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

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## Step 1: 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!

## Step 2: Enter the Sutra

The Vedic method (the general method, at least) is based on the Urdhva-Tiryagbhyam sutra. A very terse sutra, it simply translates in English to say "vertically and crosswise". The sutra is rather vague, so the technique, as well as an algebraic analysis of the technique, is presented in the following steps.

## Step 3: An Algebraic Perspective

As most readers are hopefully aware, all numbers n the base 10 number system (and number systems of any other base, for that matter) consist of a number of digits. Each digit represents a multiple times a power of 10 (or whatever the number system's base is). So, for example, given a number like 52, we could rewrite it as 5*10+2.

Algebraically speaking, we can express any 3-digit number as:

ax+b (where a, and b are integers).

So, suppose we wanted to multiply 2 2-digit numbers. We can express them in polynomial form. Then, by foiling:

(ax+b)(cx+d) = acx^{2}+(ad+bc)x+bd

Read on to see how we take the algebraic multiplication to higher numbers of digits.

## Step 4: Algebraic Multiplication for Higher Numbers of Digits

So, for brief review, 2-digit by 2-digit algebraic multiplication goes as follows (x in all of the following examples is the base of the number system being used, which is usually 10):

(ax+b)(cx+d) = acx^{2}+(ad+bc)x+bd

Expanding to 3-digit by 3-digit algebraic multiplication:

(ax^{2}+bx+c)(dx^{2}+ex+f) = adx^{4}+(ae+bd)x^{3}+(af+be+cd)x^{2}+(bf+ec)x+cf

Now, 4-digit by 4-digit algebraic multiplication:

(ax^{3}+bx^{2}+cx+d)(ex^{3}+fx^{2}+gx+h) = aex^{6}+(af+be)x^{5}+(ag+bf+ce)x^{4}+(ah+bg+cf+de)x^{3}+(bh+cg+df)x^{2}+(ch+dg)x+dh

Finally, 5-digit by 5-digit multiplication:

(ax^{4}+bx^{3}+cx^{2}+dx+e)(fx^{4}+gx^{3}+hx^{2}+ix+j) = afx^{8}+(ag+bf)x^{7}+(ah+bg+cf)x^{6}+(ai+bh+cg+df)x^{5}+(aj+bi+ch+dg+ef)x^{4}+(bj+ci+dh+eg)x^{3}+(cj+di+eh)x^{2}+(dj+ei)x+ej

By looking carefully at the products of the above multiplications, a pattern becomes noticeable. The pattern looks something like this:

1) For the highest digit of the the product, simply write the product of the highest digits of the 2 numbers.

2) The process for the middle digits is a little tricky to explain using words. For each successive digit, you cross-multiply the highest digit of the top number by the next highest digit of the bottom number and the highest digit of the bottom number by the next highest digit of the top number and add these numbers together. You continue on for all subsequent digits by decrementing the digit on the bottom number, cross-multiplying, incrementing the bottom digit while decrementing the top digit, and summing the products. If this confuses you, fear not, it will be explained via examples on the following pages.

3) For the lowest digit of the product, multiply the lowest digits of the numbers being multiplied.

4) Throughout this process, only one digit can be kept per place, thus if your sum is higher than 9, you will have to write the ones digit only and implement a carry to the next higher place in the number (or carry to several higher places if your sum exceeds 99).

Coming up, examples!

## Step 5: Vedic Multiplication Examples: 2-Digit by 2-Digit

In this step (and the following steps) we will attempt to clarify the vedic procedure through the use of several examples. Under each example, there will be a number of steps (indicated by numbers). Each step has a separate picture, so feel free to use those to follow along with what's going on.

Example 1: 42*21

1. Multiply the 2 highest digits (4 and 2), resulting in an 8.

2. For the next higher digit, cross multiply 4*1 (4) and 2*2 (4), and add together, producing the middle digit of 8.

3. For the lowest digit, multiply the 2 lowest digits (1*2) together, resulting in a 2.

4. Put all of the digits together to produce your answer (882)

A quick check on the calculator indicates our answer is indeed correct.

One thing that can and should be noted here is that the order in which you go through for the vedic process does not actually matter. So, we can similarly start with the lowest digit and work our way up to the highest digit.

Example 2: 67*23 (Starting with the lowest digit and working our way left)

1. Lowest Digit: Multiply the 2 lowest digits together (7*3), to get 21. Only one digit can be kept per place, so write a 1 and carry a 2 over to the next position.

