Wild About Math!

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, 92×96, we do two things. First, we calculate 8×4=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: 87×93. We write this:

87 — 13
93 — 7

13×7=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 11×12= 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 45×47, 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 5×3=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, 45×47. 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, 5×3=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 5×3=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 249×244. 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)

1×6=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

If you enjoyed this post, make sure you subscribe to my RSS feed!