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.
<p>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</p>
<p>Here the unit digits are 1 and 2,,what if the unit digits when multiplied give a 2 digit result ???</p>
# 2 has a typo. 6 instead of 5.
Seems like a far easier method is the &quot;SAT&quot; approach...that they unfortunately teach only in prep for the SAT's rather than earlier, when it might do some good.<br/><br/>67x23 =<br/>67*2 = 134<br/>*10=1340<br/>+67*3=201<br/>=1541<br/><br/>or 67*25=<br/>67*100=6700<br/>/2=3350<br/>/2=1675<br/>
The method you refer to I have frequently heard referred to as &quot;cluster problems&quot;, 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.<br/>
&quot;Vedic&quot; maths is very very similar to algorithms developed by a prisoner of war called Trachtenberg <a rel="nofollow" href="http://en.wikipedia.org/wiki/Jakow_Trachtenberg">http://en.wikipedia.org/wiki/Jakow_Trachtenberg</a><br/><br/>Steve<br/>
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.
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!!
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)
Hello!<br/>I tried what you said and i got it! Thank you very much again <sup></sup>. I've already showed this multiplication method to my friends and they all thought it amazing xD. Very good Instructable! Good luck to you!<br/>
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.
Hey, something I just figured out: Vedic Multiplication with scientific notation:<br/><br/>For example, what happens if you want to calculate e=mc<sup>2</sup> 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:<br/><pre> 392 3.92x10^2 123x 1.23x10^2 35916 123 48216 4.8216x10^4</pre>So basically, you just add the powers of ten together (same as multiplying in normal algebra: x<sup>3</sup> * x<sup>2</sup>=x<sup>5</sup> ) - or you just count the number of places between the first digit and the decimal place (4 in this case)<br/><br/>So for the speed of light (approximately), we have:<br/>(2.998x10<sup>8</sup>)<sup>2</sup><br/><pre> 2.998 x 10^8 2.998 x 10^8 x 4674544 1121 &lt; second row carried tens 319246 8.988004 x 10^16</pre><br/>Feel free to use this in the instructable :) speed of light squared probably isn't the best example, but it works <br/>
haha, thanks man, I really appreciate it!
great tut, very useful.<br/><br/>It would probably be a good idea to have a whole step on multiplying numbers with different lengths, ie:<br/><pre> 34213x 23</pre>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). <br/><br/>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<br/><br/>= (ax + b) (cx<sup>4</sup> + dx<sup>3</sup> + ex<sup>2</sup> + fx + g)<br/>= acx<sup>5</sup> + bcx<sup>4</sup> + adx<sup>4</sup> + bdx<sup>3</sup> + aex<sup>3</sup> + bex<sup>2</sup> + afx<sup>2</sup> + bfx + agx + bg<br/>=(ac)x<sup>5</sup> + (bc + ad)x<sup>4</sup> + (bd + ae)x<sup>3</sup> + (be + af)x<sup>2</sup> + (bf + ag)x + (bg)<br/><br/>dunno how useful that is :)<br/>
Nice instructable! Can you please give me the name of the book you got for christmas? Thanks !
<a rel="nofollow" href="http://www.vedicmaths.org/Bookstores/BookstoreWorld/05%20Vedic%20Maths/VM%20details.asp">http://www.vedicmaths.org/Bookstores/BookstoreWorld/05%20Vedic%20Maths/VM%20details.asp</a><br/><br/>hope that helps :-)<br/>

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