## Saturday, July 25, 2009

### Vedic Mathematics Lesson 9: Multiplication Special Case 2

As mentioned in the previous lesson, it is possible to make some multiplication problems much easier than even the procedures make them by taking advantage of some special properties of the numbers involved in the multiplication. In this lesson, we will talk about another special case that occurs somewhat frequently. Consider products such as 23 x 27. Notice that the last digits add up to 10 (7 + 3 = 10) and the digits other than the last digit are the same in both numbers.

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

Let us try to work out 23 x 27 using techniques from our previous lessons. We will use 10 as our base and 20 as our working base. We get the figure below:

23 + 3
27 + 7
--------
30 | 21

We now have to multiply the left hand side by our multiplier, 2 (20 / 10 = 2). This gives us a final answer of 621. The important thing to notice is that when the final digits add up to 10, the left hand side becomes 10 more than our working base (30 is 10 more than our working base of 20). Multiplying the result by the multiplier leads to the same result as taking the product of the first digit by a number one more than it and using it as the left hand side of the answer.

In fact, that is precisely this short cut that makes working out problems like the above very easy. The procedure is actually another application of a sutra we are familiar with from another lesson (this is another feature of Vedic Mathematics: many sutras are used in several contexts and the specific application of the sutra depends on the context): Ekadhikena Purvena, which means literally By One More Than The Previous One. In this case, the application of the sutra is as below:

Multiply the last digits of the numbers and use as the right hand side of the answer after padding with zeroes to make it 2 digits long
Multiply the left-over part of the number (after removing the last digits) by one more than itself and use this as the left hand side of the answer.

Let us practice the application of this procedure using the examples below:

31 x 39 = 1209 (09 is the product of 1 and 9, and 12 is 3 x (3 + 1))
55 x 55 = 3025 (25 is the product of 5 and 5, and 30 is 5 x (5 + 1))
103 x 107 = 11021 (21 is the product of 7 and 3, and 110 is 10 x (10 + 1))

The second case above, which is actually the square of a number ending in 5, is much better known than the other applications of this special case. You can find the shortcut explained as below:

Use 25 as the last two digits of the answer
Multiply the part of the number to be squared that is left over after taking away the 5 by the next number and use as the left hand side of the answer

The algebraic explanation of why it works as it does may or may not be provided along with the shortcut (in most cases I have seen, the result is presented more as a mathematical curiosity rather than as part of a coherent discourse on mental arithmetic and its roots), leading one to believe that Vedic Mathematics is nothing but some arithmetic tricks discovered by accident by idle people who had too much time on their hands! The truth is quite the contrary, and it is easy to see that these results are not coincidental or discoveries of chance, but obtained by the careful application of the fundamentals of algebra to the solution of arithmetic problems.

Let us extend this method to the case where the last digits of the numbers do not add up to 10, but to a higher power of 10, such as 100. Let us take the case of 103 x 197. Let us use 100 as the base in this case, 100 as the working base and 1 as our multiplier. This leads to the figure below:

103 + 3
197 + 97
----------
200 | 291

Since our multiplier is 1, the final answer become 20291. Notice that the left hand side of the answer is the product of 100 and one more than the multiplier. This property can be taken advantage of by formalizing this to the procedure below:

Multiply the final parts of the number that add up to a power of 10, and make the result twice as long as the number of zeroes in the power of 10, by padding to the left with zeroes as necessary. Make this the right hand side of the answer
Multiply the initial part of the number left over by one more than itself and make this the left hand side of the answer

Using this procedure, let us take a crack at the following examples:

395 x 305 = 120475 (0475 is 5 x 95 expanded to twice of 2 digits, which is 4 digits, and 12 = 3 x (3+1))
450 x 450 = 202500 (2500 is 50 x 50 (it does not have to padded with zeroes because the result is already 4 digits long) and 20 = 4 x (4 + 1))
5001 x 5999 = 30000999 (000999 is 999 x 1 expanded to twice of 3 digits, which is 6 digits, and 30 = 5 x (5 + 1))

Hopefully, the applications of this special case are clear from the explanations and examples above. Practice, as always, is the key to identifying the numerical patterns that this special case is most applicable to. So, continuous practice will make one perfect at identifying when this special case is applicable and at applying the special case quickly and correctly. Happy computing and good luck!

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