**Multiplying two numbers whose last digits add up to powers of 10**

This is my first post in the series of Non-Vedic maths, where I will be explaining speed calculation techniques. These techniques are NOT based on the ancient Indian Vedas but can be derived using elementary maths.

In this first instalment of the series I will be looking at how to multiply two numbers whose last digits add up to powers of 10 and whose leading digits are identical. To explain the method let us first look at two digit numbers. As an example, we multiply 63 with 67. For this method to work the first digit has to be the same for both numbers and the last digits have to add up to 10. Then we can write our to numbers as

10

*a*+

*b*and 10

*a*+

*c*

where

*b*+

*c*= 10 and

*a*is the leading digit. In our example we have

*a*= 6,

*b*= 3 and

*c*= 7. Multiplying the two numbers gives

In the third step we used that b+c=10. The product bc will always be less than 100 so the digits will not interfere with the first term, which is a multiple of 100. In our example we have

63 * 67 = 100 * 6 * 7 + 3 * 7 = 100 * 42 + 21 = 4221

The rule that can be extracted from this is:

**Take the first digit and multiply with one more than itself. Multiply the last two digits. The final answer is made up of the two results by joining the results together.**

We can represent this graphically in the following way.

The method clearly also works if a is not a single digit number, but it can have any number of digits. If we wanted to multiply 116 with 114 we will get

100 * 11 * 12 + 6 * 4 = 13200 + 24 = 13224.

The method can be generalised if the trailing two digits add up to 100 or the trailing three digits add up to 1000 and so on. The maths is similar to the previous case.

Again,

*b*

*c*will always be less than 10000 because each of them is less than 100. To see an example, lLet's multiply the numbers 436 and 464. Here the last two digits add up to 100 and the leading digit is the same in both numbers.

Of course one has to multiply two digit numbers, which can be a little more complicated. In another post I will present methods that can make this multiplication easier.

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