There are methods, which are totally unconventional but help in Carrying out tedious and cumbersome arithmetical operations, easily, speedily and in some cases executing them mentally.

People who are deeply rooted in the conventional methods, may find it difficult, at first reading, to understand the methods.

But, these methods, based on the Sutras (aphorisms or Formulas) of Ancient Indian Vedic Mathematics are simple and easy to understand, remember and apply, even by little children.

Here, we will apply a sutra of Vedic Math as applied to multiplication of two digit numbers.

THE URDHVA TIRYAK SUTRA (meaning : Vertically and cross-wise) :

This is a general Formula applicable to all cases of multiplication.

Using this principle, we can find the product of two numbers easily.

Multiply vertically and crosswise to get the digits of the product.

Examples will clarify the method.

Before seeing the examples, let us see the formula for finding the product of two digit numbers.

Let us say the two digits of the first number be 'a' (tens' digit) and 'b'(units' digit).

And those of the second number be 'p' (tens' digit) and 'q'(units' digit).

Write the digits of the two numbers one below the other as follows.


a b
p q


The product of these numbers has three parts which are given below seperated by '/'.


a b
p q
------
ap/(aq+pb)/bq
------


'ap' is the Hundreds' part which is the vertical product of the first column.

(aq+pb) is the Tens' part which is the sum of the cross-wise products 'aq' and 'pb'.

'bq' is the units' part which is the vertical product of the second column.

Let us see the method by examples.

EXAMPLE 1 :

To find 21 x 13


21
13
-----
2x1/2x3+1x1/1x3
= 2/7/3
21 x 13 = 273


The product has three parts and each part is seperated by slash (/).

First part of the product = product of first vertical digits = 2x1 = 2

Second part of the product = sum of the cross-wise products = 2x3+1x1 = 6+1 = 7

Third part of the product = product of second vertical digits = 1x3 = 3

Thus 21 x 13 = 273.

EXAMPLE 2 :

To find 41 x 51


41
51
----
4x5/4x1+5x1/1x1
=20/9/1
41x51 = 2091


The product has three parts and each part is seperated by slash (/).

First part of the product = product of first vertical digits = 4x5 = 20

Second part of the product = sum of the cross-wise products = 4x1+5x1 = 4+5 = 9

Third part of the product = product of second vertical digits = 1x1 = 1

Thus 41 x 51 = 2091.


EXAMPLE 3 :


To find 23 x 42


23
42
----
2x4/2x2+4x3/3x2
=8/₁6/6
=8+₁/6/6
=9/6/6
23x43 = 966


The product has three parts and each part is seperated by slash (/).

First part of the product = product of first vertical digits = 2x4 = 8

Second part of the product = sum of the cross-wise products = 2x2+4x3 = 4+12 = 16

This is a two digit number.

Units digit (6) is retained and tens' digit (1) is carried over to the immediate left place.

See that in 16, 1 is written in small letters indicating its carry over to the immediate left place and 6 is written normally indicating it is retained in its place.

Third part of the product = product of second vertical digits = 3x2 = 6

Thus 23 x 42 = 966.

To explain the procedure clearly, so many steps are shown.

In practice we can do the calculations mentally and write the answer in fewer steps as follows.



23
42
----
8/16/6
=9/6/6 (worked out from right)

23 x 42 = 966.



EXAMPLE 4 :

To find 37 x 29


37
29
----
3x2/3x9+2x7/7x9
=6/27+14/₆3
=6/27+14+6/3
=6/₄7/3
=6+₄/7/3
=10/7/3
37x29=1073


The product has three parts and each part is seperated by slash (/).

First part of the product (before carry over from second part) = product of first vertical digits = 3x2 = 6

Second part of the product (before carry over from third part) = sum of the cross-wise products = 3x9+2x7 = 27+14


Third part of the product = product of second vertical digits = 7x9 = 63

This is a two digit number.

Units digit (3) is retained and tens' digit (6) is carried over to the immediate left place.

See that in 63, 6 is written in small letters indicating its carry over to the immediate left place and 3 is written normally indicating it is retained in its place.

Now,

Second part of the product (after carry over from third part) = 27+14+6 = 47

This is a two digit number.

Units digit (7) is retained and tens' digit (4) is carried over to the immediate left place.

See that in 47, 4 is written in small letters indicating its carry over to the immediate left place and 7 is written normally indicating it is retained in its place.

Now,

First part of the product (after carry over from second part) = 6+4 = 10

Thus 37 x 29 = 1073.

To explain the procedure clearly, so many steps are shown.

In practice we can do the calculations mentally and write the answer in fewer steps as follows.


37
29
----
6/41/63
=10/7/3 (worked out from right)

37 x 29 = 1073.


EXAMPLE 5 :

To find 68 x 79


68
79
----
6x7/6x9+7x8/8x9
=42/54+56/₇2
=42/110+₇/2
=42/₁₁7/2
=42+₁₁/7/2
=53/7/2
68x79=5372


By this time, you are familiar with the procedure and you can understand the above problem.

Thus 68 x 79 = 5372.

To explain the procedure clearly, so many steps are shown.

In practice we can do the calculations mentally and write the answer in fewer steps as follows.


68
79
----
42/110/72
=53/7/2 (worked out from right)
68 x 79 = 5372.



PROOF OF THE METHOD ADOPTED :

We know,

(ay + b)(py + q) = y²(ap) + y(aq + bp) + bq

Multiplying two digit numbers is similar with y = 10

10² term and units term are vertical products and 10 term is cross-wise products' sum.

[a(10) + b][p(10) + q] = 10²(ap) + 10(aq + bp) + bq

THE DIGITS OF THE TWO NUMBERS :


a b
p q


THE THREE PARTS OF THE PRODUCT :

ap/(aq + pb)/bq (proved.)

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