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Saturday, July 25, 2009

Vedic Mathematics Lesson 12: Vertically and Cross-wise II

In the previous lesson, we learnt the basic application of the Urdhva Tiryagbhyam sutra to simple problems of multiplication.

You can find all the 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

Multiplication Special Case 3

Vertically And Crosswise I

Before we dive into the application of the method to products of numbers with more than two digits, let us study the pattern of vertical and cross-wise multiplications the technique in the previous lesson involved.

The procedure started out using just the right-most digits of the numbers in a vertical multiplication. Then the method expanded to the left and involved the next column of the numbers as well as the right-most column. This was a cross-wise multiplication. Then the method proceeded still further to the left, but since there were no more digits to the left, the right-most digits were excluded from the multiplication, leaving us with just the left-most digit in a vertical multiplication. Thus the method expanded from right to left, starting with a vertical multiplication of the right-most digits, until the left-most digit was involved in a cross multiplication with the right-most digit, then contracted from right to left, leaving us in the end with just the left-most digits involved in a vertical multiplication. There is an inherent symmetry in the procedure and this symmetry will help us remember the procedure very easily.

Now let us start by working out a simple problem: 123 x 321. After that, we will expand on the algebraic explanation of the technique to provide a general formula, and then finish with more examples that illustrate the expansion and contraction of the cross-products and the symmetry inherent in the procedure. To find 123 x 321, we will start with the figure below:




As in the previous lesson, we will start by vertically multiplying the right-most digits of the numbers and putting the product down as the right-most digit of the answer. The product in this case is 3 x 1 = 3.





The numbers involved in the multiplication and the corresponding product are denoted by red letters in the figure above.

Now, we expand the multiplication from right to left, thus involving the second and third digits of both numbers in a cross-multiplication. The cross-multiplication produces the answer 2 x 1 + 2 x 3 = 8.





The numbers involved in the multiplication and the corresponding product are denoted by orange letters in the figure above.

Now, we expand the multiplication still further to the left, thus involving all three digits of each of the two numbers. The cross-multiplication in this case will involve the product of the first and third digits, the second and second digits and the third and first digits of the two numbers. Notice the symmetry of the cross-multiplication. Note also, that when the cross-multiplication involves an odd number of digits, the middle digits are multiplied vertically with each other and added to the cross-product. The cross-product we get is 1 x 1 + 2 x 2 + 3 x 3 = 14. Since this product consists of two digits, we are going to write the right digit on the first row of the answer and the left digit on the carryover row as shown below:






The numbers involved in the multiplication and the corresponding product are denoted by light blue letters in the figure above.

Since we have now expanded the cross-multiplication all the way to the left-most digits of the numbers, we start the contracting from right to left. We exclude the right-most digits from the next cross-product, giving us the cross-product 1 x 2 + 3 x 2 = 8.






The numbers involved in the multiplication and the corresponding product are denoted by dark blue letters in the figure above.

We now contract further by excluding the middle digits and perform a vertical product of the left-most digits of the numbers. We get 1 x 3 = 3.






The numbers involved in the multiplication and the corresponding product are denoted by violet letters in the figure above.

Adding up the two rows of the answer gives us 39483, which can be verified to be the correct answer.

Let us now examine the algebraic basis for the technique with respect to 3-digit numbers in particular. Any 3-digit number can be expressed as az2 + bz + c, where z = 10. So, the multiplication of 2 3-digit numbers can be expressed as the product (az2 + bz + c) x (dz2 + ez + f). Expanding out the terms, we immediately see our vertical and cross-products clearly. The answer has deliberately been written with the constant term first and the powers of z increasing to mimic our performance of the multiplication from right to left (even though numbers are normally written with the powers of 10 decreasing from left to right by convention).

cf + (bf + ec)z + (af + be + cd)z2 + (ae + bd)z3 + adz4

The cross and vertical products we performed are readily apparent in the algebraic formula above.

The actual algebraic expansion of the product of any two numbers of any length can be worked out easily and it can be verified that the pattern of vertical and cross-wise multiplications follows the expansion and contraction from right to left as we mentioned earlier. The following sequence of figures below shows the pattern of multiplications for a product of 2 six-digit numbers. The numbers are abcdef and ghijkl (where each letter actually denotes a numerical digit). Similar figures can be drawn for any number of digits easily.