2. Now, move over by 1 position on top, (3 will be multiplied by 6 now) and cross multiply pairs, decrementing the position on top while incrementing the position on the bottom. That is to say, multiply 3*6 (18) and 2*7 (14) and add them together (32). Write a 2, carry a 3 to the next position.

3. At this point, we have reached the end of the number, and thus incrementing the position on top is not possible. Thus, we simply increment the position on the bottom (2 will be multiplied by 6) and no cross multiplication will occur (as incrementing the bottom digit is not possible, we have reached the end of the number). So, 2*6 will be 12, so write a 2 and carry a 1.

4. Now that the multiplication has been completed, we have 2 numbers. The one on top is all of ones digits from the multiplications, and the one on the bottom is all of the carries that occurred. So, simply add these 2 numbers (221 and 1320) together, resulting in the answer (1541).

In the following steps, we will perform examples to expand the idea to higher orders of digits.

## Step 6: Vedic Multiplication Example: 3-Digit by 3-Digit

Example 3: 341*562

At this point, some of the faint of heart may be reaching for their calculators. Be strong, you can handle this using the vedic technique with relative ease (and considerably less writing) than that which you probably learned in elementary school.

1) Ones Digit: 2*1 = 2

2) Tens Digit: 2*4+6*1 = 8+6 = 14 (write a 4, carry a 1)

3) Hundreds Digit: 2*3+4*6+5*1 = 6+24+5 = 35 (write a 5, carry a 3)

Notice at this point that we have reached the end of the top number, that is to say, 2 can not be multiplied by anything else, as there are no more digits. Thus, for the thousands digit we shall multiply 6 and 3 first, and then for the ten thousands digit, 5 and 3. (read on for the steps)

4) Thousands Digit: 6*3 + 5*4 = 18+20 = 38 (write an 8, carry a 3)

5) Ten Thousands Digit: 5*3 = 15 (we've reached the end of the multiplication, so feel free to simply write 15 rather than carry a 1)

6) Add in the carries to produce the final answer: 191642 (verified via calculator)

At this point it should be noted that the 2 numbers need not have the same number of digits. For example, if a 3-digit number were to be multiplied by a 2-digit number, just follow the formula for 3-digit by 3-digit multiplication and just pad out the missing digits with zeros.

## Step 7: More Examples and Practice

This step is dedicated to a few additional examples. The viewer is encouraged to work these out on his or her own and verify that they have done the technique correctly. Attached are images showing my work for each of the practice problems below.

All of the problems below have been verified by calculator. Also, the vedic method should be executable much more quickly than the standard multiplication algorithm (the author of this instructable performed all of the multiplications below in under a minute each, including the last one).

1) 32*63

2) 98*47

3) 392*186

4) 528*76

5) 1492*1337

6) 2603*1865

and finally, for those who are gluttons for punishment...

7) 125468*337962

You may not get the technique right away, but with a little practice you too can multiply quickly, efficiently, and like a pro. Happy hunting!

## 17 Discussions

Question 1 year ago on Step 7

Is this method is only capable doing two numbers multiplication...

When it comes to 3 numbers multiplication then... How we multiply 11×13×15 from this method...

Tell me man...??

3 years ago

what if while cross multiplying the number becomes greater than 100 ie a 3 digit number, how should i carry in that case? Example being - 856*567

4 years ago on Step 6

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

7 years ago on Step 1

# 2 has a typo. 6 instead of 5.

10 years ago on Step 5

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

*10=1340

+67*3=201

=1541

or 67*25=

67*100=6700

/2=3350

/2=1675

Reply 10 years ago on Step 5

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.

10 years ago on Introduction

"Vedic" maths is very very similar to algorithms developed by a prisoner of war called Trachtenberg http://en.wikipedia.org/wiki/Jakow_Trachtenberg

Steve

Reply 10 years ago on Introduction

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.

10 years ago on Step 7

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!!

Reply 10 years ago on Step 7

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)

Reply 10 years ago on Step 7

Hello!

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!Reply 10 years ago on Step 7

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.

10 years ago on Introduction

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

Reply 10 years ago on Introduction

haha, thanks man, I really appreciate it!

10 years ago on Introduction

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 :)

10 years ago on Introduction

Nice instructable! Can you please give me the name of the book you got for christmas? Thanks !

Reply 10 years ago on Introduction

http://www.vedicmaths.org/Bookstores/BookstoreWorld/05%20Vedic%20Maths/VM%20details.asp

hope that helps :-)