. In the middle column, put a "-" if the number is less than the base and a "+" if the number is more than the base. In our case, the middle column for both rows will be "-" since they are both less than 10.
  • If the middle column is a "-", in the right hand side, write the 10's complement of the number, or the deficit from the base.
  • If the middle column is a "+", in the right hand side, write the amount by which the number is greater than the base (we will deal with this case in a subsequent lesson).

After these 4 steps, we get the figure below:

9 - 1
8 - 2

Our first number is 9. It is less than our base, 10, so there is a "-" in the middle column. In the right hand side, we have the 10's complement of 9, which is 1. Similarly, in the second row, we have our second number, 8, a "-" because 8 is smaller than our base, 10, and the 10's complement of 8, which is 2.

Now, the product will have two parts, a left hand part and a right hand part. We can draw a vertical line on the third row to demarcate the two parts (this will not be necessary as we become better at the method with practice and learn to do the whole thing mentally). So, now we have the figure below:

9 - 1
8 - 2
-------
|

The left hand side of the answer can be found in one of 4 different ways:

  • Add the numbers on the left hand side and subtract the base from the answer. This gives us 9 + 8 - 10 = 7.
  • Add the deficiences on the right hand side and subtract that from 10. This gives us 10 - 2 - 1 = 7.
  • Cross subtract the deficiency on the first line from the second number. This gives us 8 - 1 = 7.
  • Cross subtract the deficiency on the second line from the first number. This gives us 9 - 2 = 7.

So, our figure now looks as below:

9 - 1
8 - 2
--------
7 |

The right hand side of the answer can be found with one simple calculation:

  • Simply multiply the deficits by each other to get the right hand side of the answer.

Our figure will now look as below:

9 - 1
8 - 2
--------
7 |2

The answer to the problem is the combination of the left and right hand sides of the answer above, which gives us 72. This is obviously easy to verify as the correct answer not only with a calculator, but also by using our memory of multiplication tables.

What is the algebra behind why this method works? Assume that our base is b. Let us assume that the two numbers whose product we have to find are y and z. Let us assume that d and e are the deficits of y and z from our base, b. So, we are trying to find the product y x z which is the same as (b - d) x (b - e). We see that this can be written as b x (b - d - e) + d x e.

Which is precisely what we did using the method illustrated in the steps above: We found the left hand side as the difference between the base and the sum of the deficits, then multiplied it by the base so that it becomes the left hand side of the answer. Then we multiplied the deficits by each other and made it the right hand side of the answer.

Now, let us illustrate the method on more complex problems. These problems will not only give us practice, but also illustrate how to deal with some issues we might encounter along the way.

Let us take 96 x 95. First we choose 100 as the base. Then we write the problem out as below:

96 - 4
95 - 5
--------
|

Note that both numbers are below the base, hence the "-" sign in the middle column. 4 and 5 are the 10's complements of 96 and 95 respectively (their deficits from our base of 100).

Now, we find the left hand side of the answer using any one of the 4 methods outlined earlier. We get either 96 + 95 - 100 = 91, or 95 - 4 = 91, or 96 - 5 = 91 or 100 - 4 - 5 = 91. Our figure becomes:

96 - 4
95 - 5
---------
91 |

Now we multiply the deficits by each other to find the right hand side of our answer. It is 4 x 5 = 20. So, our final answer is 9120, and our figure looks as below:

96 - 4
95 - 5
---------
91 | 20

Now, let us apply this method to problems which present special cases we have not encountered so far.

First, let us try to do 99 x 99. Following the methodology outlined above, we get the figure below:

99 - 1
99 - 1
--------
98 | 1

We quickly realize that 981 is not the answer to the problem. Is there something wrong with the Vedic method? Not really. Looking at the algebraic explanation of the method, we find that the answer is actually the left hand side of the answer multiplied by the base, added to the right hand side of the answer. This means that the right hand side of the answer has to have exactly as many digits as the number of zeroes in the base (when we multiply the left hand side by the base, in this case, 100, we get 9800. The right hand side is actually added to 9800, not just appended to 98 though that is the practical effect when the right hand side contains exactly two digits). In this case, the base is 100, so the right hand side has to have 2 digits. We accomplish this by padding the answer with zeroes to the left until we get the requisite number of digits. Thus our figure becomes:

99 - 1
99 - 1
---------
98 | 01

Now our answer is 9801 and this can be verified to be correct using a calculator, or by long multiplication.

