Vedic Mathematics Simplified-Part 1

In my previous post I had promised to start writing lessons on vedic mathematics and here is the first lesson.

Vedic mathematics contains sutras (formulas) which are simple and straightforward, even though they are based on algebra, these sutras work like magic and unlike modern algebraic formula, they are easy to remember.

More importantly, vedic mathematics means almost zero paper work, and you can come out with answers to a mathematical problem, as if you are spelling out the alphabets of an English word!

Unlike how conventional books on vedic mathematics go, I am starting out with one of the multiplication methods in the first lesson today. Remember vedic mathematics is flexible, and for each problem there are multiple ways to solve it and you can also come out with your own method to solve it! So this is just one of the methods to do multiplication.

Multiplication Formula

The formula is not something complicated to remember like (a+b) 3 =a 3 +b 3 +3ab(a+b)

Its as simple as Urdhva Tiryagbhyam in Sanskrit, which in English simply means vertically and crosswise

Let us see how.

Conventional Method

In our schools the multiplication method taught is very complicated and paper work increases as the number of digits increase in the numbers being multiplied. See below for some examples:

Example 1 of Conventional Method

75

X 32
_______
150
225
_______
2400
_______

Example 2 of Conventional Method

745
x 925
________
3725
1490
6705
________
689125
________

As you can see above, as the number of digits increases, the method gets more and more complicated, involves more paper work, is more elaborate, takes more time to solve, is more prone to human errors while adding, placing digits one below other etc.

Vedic Method

Now let us see how simple it is to do the same arithmetic using the vedic mathematics sutra Urdhva Tiryagbhyam mentioned above i.e vertically and crosswise

The same problems when solved using vedic method become as simple as

75

X 32
_______
2400
_______

and

745
x 925
________
689125
________

The formula is pretty simple, and before explaining it in words, let me show it pictorially to you, not just because a picture is equivalent to some hundred words or so, but because it also takes a lot of effort to type it out :)

Two digit multiplication

For a two digit multiplication the formula is as follows with simple three mental steps!

Suppose BC X EF is the problem, then

vedic mathematics formula

Let me explain it for 7532

Step 1 : 52=10, write down 0 and carry 1

Step 2 : 72 + 53 = 14+15=29, add to it previous carry over value 1, so we have 30, now write down 0 and carry 3

Step 3 : 73=21, add previous carry over value of 3 to get 24, write it down.

So we have 2400 as the answer.

Three digit multiplication

The method for 3 digits is an extension of the 2 digit method as follows

vedic mathematics urdhva thiryagbyam formula

Let me explain this method for 745 x 925

Step 1 : 55=25, write down 5 and carry over 2

Step 2 : 45 + 52 =20+10=30, add previous carry over 2, so we have 32, write down 2 and carry over 3

Step 3 : 75 + 42 + 59 = 35+8+45 = 88, add previous carry over 3 to get 91, now write down 1 and carry over 9

Step 4: 72 + 49 = 14+36 = 50, add previous carry over 9 to get 59, write down 9 and carry over 5

Step 5: 79= 63, add previous carry over 5 to get 68, write down 68

So we have 689125 as the answer

Conclusion

No matter how many digits are involved, you can do all the calculations directly in your mind, by doing carry overs and one digit multiplication and keep simply writing down the answer!!

If children are taught this method right from their childhood to do multiplication, they will get used to doing super fast multiplication without having to use pen and paper!

I suggest that you try out the above method to do some really large multiplications, and you will be amazed at your own speed!

Just look at the problem and start writing the answers! Which is what vedic mathematics is all about. It is mathematics simplified!

Next Article >
The Eternal Cosmos
  • Srikanth Rangdal

    Hi gurudev,

    Vedic mathematics always mesmerised me with its short techniques. I have 2 books in the subject which i bought years back with many more techniques.

    My question here is about its origins. I have heard about the proficient calculation skills of Gurukul scholars in speeches of Rajiv dixitji. But i remeber having read somewhere that the Vedic maths skills as we know them now were only put down recently & do not have origins in the Vedic era.

    You seem to be researching a lot on the matter. So what do you think?

