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Vedic multiplication using bases: an introduction

There are numerous introductions to Vedic mathematics on the web. I won't be doing a general introduction to Vedic Math now. In this article I want to explore one particular Vedic mathematics technique, using something called bases, to optimize certain multiplication problems.

This technique is extremely powerful and it takes getting used to. It's partially a cookie cutter technique but there's also some thinking involved in selecting proper bases for performing multiplication. Don't get frustrated if you can't understand this technique in one reading. It took me a fair amount of focused attention and practice to understand and appreciate the power of this approach. If there's enough interest I'll produce some videos explaining this Vedic technique.

Let's begin.

Let's say that you want to multiply together two numbers both of which are a little bit less than 100. We'll use 92 and 96 as the numbers to multiply. 100 is our "base" for this particular multiplication. We write the following:

92 -- 8
96 -- 4

The 8 in the first row is 100-92. The 4 in the second row is 100-96.

Now, to get the product, 92x96, we do two things. First, we calculate 8x4=32. 32 becomes the right part of the product. Second, we calculate 100-(8+4)=100-12=88. 88 is the left part of the product. We put together 88 and 32 and get 8832, which is the product of 92 and 96. Pretty easy, huh?

Let's do another example: 87x93. We write this:

87 -- 13
93 -- 7

13x7=91. 100-(13+7)=100-20=80. Our answer is 8091.

Let's do a third example, this one involving a carry. We'll multiply 89 and 88. We write:

89 -- 11
88 -- 12

We multiply 11x12= 132, which is larger than 100. We calculate 100-(11+12)=100-23=77. We use 32 as the right part of the product, carrying the 1 and adding it to 77 to get 78. 7832 is our product.

Why does this technique work? Let's look at the algebra. We're multiplying 100-a by 100-b.

(100-a)x(100-b)=10000-100b-100a+ab = 10000-100(b+a)+ab. ab represents the product of the those two numbers we wrote on the right of each row. 10000-100(b+a) = 100(100-(b+a)). The 100 outside of both parentheses represents the left part of the product since multiplying by 100 moves a number two places to the left. 100-(b+a) is the arithmetic we did to get that left part of the product.

I won't discuss cases where one or both numbers are a little over 100 but if you do the algebra you should be able to derive the proper steps for those kinds of arithmetic.

Let's look at a case where the base is not 100. Let's multiply 45x47, both of which are near 50. We introduce two terms, theoretical base and working base. The theoretical base is always a power of 10. In this case the theoretical base is 100. This will become clear later. 50 is our working base since both numbers we're multiplying are near 50. Note that 50=100/2. You'll want your working base to be a factor or multiple of your theoretical base. We write, in similar fashion to earlier examples:

45 -- 5 (5 is 50-45)
47 -- 3 (3 is 50-47)

We multiply 5x3=15. That's the right part of the product just as in all the prior examples. Next, we calculate 50-(5+3)=42. We take 42 and divide it by 2 which is the ratio of our theoretical base to our working base = 100/50 =2. So, 42/2 = 21, which becomes the left part of our answer which is 2115.

Now, here's where things get interesting. For any given multiplication you can pick the working base and theoretical base that are most convenient for the problem at hand, remembering that the theoretical base is always a power of 10 and the working base is always a factor or multiple of the base. Let's look at the same product, 45x47. We'll still use 50 as our working base but let's use 10 rather than 100 as our theoretical base. As before, we write the following:

45 -- 5 (5 is 50-45)
47 -- 3 (3 is 50-47)

As before, 5x3=15 is the right part of the answer except for an adjustment which we'll make soon to it. As before, we calculate 50-(5+3)=42. This time, though, we note that the ratio of our theoretical base, 10, to our working base, 50, is 1/5. So, we divide 42 by 1/5, or multiply it by 5 and get 210. Now, since our theoretical base is 10 and not 100 we only get to have a 1 digit number as the right part of our answer. So, the 1 from 5x3=15 is actually a carry. So, the right digit of the product is still a 5 and we add the carry of 1 to 210 to get 211. So, 2115 is our product.

What's probably not clear at this point is, which theoretical base do you pick? The answer is, it takes practice and experience to know whether to pick 10, 100, or 1000, etc. for the theoretical base. Sometimes the arithmetic is easier if you pick the lower power of 10, and sometimes it's easier if you pick the higher power of 10. I encourage you to try different theoretical bases for each practice problem until you develop a sense of which to use.

Let's do one final example, to illustrate a couple of points you'll need to address in this technique of Vedic multiplication. Lets look at 249x244. We pick 1000 as our theoretical base and 250 as our working base. We write:

249 -- 1 (1=250-249)
244 -- 6 (6=250-24)

1x6=6. This will be used as the right part of the product with an adjustment we'll encounter in the next paragraph. Note that the right part of the product, so far, is not 6 but 006, since 1000 is our theoretical base. The number of digits in the right part of the product always corresponds to the power of 10 in the theoretical base.

