In the previous lessons we learnt how to do multiplication of numbers that are close to a power of 10, or close to a multiple of sub-multiple (factor) of a power of 10. Unfortunately, we know that not all products involve numbers that are so well-behaved. If the numbers are very far apart, then there will obviously be no powers of 10 or multiples or sub-multiples thereof that will be close enough to both numbers. This will make the use of the techniques described in the previous lessons inconvenient, and it may not be possible to use them mentally at all.

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

Multiplication Part 5

Multiplication Special Case 1

Multiplication Special Case 2

In this lesson, we will learn the application of a more general mental multiplication technique that is described in a Vedic Sutra in the following words: Urdhva Tiryagbhyam. Literally, it means Vertically and Cross-wise. This sutra is used in various contexts, for various applications, but its most common and elementary application is in the context of multiplication.

To illustrate the application of this technique, we will start with a simple example. Then, we will examine the algebraic basis for the technique, which will clarify the application of the technique to some special cases we may encounter in practice. The algebraic basis of the technique will also allow us to appreciate the full power and potential of the technique for all kinds of multiplication tasks.

Let us see how we would compute the product 52 x 21 using this method. First write 52 and 21 on two rows, with the digits aligned vertically. This gives us:

52

21

---------

Now, start from the right hand side. Multiply the right-most digits of the two numbers vertically (as the name suggests), and write the product down as the right-most digit of the answer. We get 2 as the right-most digit of the answer since 2 x 1 = 2:

52

21

----------

••2

Now, cross multiply the first digit of the first number with the second digit of the second number and add the result to the cross product of the second digit of the first number and the first digit of the second number (this is the cross-wise in the name of the technique). Write this number down as the second digit of the answer. We get 9 since 5 x 1 + 2 x 2 = 9

52

21

----------

•92

As the last step, multiply the first digits of the two numbers vertically, and put the product down as the left-most digit of the answer. We get 10 since 5 x 2 = 10.

52

21

-----------

1092

We can verify with a calculator that the product is indeed 1092. Because we perform either vertical multiplication or cross-products for the different digits of the answer, the name of the method makes sense and is easy to remember.

Let us now examine the algebraic basis for the method. To do so, we first note that any two-digit number can be written as az + b where z = 10. So the two numbers whose product is to be found can be expressed as az + b and cz + d. Notice that (az + b) x (cz + d) = acz^{2} + (ad + bc)z + bd.

It is immediately obvious that ac is the vertical product of the first digits of the numbers to be multiplied, bd is the vertical product of the last digits of the numbers to be multipled, and (ad + bc) is the sum of the cross-products of the digits in the two numbers. Notice the symmetry of the technique as we start with a vertical product at the right, proceed to a cross-product involving both right and left hand sides, and then continue on to a vertical product at the left. This symmetry will be expanded upon when we extend this method to deal with products of numbers that are more than two digits each.

What is also immediately obvious is that the right-most vertical product represents the units digit of the answer, the sum of the cross-products represents the tens digit of the answer, and the left-most vertical product represents the hundreds digit. Because each of them represents one place in the final answer, if any of these numbers have more than one digit, we have to carry over to the left any excess digits. That is one of the reasons we work the problem out starting from the extreme right so that we can keep track of the carryovers, if there are any, when we find the digits towards the left.

To illustrate the concept of carryover, let us work out a few examples. We will start with 75 x 92. We will use colors in the figure below to make sure the technique is well-understood since the carryovers can cause some confusion. Let us first draw the figure below:

75

92

--------

Now, we multiply the right-most digits vertically and find that the product is 10. Since this is 2 digits long, we write the right-most digit on one line and the carryover digit (the 1) on the next line to make sure we understand that it is carried over:

75

92

---------

••0

•1

Both the red numbers with yellow backgrounds in the figure above are derived from the first vertical multiplication. Now, let us perform the cross-multiplication of the digits. We get 7 x 2 + 9 x 5 = 14 + 45 = 59 as the cross-product. Just like we did for the vertical product, we will write the two digits of 59 on different lines to distinguish the carryover from the non-carryover number:

75

92

---------

•90

51

Both the green numbers with violet backgrounds in the figure above are derived from the cross-multiplication. Now, let us perform the final vertical multiplication of the left-most digits. We obtain 63 since 7 x 9 = 63. Since there are going to be no more digits to the left, we can put down the 63 as it is on the first line instead of spreading it over 2 lines. We therefore, get the figure below:

75

92

---------

6390

•51

Both the dark blue numbers with light blue backgrounds in the figure above are derived from the last vertical product. Now add up the two rows of numbers to derive the final answer:

75

92

---------

6390

•51

---------

6900

We need to initially resort to the procedure of writing the answer down in two lines and adding them up to get the final answer to keep the carryovers from becoming too confusing. But as we practice the technique over and over again, it will become easy to keep the carryovers straight mentally, so that we can reel off the answer in one step for any multiplication involving two digits. To illustrate the technique further, I have worked out a few more problems. The color schemes are the same as what was used in the above example so that it is easy to follow the different steps all the way to the derivation of the final answer.

84

29

---------

1606

•83

---------

2436

99

99

---------

8121

168

---------

9801

79

08

---------

062

57

---------

632

Notice that we padded the second number in the product above with a zero to the left so that it is 2 digits long. When we extend this technique to products of numbers with more digits, we will see that the padding of numbers with zeroes to the left is a common and needed technique to make the numbers the same length so that this technique can be applied mechanically.

And finally:

50

50

---------

2500

Note that even though this lesson has explained the procedure starting with the vertical multiplication of the right-most digits, we can proceed in the opposite direction, starting with the vertical product on the left-hand side, the cross-product and then the vertical product on the right-hand side. It is more difficult to keep track of the carryovers from the right to the left while at the same time solving the problem from left to right, especially when we are trying to solve the problem mentally, that is why solving from left to right has been avoided. But if we write the answer down on paper using rows of numbers as in the figures above, there is no difference between starting from the left and starting from the right. You just have to remember that carryover always proceeds from right to left and only the right-most digit of each product is retained while the rest of the digits are carried over to the left.

Hopefully, the examples above have provided a good sample of what to expect during the application of this method to 2-digit numbers. Practice will make the application of the technique very easy and quick. In the next lesson, we will extend this method to multiplication involving numbers that are more than 2 digits long. Happy practicing, and good luck!

## 4 comments:

I love the vertically and crosswise method, but I must correct you in that 52 x 21 = 1092, by the method, not 1091!!!

Of course, you are absolutely right. I must have started out with a different example and changed some numbers but forgot to change some others. Thanks for catching it! I have now corrected it. Please don't hesitate to let me know if you spot any other errors in these posts.

I 'm not getting this 658*795 by criss cross method...can u pls teach me

This post only deals with using this method for 2-digit numbers? Did you check out the next lesson in the series (http://blogannath.blogspot.com/2009/07/vedic-mathematics-lesson-12-vertically.html)? That one deals with the method to use when you are dealing with more than 2 digits.

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