## Saturday, July 4, 2009

### Vedic Mathematics Lesson 5: Multiplication Part 3

In this lesson, we are going to explore multiplication even 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
Multiplication Part 2

In the previous lessons, we dealt with the multiplication of large numbers that were both either just under a power of or just above a power of 10. Thus, we solved problems like the finding the product 9986 x 9997, and 1002 x 1005 as illustrations of the techniques we learned. In this lesson we will deal with multiplication of two numbers one of which is just below a power of 10, with the other being just above that power of 10. In short, by the end of this lesson, we should be able to solve problems like 996 x 1006 in 3 to 5 seconds mentally, without the need for calculators or a computer.

Just like in previous lessons, let us start with a simple problem to illustrate the approach and then examine the algebraic basis of the method, so that we can then apply it with confidence to larger problems. Let us solve the problem of finding the product 8 x 13. 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 8 x 13:

• 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.
• 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:

8 - 2
13 + 3

Our first number is 8. It is less than our base, 10, so there is a "-" in the middle column. In the right hand side, we have the deficit of 8 from 10 (also known as the 10's complement of 8), which is 2. Similarly, in the second row, we have our second number, 13, a "+" because 13 is greater than our base, 10, and the excess of 13 over 10, which is 3.

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:

8 - 2
13 + 3
-------
|

The left hand side of the last row of the figure (I am not going to call it the left hand side of the answer, and the reason will become clear as we proceed) can be found in one of 4 different ways:

• Add the numbers on the left hand side and subtract the base from it. We get 8 + 13 - 10 = 11.
• Cross add the number in the left column of the first row with the signed number in the right column of the second row (by signed number, we mean, prepend the sign in the middle of the second row to the number in the right column of the second row). We get 8 + 3 = 11 (note that +3 is the signed number in the second row because the middle column of the second row is "+" and the number in the right column of the second row is 3).
• Cross add the number in the left column of the second row with the signed number in the right column of the first row. This time, we get 13 - 2 = 11. The number to be added to 13 is -2 because the middle column of the first row is "-" and the number in the right column of the first row is 2.
• Add the signed numbers in the first and second rows and add 10 to the result. In this case we get -2 + 3 + 10 = 11. The signs in front of the numbers are derived by the same logic as above.

Our figure now becomes:

8 - 2
13 + 3
----------
11|

Now find the right hand side of the last row of the figure (once again, the reason why I don't call it the right hand side of the answer will become clearer as we proceed) as the product of the numbers in the right hand column of the 2 rows. Our figure now becomes:

8 - 2
13 + 3
---------
11| 6

At this point, we are going to deviate from the method in the previous lessons. We are not going to declare 116 to be the answer. Instead, do the following three steps to convert the left and right hand sides of the last row into the left and right hand sides of the answer (we will refer to this in subsequent parts of the lesson as the three-step procedure):

• If the right hand side consists entirely of zeroes, the left and right hand sides of the last row are the left and right hand sides of the answer. Skip the next two steps.
• Subtract 1 from the left hand side, and make the result the new left hand side of the answer.
• Take the 10's complement of the right hand side, and make that result the new right hand side of the answer.

After the above two steps, we get the figure below:

8 - 2
13 + 3
---------
10 | 4

We subtracted 1 from the previous left hand side of the answer (11) to get 10 as the new left hand side. We also took the 10's complement of 6 (which was our previous right hand side) to get 4 as the new right hand side. The answer to our problem is 104, which can be verified with a calculator or our memory of multiplication tables.

Examining the algebraic basis of the method will reveal why we had to perform the three-step procedure to arrive at the final answer. Assume that we have find the product y x z. Let y be less than our base, b, and z be greater than our base b. Moreover, let d be the deficit of y from b, and e be the excess of z over b. So, we are interested in finding the product y x z, which can be rewritten as (b-d) x (b+e).

Expanding the above product gives us b x (b-d+e) - d x e. We performed b-d+e using one of the 4 methods provided for finding the left hand side of the last row. We then made it the left hand side of the last row by multiplying it by our base, b. The problem arises because we have to subtract d x e from this answer rather than adding, as we performed in previous lessons. If d x e is zero, then nothing more needs to be done. That is covered by the first step of the three-step procedure outlined above. If it is not zero, that leads to the 2 extra steps of the three-step procedure that result in subtracting the right hand side from the left-hand side multiplied by the base.

This may be a little confusing initially, so let us work out a few more problems to make sure we understand the method, not only for the standard case, but also when some special cases arise, as we shall see shortly.

Let us solve 95 x 106 using the method outlined above. First we see that we have to use 100 as our base. Then we draw the figure below:

95 - 5
106 + 6
------------
|

In the first row, we have our first number, 95, a "-" because 95 is less than our base, 100, and the deficit of 95 from 100, which is 5. In the second row, we have our second number, 106, a "+" because 106 is more than our base, 100, and the excess of 106 over 100, which is 6.

