So, without further ado, I present to you Vedic Multiplication

**Signing Up**

## Step 1: Multiplication...The Standard Algorithm

1) 5 times 2 is 10 (put down a 0, carry the 1)

2) 5*5 is 25, add the carry (put down a 6, carry a 2)

3) On a second line, add a 0 to the 1's place (4 has place value of 10)

4) 4 times 2 is 8 (put down an 8)

5) 4 times 5 is 20 (put down a 0, carry a 2)

6) Add together the 2 numbers (260 + 2080) to get the product (2340).

This method is general and efficient, to be sure. However, it is not hard to see that this method can involve considerable tedium and rigor when you start doing more involved multiplications (3 digits by 3 digits or more).

Read on to see how we can more efficiently multiply, the Vedic way!

67x23 =

67*2 = 134

*10=1340

+67*3=201

=1541

or 67*25=

67*100=6700

/2=3350

/2=1675

Steve

I tried what you said and i got it! Thank you very much again . I've already showed this multiplication method to my friends and they all thought it amazing xD. Very good Instructable! Good luck to you!

For example, what happens if you want to calculate e=mc

^{2}to four significant places? You need to calculate the square of the speed of light as part of that. But the speed of light is 299,792,458 m/s. 9 digit vedic multiplication would definitely be be intersting, but probably not fun. But if you just want an approximation, you could simply use scientific notation. First a simpler example:So basically, you just add the powers of ten together (same as multiplying in normal algebra: x

^{3}* x^{2}=x^{5}) - or you just count the number of places between the first digit and the decimal place (4 in this case)So for the speed of light (approximately), we have:

(2.998x10

^{8})^{2}Feel free to use this in the instructable :) speed of light squared probably isn't the best example, but it works

It would probably be a good idea to have a whole step on multiplying numbers with different lengths, ie:

I worked it out myself, but then I noticed you'd done it a different way (by adding 0s to the start of the lower number).

For anyone else wondering, it can also be done without the zeros, basically by exactly the same method as two 5 digit numbers, but by shifting the lower numbers until you hit the end. The equation is

= (ax + b) (cx

^{4}+ dx^{3}+ ex^{2}+ fx + g)= acx

^{5}+ bcx^{4}+ adx^{4}+ bdx^{3}+ aex^{3}+ bex^{2}+ afx^{2}+ bfx + agx + bg=(ac)x

^{5}+ (bc + ad)x^{4}+ (bd + ae)x^{3}+ (be + af)x^{2}+ (bf + ag)x + (bg)dunno how useful that is :)

hope that helps :-)