## Vedic Multiplication

On Friday, timehop told me it was a year since I wrote this post on multiplication methods. I’d forgotten about the post, and tweeted a link to it. A number of people commented and tweeted about it and a nice discussion ensued. Part of the discussion moved to Vedic Multiplication.

I know a few Vedic Multiplication methods, and believe there to be many more. Most of the ones I know link to the most common algorithms but there is this curious one used for numbers close to one hundred.

Start with a multiplication problem:

First, take bother numbers away from 100:

Multiply those numbers and make them your last two digits:

Take one of the differences from 100 away from the other number (it should be the same):

That becomes your first two digits:

It’s an interesting little trick. I don’t see it as something that there is any reason to teach, and I don’t think it promotes understanding at all, but it’s interesting nonetheless. I think it may have a use in lessons, as an interesting introduction to algebraic proof.

**How would you prove it?**

First, consider the product ab, and apply the same steps:

The last two digits are (100-a)(100-b). The first two are a-(100-b) which equals a+b-100. Or b-(100-a) which also equals a+b-100. To make these the first two digits of a four digit number we need to multiply the expression by 100.

This gives:

**(100-a)(100-b)+100(a+b-100)**

Which expands to:

**10000-100a-100b+ab+100a+100b-10000**

Which cancels to:

**ab** As required.

A nice, accessible, algebraic proof that proves this works. It works for all numbers, not just those close to 100, but if your product (100-a)(100-b) > 99 (ie more than 2 digits) you need to carry the digits.