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

The corollary deals with the multiplication of numbers by numbers that consist entirely of 9's, such as 9, 99, 999, etc. Notice that all these numbers have a deficit of 1 with respect to a power of 10. So, this corollary takes advantage of that property in an obvious way. The sub-sutra that deals with this property reads Ekanyunena Purvena. It sounds very similar to the sutra we studied earlier, Ekadhikena Purvena, and is actually the converse of that. Instead of increasing by one, we reduce by one, as will become obvious when we work out the details. Let us take the case of a simple example to illustrate. Let us try to compute 79 x 99 using the techniques we have learnt so far.

Using the methods in this previous lesson, we obtain the figure below:

79 - 21

99 - 1

---------

78 | 21

The answer is 7821. But it is obvious that the answer can be derived without actually drawing the figure above. We simply reduce the number to be multiplied by the number that contains all 9's by 1, to use as the left hand side of the answer. On the right hand side, we simply write down the 10's complement of the number. Let us make this clearer by working out a few more examples:

65 x 99 = 6435 (64 is one less than 65, and 35 is 65's 10's complement)

578 x 999 = 577422 (577 is one less than 578, and 422 is 578's 10's complement)

8621 x 9999 = 86201379 (8620 is one less than 8621, and 1379 is 8621's 10's complement)

In all the examples above, the numbers to be multiplied together both have the same number of digits. What happens when the number of digits is different between the two numbers to be multiplied with each other? To explore, let us take the example of 7 x 99. Using 100 as the base, it is possible to draw the figure below:

7 - 93

99 - 1

---------

6 | 93

The rule then becomes obvious. If the number to be multiplied with the number that contains all 9's has fewer digits, use one less than that number as the left hand side of the answer. For the right hand side, pad the number with zeroes to the left and take its 10's complement with respect to the power of 10 just above the number that contains all 9's. To illustrate this rule, let us consider the following additional examples:

56 x 999 = 55944 (55 is one less than 56, and 944 is the 10's complement of 56 with respect to 1000)

78 x 9999 = 779922 (77 is one less than 78, and 9922 is the 10's complement of 78 with respect to 10000)

987 x 999999 = 986999013 (986 is one less than 987, and 999013 is the 10's complement of 987 with respect to 1000000)

Now, let us explore what happens when the number of digits in the number that consists of all 9's has fewer digits? To explore, let us work out an example such as 82 x 9. Using 10 as the base, we can draw the figure below:

82 + 72

9 - 1

-----------

81 | 72

Using the 3-step procedure from this previous lesson (notice that this is a case of the numbers being on either side of our base, so we have to use the 3-step procedure to derive the final answer), we get the final answer as 738. Obviously, it is still easy to work out the answer using the lessons we have learnt earlier, but it may not be as mentally intuitive as the previous cases. Instead of going through the procedure of drawing the figure above to derive the answer (after the application of the 3-step procedure), we can formalize the derivation of the final answer as below:

Divide the number to be multiplied (multiplicand) by the number consisting of all 9's (multiplier) into two parts: a right-hand part that contains as many digits as the multiplier and a left-hand part consisting of the extra digits

Subtract from the multiplicand one more than the left-hand part of the multiplicand and use as the left hand side of the answer

Take the 10's complement of the right-hand part of the multiplicand and use as the right hand side of the answer

To illustrate this procedure, let us consider the following examples:

7437 x 99 = 736263 (7362 is 7437 - (74 + 1), and 63 is the 10's complement of 37)

3948 x 999 = 3944052 (3944 is 3948 - (3 + 1), and 052 is the 10's complement of 948)

73367 x 999 = 73293633 (73293 is 73367 - (73 + 1), and 633 is the 10's complement of 367)

Another way to handle this kind of problem is also presented below. Depending on personal tastes, it is possible to use either method. Let us take the example of 82 x 9 once again. This time, let us reformulate 82 x 9 as 82 x 90 / 10. 82 x 90 can be solved using the figure below:

82 - 18

90 - 10

---------

72 | 180

Carrying the 1 from the left hand side to the right hand side, we get 7380. Dividing this answer by 10 gives us 738, as before. To make it easier to solve mentally, the following steps can be formalized:

Let the difference in digits between the multiplicand and multiplier be d

Subtract the power of 10 with the d zeroes from the multiplicand and make this the temporary left-hand side of the answer

For the temporary right-hand side of the answer, use the 10's complement of the multiplicand

To derive the final left and right-hand sides of the answer, retain as many digits of the right-hand side as there are digits in the multiplier and carry over the rest of the digits to the left-hand side

To illustrate this procedure, let us use it on the same examples as before:

7437 x 99 = 736263 (7337 | 2563 is the temporary answer, leading to a final answer of 736263 after carrying over 25 to the left hand side)

3948 x 999 = 3944052 (3938 | 6052 is the temporary answer, leading to a final answer of 3944052 after carrying over 6 to the left hand side)

73367 x 999 = 73293633 (73267 | 26633 is the temporary answer, leading to a final answer of 73293633 after carrying over 26 to the left hand side)

Hopefully, this lesson has provided an insight into the kinds of simplifications that result when special properties of certain numbers are taken advantage of while using the methods of the past few lessons. The next lesson will focus on some more special applications of the past few lessons. In the meantime, happy computing, and good luck!

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