We calculate 250-(1+6)=250-7=243. We divide 243 by 4 which is the ratio of our theoretical base, 1000, and our working base, 250. 243/4=60 3/4 or 60.75. What we need to do is to take the .75 from 60.75 and multiply it by our theoretical base, 1000, and we get 750, which we add to the 6 which we had on the right to get 756 as the right part of the product. 60 is the left part of the product. The whole product is 60756.

I realize that this last example, if not some of the ones before it, might be very confusing upon only one read. Read this article several times, make up some examples and work them out and if this approach still doesn't make sense leave me comments and I will clarify the approach as best I can, in videos or in text

Comments (10) Trackbacks (2)
  1. Easier to understand if presented visually-imagine the multiplication represented as an area on graph paper. Your first example, 92*96 is close to 100*100. Mark out an area of 100*100 and within this area mark out 92*96. The easiest way to mark the area 92*96 is to draw lines at 100-92 and 100-96 i.e 8 from top and 4 from side. The number of 100s in the answer can now be clearly seen from the marked paper-either 92-4 or 96-8. This visual representation also reveals that by deducting 8+4 rows from 100 rows of 100 that we have deducted the overlap twice. We have to compensate for this by adding the overlap back to the first part of our answer of 88–. The overlap is 8*4 so our answer is 8832. To do mentally the calculation should be carried out left to right as follows: 92*96. Start by subtracting either 92 or 96 from 100 and deduct the result from the other number e.g 100-96=4 92-4=88– or 100-92=8 96-8=88– this gives the number of 100s in the answer. Now we need to add 8*4 to get 8832. I hope that my explanation helps. As a guide, i would use base 50 for numbers between 30 to 70, base 100 for number above 70 and either base 10 or 20 for numbers below 30. Remember that the difference from the base tells you how many of that base number there is in the answer.

  2. will be able to tell me how do i multiply 38 x 23 using vedic maths. what are the theoretical base and working base which i need to use to solve this???

  3. Reply to Smitha, Re your question 38×23. The base method is best applied to numbers close to a suitable base. Forget about theoretical and working bases which confuse the issue. In the sum presented i would recommend the following approach which can be applied to any multiplication. Working from left to right multiply 3×2 (ie 30×20)to obtain the number of hundreds in the answer. To 600 we add the number of tens in the answer by cross multiplying the units with the tens. In the example we multiply as follows 3×3 and 8×2 (ie 3×30 and 8×20). This gives 25 tens which we add to 600 giving a total of 850. Lastly we determine the number of units in the answer by multiplying 8×3 which when added to 850 gives the answer to your sum. With a little practice you will be able to solve any 2 digit by 2 digit multiplication mentally with ease. Obviously use the base method on suitable sums. The base method can be extended as follows-simple example which demonstrates principal, 38×23, double 23 so that sum is now 38×46 which are close to base 50. solve as follows 50-46=4 subtract 4 from 38 to obtain number of 50s in answer. obviously if you half this you get 17 which is the number of hundreds in answer. To 1700 add 4×12 (50-46=4 50-38=12 difference from base) to get 1748. Finally you will need to half this to arrive at final answer. Halving is an easy process if carried out left to right. Take the 17– and half to get 8– (disregarding the .5 at this stage) then half 48 to get 24 and add 50 ie the .5 from above. this gives a final result of 874. I hope that this has answered you enquiry

  4. Umm… The method doesn’t work for 7*6… A glitch, maybe.

  5. Sarah,

    Actually, I believe it does work for those numbers. This is the first I’ve heard of this technique and I’m trying to wrap my mind around it, but for 7 x 6, here’s how I think you would do it.

    Using a base of 10:

    7 – 3 (10-7)
    6 – 4 (10-6)

    So the right hand side will be 3 x 4 = 12, but remember because we are using 10 as our base, we only have “room” for one digit: 2. 2 will be the right hand side of our answer, the 1 will be a carry.

    Then for the left hand side, we have 10 – (3+4) which equals 3. But let’s not forget the carry from before! So, we add that 1 to the 3 which gives us the left side of our answer: 4.

    Which gives us the expected answer of 7 x 6 = 42.

    Now to really bake your noodle, try 3 x 2. Using a base of 10:

    3 – 7 (10 -3)
    2 – 8 (10 – 2)

    So for the right hand side, we multiply 7 x 8 — which give us 56! Yikes! But wait, because we used 10 for our base, we only have room for one number for our right hand side, so only that 6 will be on the right hand side, making the 5 a carry.

    And for the left hand side we get 10 – (7+8) thus 10 -15 which equals -5. When we add the carry -5 + 5 = 0. So we wind up with 0 for the left hand side.

    Which gives us 06 for the answer or just 6, which is just what we’d expect for 2 x 3.

    Pretty cool. 🙂

  6. How do you use the vedic multiplication method for numbers that quite distant in terms of the closest base number. e.g. 35 x 85, or 231 x 33?

    Would love to see your thoughts. Thanks.

  7. very effective even for bigger numbers i.e. working base 10, 20, 50, 100, 200, 250, 500…. need more practice store values for bases beyond 100.

    Pretty cool though. Thanks mate!

  8. Thank you its very intresting is there any other method to calculate 37×83

  9. How can I solve this problem using base multiplication method? 13×9

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