Next, let us see what happens when we try to multiply 90 x 88. We get the figure below:

90 - 10
88 - 12
---------
78 | 120

Obviously 78120 is not the correct answer to the problem. But the algebraic explanation of the problem comes to the rescue again. We see that the answer, algebraically, is actually 78 x 100 + 120. The right hand side has to be restricted to two digits. In our case, we have 3 digits, so retain the right most 2 digits and use any leftover digits as carryover to increase the left hand side by. This leads, in this case, to the right hand side becoming 20 and the 1 becoming a carryover digit. Adding 1 to the left hand side gives us 79. Our figure now becomes:

90 - 10
88 - 12
----------
78+1 | 20

The answer is 7920, which can be verified to be correct.

Now, let us tackle the problem we initially posed in this lesson: what is 9986 x 9997? We quickly realize that the base required for this problem is 10000. Calculating the deficits of the numbers from 10000, we draw the figure below:

9986 - 14
9997 - 3
--------------
|

Next we find the left and right hand sides of the answer as below:

9986 - 14
9997 - 3
--------------
9983 | 42

Now we note that the right hand side contains only 2 digits whereas 10000 has 4 zeroes. So, we need to pad 42 out to 4 digits, giving us 0042. The figure now looks as below:

9986 - 14
9997 - 3
--------------
9983 | 0042

This automatically leads to the answer 99830042 which can be verified using a calculator.

In the interest of a little more practice, and the application of another of the special cases we are likely to encounter, let us try to find 9900 x 9900. We use 10000 as the base, which leads to the figure below:

9900 - 100
9900 - 100
-------------
9800 | 10000

Immediately, we see that the right hand side actually has 5 digits, one more than it should have. Using the rule regarding carryovers, we modify the diagram as below:

9900 - 100
9900 - 100
--------------
9800 + 1 | 0000

This leads to the answer 98010000 which can be verified to be correct.

Now, let us do 9000 x 9000. Again the base is 10000, which leads to the figure below:

9000 - 1000
9000 - 1000
----------------
8000 | 1000000

Our right hand side has 3 extra digits. Moving them to the left hand side using the rule regarding carryovers, we get:

9000 - 1000
9000 - 1000
----------------
8000 + 100|0000

This gives us the correct answer of 81000000.

In subsequent lessons, we will expand on what we learned in this lesson so that we can handle a wider variety of problems. After all, not all the numbers we need to multiply are just under a power of 10! Practice makes perfect, so happy practicing and good luck!!

10 comments:

Dhwani Kapoor said...

very helpful..
the simple language and examples are good for a new learner.
thanks!

Blogannath said...

Thank you. I am glad you found the post simple and helpful.

Anonymous said...

it will be helpful if you post the problems over here....or suggest me a good book to do the same...where i can apply some of these tricks very easily...

Blogannath said...

I have provided only the method along with some examples in my blog, not actual exercises with sample problems. If you are interested in sample problems, you can try the series of books called "Vedic Mathematics For Schools" by James T. Glover. One of them is available on Amazon.com, and the whole series may be available in your local bookstore. Good luck.

Anonymous said...

Nice post...
How about doing multiplication of 11x88, using base as 100..??

Blogannath said...

As the second paragraph of my post explains, this post is "to learn to multiply large numbers that are just under a power of 10". As such, 11 and 88 are not two numbers that are just below a power of 10, so you have to read other posts in this series to figure out how to do that without long multiplication. This post can not help you with that.

Ketul Patel said...

very detailed explanation. Thank you very much. if you consider twisting this info in a manner more helpful for those preparing for competitive exams, it would be even better. The present one is excellent, but this is just a suggestion.

Anonymous said...

Superb blog !

venkatesh said...

have u mentioned anythng regarding multiplication of numbers with different base... for example 125*525? pls help me...

Shanthi said...

Great Article thanks a lot for sharing with us

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