## Friday, July 3, 2009

### Vedic Mathematics Lesson 4: Multiplication Part 2

In this lesson, we are going to explore multiplication further and learn a few more techniques for mentally working out large multiplication problems without having to do long multiplication, or relying on 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

In the previous lesson we dealt with large numbers that are just under a power of 10. We solved problems like 9986 x 9997 as illustrations of the technique. Today, we will modify the method slightly to deal with problems such as 1002 x 1005. In short, we will learn to multiply large numbers that are just above a power of 10.

As in the previous lesson, it is best to start from the basics and build up from that. As a basic illustration, let us solve the problem of finding the product 11 x 12. This is a very basic problem that most people will know the answer to based on their memory of multiplication tables. 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 11 x 12:

• 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 contain "+" because the numbers are larger than our base, 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 (we dealt with this case in the previous lesson. This may be a good time to review that lesson and refresh your memory before proceeding).
• If the middle column is a "+", in the right hand side, write the amount by which the number is greater than the base.

After these steps, we get the figure below:

11 + 1
12 + 2

Our first number is 11. It is more than our base, 10, so there is a "+" in the middle column. In the right hand side, we have the excess of 11 over 10, which is 1. Similarly, in the second row, we have our second number, 12, a "+" because 12 is greater than our base, 10, and the excess of 12 over 10, 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:

11 + 1
12 + 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 11 + 12 - 10 = 13.
• Add the excesses on the right hand side and add that to 10. This gives us 10 + 2 + 1 = 13.
• Cross add the excess on the first line to the second number. This gives us 12 + 1 = 13.
• Cross add the excess on the second line to the first number. This gives us 11 + 2 = 13.

So, our figure now looks as below:

11 + 1
12 + 2
--------
13 |

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

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

Our figure will now look as below:

11 + 1
12 + 2
--------
13 |2

The answer to the problem is the combination of the left and right hand sides of the answer above, which gives us 132. This is obviously easy to verify as the correct answer not only with a calculator, but also by using our memory of multiplication tables.

The algebra behind this is very similar to the algebra behind why the method we covered in Part 1 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 excesses of y and z over 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 sum of the base and the excesses, then multiplied it by the base so that it becomes the left hand side of the answer. Then we multiplied the excesses 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. Note that the solutions to these issues are going to be very similar to the solutions we uncovered in Part 1 when we were multiplying numbers just below our base.

Let us take 104 x 105. First we choose 100 as the base. Then we write the problem out as below:

104 + 4
105 + 5
--------
|

Note that both numbers are above the base, hence the "+" sign in the middle column. 4 and 5 are their excesses from our base of 100 respectively.

Now, we find the left hand side of the answer using any one of the 4 methods outlined earlier. We get either 104 + 105 - 100 = 109, or 105 + 4 = 109, or 104 + 5 = 109 or 100 + 4 + 5 = 109. Our figure becomes:

104 + 4
105 + 5
---------
109 |

Now we multiply the excesses by each other to find the right hand side of our answer. It is 4 x 5 = 20. So, our final answer is 10920, and our figure looks as below:

104 + 4
105 + 5
---------
109 | 20

As mentioned earlier, the special cases we might encounter are identical to the special cases we encountered in Part 1. Our right hand side could contain too few or too many digits. In both cases, the solution in this case is identical to how we dealt with the problem in Part 1.

The right hand side of the answer has to contain the same number of digits as there are zeroes in the base. That follows directly from the algebraic basis of the method. So, if the right hand side contains too few digits, we pad the right hand side with zeroes to the left until it has the appropriate number of digits. This is illustrated in the problem below:

101 + 1
102 + 2
------------
103 | 02

The answer is 10302 rather than 1032.

Similarly, if we were to work out 110 x 112, we would get the figure below:

110 + 10
112 + 12
--------------
122 + 1| 20

The answer is 12320 rather than 122120 because the extra digit on the right hand side (1) is to be carried over to the left hand side.

We will now find the answer to the problem posed at the top of the lesson: what is the product of 1002 and 1005. We draw our figure as below using 1000 as the base:

1002 + 2
1005 + 5
---------------
1007 | 010

We note that since our base is 1000, we need to have 3 digits in the answer on the right hand side. Since the product of the excesses (5 x 2 = 10) has only 2 digits, we pad it with a zero to the left to get 010. The answer to our product thus becomes 1007010.

We will conclude the lesson by dealing with the other type of special case with a couple of large numbers. Let us find the product 1100 x 1100. We draw the figure below:

1100 + 100
1100 + 100
---------------
1200 | 10000

We see that the right hand side contains 5 digits, so the left-most 2 digits become a carryover to the left hand side, leaving us with the figure as below:

1100 + 100
1100 + 100
--------------
1200 + 10 | 000

This gives us the answer 1210000 which can be verified to be correct using a calculator or computer.

In subsequent lessons, we will expand this method even further by talking about how to multiply numbers that are on either side of a power of 10, and numbers that are close to some multiple or factor of a power of 10. Until then, happy computing and good luck!

Anonymous said...

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SR lakshmi said...

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