Now, we calculate the left hand side of the last row using any of the four methods outlined earlier. We get 95 + 106 - 100 = 101, 95 + 6 = 101, 106 - 5 = 101 or -5 + 6 + 100 = 101. The right hand side of the last row becomes 30, the product of 5 and 6. Our figure thus becomes:

95 - 5
106 + 6
----------
101 | 30

Now, we perform the three-step procedure to complete the method. The right hand side is not all zeroes, so we do the additional two steps of the three-step procedure. The left hand side of the answer becomes 101 -1 = 100. The right hand side of the answer becomes the 10's complement of 30, which is 70. We then get our final answer as 10070. It is easy to verify that this is indeed the correct answer.

The special cases that may arise are similar to the special cases we dealt with in the previous lessons: The right hand side of the last row may contain too few or too many digits. In either case, we can easily deal with the situtation if we keep the algebraic basis of the solution method in mind. The rules for dealing with these special cases are outlined below:

• The right hand side of the last row should have the same number of digits as the number of zeroes in the base. If this is already the case, nothing additional needs to be done. We just follow the three-step procedure outlined earlier for converting the numbers on the left and right hand sides of the last row into the left and right hand sides of the answer.
• If the right hand side contains too few digits, pad it to the left with zeroes. Then take the 10's complement of that number from the base. Do not take the 10's complement of the original right hand side (with its fewer-than-required digits) from its next higher power of 10 and then pad the result with zeroes. We will illustrate this in an example below.
• If the right hand side contains too many digits, retain as many digits on the right hand side as there are zeroes in the base. Move the remaining digits to the left as a carryover. However, instead of adding the carryover to the left hand side, subtract the carryover from the left hand side. Then follow the three-step procedure outlined earlier for converting the left and right hand sides of the last row into the left and right hand sides of the answer.

To illustrate the case of the right hand side containing too few digits, consider the product 99 x 101. We get the figure below after following all the steps before the special procedure needed to convert the left and right hand sides of the last row into the left and right hand sides of the answer:

99 - 1
101 + 1
-------------
100 | 1

Now, we notice that the right hand side contains a single digit, 1. Since our base is 100, which has 2 zeroes, we need to pad 1 with zeroes to the left until it becomes 2-digits long. We therefore get 01 as the right hand side of the last row. We then proceed to subtract 1 from the left hand side to get 99 and then take the 10's complement of 01 with respect to 100 (our base) to get 99 (if we take the 10's complement of 1 with respect to its next higher power of 10, we would get 9, and padding it with zeroes to the left would give us 09, which is not the correct answer. So, it is important to follow the steps exactly as outlined, and not do the complementing/padding in arbitrary order). This gives us the final answer of 9999 which can be verified to be correct.

Next let us work out the product 88 x 110. We get the figure below after following all the steps before the three-step procedure:

88 - 12
110 + 10
---------------
98 | 120

We see that the right hand side contains 3 digits which is one more than the number of zeroes in our base, 100. So, we carry over the left most digit to the left hand side and instead of adding it, we subtract it from what is already on the left hand side. We are left with 97 and 20 on the left and right hand sides of our figure. Now, we apply the three-step procedure to convert these numbers into the left and right hand sides of our answer. We get 96 and 80, leading to the answer of 9680, which can be verified to be correct.

It is important to practice this thoroughly to become fluent at it so that we can do this correctly in our mind without missing some important subtraction or addition. But before, I leave you, let me solve the problem we started this lesson with: 996 x 1006. We draw the figure below to begin with:

996 - 4
1006 + 6
------------
1002 | 24

We see that our number on the right hand side of the last row, 24, needs an extra digit to be the same length as the number of zeroes in our base, 1000. Once we pad the right hand side with zeroes to the left, we get the answer quickly as 1001976. This can be verified to be true.

Let us solve 900 x 1100 as our final illustration. We get the figure below:

900 - 100
1100 + 100
--------------
1000 | 10000

Since 10000 has 2 digits more than the number of zeroes in our base, 1000, we need to move them to the left hand side and subtract them from 1000. This leads to a left hand side of 990. Now, note that the right hand side consists entirely of zeroes. Therefore, nothing more needs to be done to derive the answer, according to the first step of the three-step procedure outlined earlier. The answer is simply 990000, which can be verified accurate. The case with all zeroes in the right hand side of the final row therefore has important implications, and needs to be reviewed carefully.

Practice is particularly important for this lesson. Practice will not only make you more proficient, but also get you into the habit of working out mathematical problems using the new techniques I introduce here. This is going to be important as we proceed deeper and deeper into the jungle of arithmetic, and the techniques become more and more involved. Hope you will make the time to do so. In the next lesson we will deal with products of numbers that are close to a multiple or submultiple of a power of 10. Happy computing and good luck!

Anonymous said...

how to do 23 x 33

Blogannath said...

Actually, 23 x 33 is not easy to do using the procedure explained in this lesson because both of the numbers are not near any number that is a power of 10. The easiest way to do 23 x 33 is to use the vertically and crosswise method explained here.

SR lakshmi said...

Thanks. Great blogs. For best vedic math solution please visit http://help-homework-math.com/

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