Units digit (100)

Tens digit (101)

Hundreds digit (102)

Thousands digit (103)

Ten Thousands digit (104)

Hundred Thousands digit (105)

Millions Digit (106)

Ten Millions Digit (107)

Hundred Millions Digit (108)

Billions Digit (109)

Ten Billions Digit (1010)

Now, we will conclude this lesson by working out a couple more examples to solidify our understanding of the concepts covered in this lesson. In the following examples, each vertical and cross-product is written on a separate line so that explanations can be provided. Note that this is done only for clarity, and in general, even with carryovers, most problems can be solved by writing the numbers involved in two or three lines at most (note that even if it is not possible to solve the problem mentally, this technique will involve much fewer lines of numbers to calculate the final answer than will usually be the case with full long multiplication).




••••••18 (vertical multiplication, 2 x 9)

•••••95• (cross-wise multiplication, 9 x 9 + 7 x 2)

•••141•• (cross-wise multiplication, 8 x 9 + 9 x 7 + 3 x 2)

••118••• (cross-wise multiplication, 3 x 9 + 8 x 7 + 3 x 9 + 4 x 2)

••81•••• (cross-wise multiplication, 3 x 7 + 8 x 3 + 4 x 9)

•41••••• (cross-wise multiplication, 3 x 3 + 4 x 8)

12•••••• (vertical multiplication, 3 x 4)






•••••••••6 (vertical multiplication, 3 x 2)

••••••••4• (cross-wise multiplication, 2 x 2 + 0 x 3)

••••••16•• (cross-wise multiplication, 8 x 2 + 2 x 0 + 0 x 3)

•••••24••• (cross-wise multiplication, 9 x 2 + 8 x 0 + 0 x 2 + 2 x 3)

••••42•••• (cross-wise multiplication, 7 x 2 + 9 x 0 + 8 x 0 + 2 x 2 + 8 x 3)

•••32••••• (cross-wise multiplication, 7 x0 + 9 x 0 + 2 x 8 + 8 x 2)

••82•••••• (cross-wise multiplication, 7 x 0 + 9 x 2 + 8 x 8)

•86••••••• (cross-wise multiplication, 7 x 2 + 8 x 9)

56•••••••• (vertical multiplication, 7 x 8)



Hope these samples clarify the application of this method to multiplication problems of all sorts. Multiplying long numbers with each other is always a challenge using any method. Hopefully this lesson has provided a new perspective on how to tackle such problems. Practice will enable one to master this technique so that it becomes easy to apply when the need arises. Happy computing, and good luck!


Anonymous said...

I didn't understand dat 4-digit multiplication(3892*4379),can u pls tell me how to place numbers after multiplying each time..

Blogannath said...

Well, the first multiplication is of the last digits. This is 2x9 = 18. Now, the second step is to expand the multiplication to the last two digits. So, you get 9x9 + 2x7 = 95. Write this one digit to the left of the previous answer. So, you would write 95 such that the 5 is below the 1 of the 18. Then expand the multiplication to the last 3 digits. Write that answer one more digit to the left, and so on. Keep moving the last digit of each answer one digit to the left at each step.

Rahul Bhangale said...

This is a great post of UrdhvaTiryak Sutra.

Even I like Urdhva Sutra a lot. Its a great shortcut to multiply any digits of numbers.

But if the numbers have digits greater than 5 then Urdhva Sutra becomes little difficult. Hence if we use it with Vinculum Process, UrdhvaTiryak for such numbers will be very effective.

For such numbers, I have solved Urdhva Tiryak Sutra with Vinculum Process
Checkout here -> Urdhva Tiryak Sutra & Vinculum Process

lim lin said...

Can you pls tell me why when i do 1201 X 2012 I separate the number into

12 01
x20 12

24 164 12

Why the 1 from the 164 does not need to carry to 24? I would like to know the reason behind as i know the base is 1000

Blogannath said...

20x12 is 240, not 24. So, the 1 from 164 *is* carried over, but it is carried over to 240, not to 24.

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