  • ajay

    iam finding it little difficult because this is the first time iam seeing this method i will soon be practised to this method. and gurudev once while browsing in the net  i came across this method for 93*97, 93- 7(taking compliment to 100,93+7 = and 97+3 = 100)
                                               97- 3 = 97-7 or 93-3 = 90,7*3= 21,9021 but this is valid only for problems above 50*50 can you explain this and you mentioned a very easy method but is there any proof for these methods.

  • ajay

    you told that there are different methods to solve an answer in vedic maths are there any other method to solve 75*32

  • Akshay

    I got 789299 * 56 using the same method.
    just do the same multiplication from right to left on it just taking 2 numbers at a time. So we can do any kind of multiplication.

    • http://www.hitxp.com Gurudev

      Great you found the answer yourself. Strongly recommend that you work out other formulae in Vedic Maths too, it will be very easy for you as you seem to quickly grasp the logic behind its formulae :)

    • Anonymous

      Great you found the answer yourself. Strongly recommend that you work out other formulae in Vedic Maths too, it will be very easy for you as you seem to quickly grasp the logic behind its formulae :)

      • Akshay

        I would love to. Give me some links.
        I am not that good in maths. I even suffer doing simple programming in C, Cpp. I found this quite easy.

      • Anil

        Gurudev, can u show us the steps for 789299 * 56?

        • Anonymous

          Try out the below generalization for 6 digits by making the problem as 789299 * 000056

          For two digits ab x cd it is
          ac (ad+bc) bd

          For three digits abc x def it is
          ad (ae+bd) (af+be+cd) (bf+ec) cf

          Thus for four digits abcd x efgh it would be
          ae (af+be) (ag+bf+ce) (ah+bg+cf+de) (bh+cg+df) (ch+dg) dh

          and so on

  • Akshay

    That really makes it very simple. What if the multiplication is 789299 * 56 ?

  • Suchin

    Guru, I tried this for four digits:

    4 2 7 6
    7 5 4 9

    Please tell me if the following steps is correct.

    1) 6 x 9
    2) 7 x 9 + 6 x 4
    3) 2 x 9 + 7 x 4 + 6 x 5
    4) 2 x 4 + 7 x 5
    5) 4 x 4 + 7 x 7 + 2 x 5
    6) 4 x 5 + 2 x 7
    7) 4 x 7

    • http://www.hitxp.com Gurudev

      Just a small correction in the fourth step Suchin, It would be
      4×9 + 2×4 + 7×5 + 6×7

  • Larry Smith

    The procedure you explained worked fine for 3 digit multiplication but I could not figure out the correct procedure for 4, 5, or 6 etc digit multiplication. Please reply with the procedures for 4,5 etc digit multiplication in the form ABCDXEFGH etc as you did in your 3 digit explanation eg. Step 1-Step2-Step3 etc. Thank You.

    • http://www.hitxp.com Gurudev

      You can generalize the formula as follows

      For two digits ab x cd it is
      ac (ad+bc) bd

      For three digits abc x def it is
      ad (ae+bd) (af+be+cd) (bf+ec) cf

      Thus for four digits abcd x efgh it would be
      ae (af+be) (ag+bf+ce) (ah+bg+cf+de) (bh+cg+df) (ch+dg) dh

      For five digits abcde x fghij it would be
      af (ag+bf) (ah+bg+cf) (ai+bh+cg+df) (aj+bi+ch+dg+ef) (bj+ci+dh+eg) (cj+di+eh) (dj+ei) ej

      and so on… Hope this helps.

      • Snowy1

        Thank you! Your representation of the digits along with the grouping of the digits is an excellent visual for understanding the process.

        • Anonymous

          You are welcome Snowy1

  • http://none stephen

    I beleive the above is just one formula I am sure there is a simpler method , as a child I could do theses things without thought and the answere was there, treated by the educated minds around me was forced to use their system that failed at every point and does not represent the explanation to solve universal questions,My answere to the steel pillar is the excess of negative energy below its base compensates for the positive above.even though cast iron has a grater life span than steel some how this has been understood in ancient times and is an edific to the collective science since lost

  • lokesh

    thax for the post. want to know how to find square root and cube root of two digit number.

  • melric

    this was awsomeeeeeeeeeeeeeeeeeeee superbbbbbbbbbbbbb

    rock on!