Picture of Vedic Multiplication
Some time ago (this past summer, to be exact) I created my first instructable. A fun little trick, really, about how to quickly extract the cube roots of large integers mentally. Included in the comments section of this instructable was a very useful comment from mahi16 which suggested I incorporate Vedic Mathematics methods into my technique. Having never heard of it before, I investigated it, and found myself intrigued by the clever strategies employed in the ancient Indian system. I got a book on it this year for Christmas and am researching the different techniques and methodologies presented in it.

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

Step 1: Multiplication...The Standard Algorithm

Picture of Multiplication...The Standard Algorithm
Most of us remember the standard algorithm for multiplication (well, those of us not being brought up under the hilarious joke that is modern math education in some schools, anyway). To demonstrate it, consider the multiplication problem 52x45. This is done in the following steps (which are demonstrated in photos as well):
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!
TusharC22 months ago
omkarp13 months ago
eunicorn6 months ago
gphm7 months ago

Here the unit digits are 1 and 2,,what if the unit digits when multiplied give a 2 digit result ???

kkim103 years ago
# 2 has a typo. 6 instead of 5.
igorzal6 years ago
Seems like a far easier method is the "SAT" approach...that they unfortunately teach only in prep for the SAT's rather than earlier, when it might do some good.

67x23 =
67*2 = 134

or 67*25=
purduecer (author)  igorzal6 years ago
The method you refer to I have frequently heard referred to as "cluster problems", and it can be useful for some cases. However, I am of the opinion that this method becomes less useful as you start applying it to higher numbers of digits (ex. 97623 * 85428). You quickly stretch the limits of your grey matter attempting to remember all of the little sub-problems you created to reach the final result, and you waste lots of time writing down all the little intermediate steps you had to take to get to the answer. For this reason, I recommend the vedic approach or at least the standard algorithm for multiplication, particularly if you intend to multiply something beyond 3-digits by 3-digits.
"Vedic" maths is very very similar to algorithms developed by a prisoner of war called Trachtenberg http://en.wikipedia.org/wiki/Jakow_Trachtenberg

purduecer (author)  steveastrouk6 years ago
I know of the Trachtenberg system, but have unfortunately not had time to read up on it. I'll go ahead and take your word on it for the time being.
phelippe6 years ago
Very nice instructable, it seems to be very useful. But i had problems when i multiply numbers with many "nines" or "eights" like 998 x 788. In that case i first do 8 x 8, put down 4 and carry 6, ok. But when I do (8 x 9) + (8 x 8), i get 136... how do i put this value in this algorithm?!?? I've tried to put down 6, carry 13, to put down 36 and carry 1, to put down 6, carry 3 then "carry again" 1 and nothing worked... How can i solve that kind of multiplications?!?? Thanx!!
purduecer (author)  phelippe6 years ago
in that case you should be able to put down a 6 and carry a 13 or alternatively put down a six, and carry a 3 one place to the left and then carry a 1 an additional place to the left. You cannot put down a 36 and carry a 1, however, as that is like putting down a 6 and carrying a 4, which is mathematically incorrect. Try it again and let me know what you get, and whether or not that is agreed upon by your calculator. (Depending on the calculator, there's also the possibility that the calculator is making the mistake, not you)
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!
purduecer (author)  phelippe6 years ago
Haha, thank you very much. I will try to post more mental math methods as I learn about them. If non-calculator based math is your thing, you might like some of my other instructables, so feel free to check them out if you haven't already.
naught1016 years ago
Hey, something I just figured out: Vedic Multiplication with scientific notation:

For example, what happens if you want to calculate e=mc2 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:
   392   3.92x10^2   123x  1.23x10^2  35916 123 48216   4.8216x10^4
So basically, you just add the powers of ten together (same as multiplying in normal algebra: x3 * x2=x5 ) - 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:
   2.998 x 10^8   2.998 x 10^8 x 4674544 1121     < second row carried tens 319246 8.988004 x 10^16

Feel free to use this in the instructable :) speed of light squared probably isn't the best example, but it works
purduecer (author)  naught1016 years ago
haha, thanks man, I really appreciate it!
naught1016 years ago
great tut, very useful.

It would probably be a good idea to have a whole step on multiplying numbers with different lengths, ie:
 34213x   23
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) (cx4 + dx3 + ex2 + fx + g)
= acx5 + bcx4 + adx4 + bdx3 + aex3 + bex2 + afx2 + bfx + agx + bg
=(ac)x5 + (bc + ad)x4 + (bd + ae)x3 + (be + af)x2 + (bf + ag)x + (bg)

dunno how useful that is :)
sfacs6 years ago
Nice instructable! Can you please give me the name of the book you got for christmas? Thanks !
purduecer (author)  sfacs6 years ago