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Rapid Multiplication ... Diagonally?

Date: 02/24/2010 at 07:50:32
From: George
Subject: Alternative multiplication method (fast!)

Some years ago, I saw a system of multiplication demonstrated on 
television.

As I recall, the system involved multiplying the top left digit by 
the bottom right, and continuing to multiply diagonally in this way, 
moving in opposite directions along each term.

How does this high-speed calculation technique work?



Date: 02/24/2010 at 21:53:10
From: Doctor Vogler
Subject: Re: Alternative multiplication method (fast!)

Hi George,

Thanks for writing to Dr. Math.

Here's what you probably saw:

To find the product of two multi-digit numbers, you have to keep a 
running sum in your head, which we might call your accumulator (to 
borrow a term often used in computers).  

Start out your accumulator at zero.  Compute the digits of the 
product as follows, starting with the rightmost digit and moving to 
the left:  To compute the n'th digit from the right, you have to add 
n products to your accumulator, starting with the n'th digit from the 
right of the top factor and the rightmost digit of the bottom factor, 
and ending with the rightmost digit of the top factor and the n'th 
digit from the right of the bottom factor.  After adding all n of 
those products to your accumulator, you write down the last digit of 
your accumulator and then shift the accumulator to the right by one 
digit.

For example, to compute 486 * 197, you start with the accumulator at 
0.  Then you add 6*7 = 42, giving 42.  You write down the digit 2 in 
the "ones" place (that's the rightmost digit of your product) and 
shift your accumulator, which is now waiting at 4 in the "tens" place:

    _ _ _   (4) 2

Then you add to that 8*7 + 6*9 = 56 + 54 = 110, giving 114.  You 
write down the 4 and shift the accumulator, which is now 11:

   _ _ (11)  4  2

Then you add 4*7 + 8*9 + 6*1, giving 117.  You write down the 7, and 
shift again, giving 11:

   _ (11) 7  4  2


Then you add 4*9 + 8*1, giving 55.  You write down the 5, and shift, 
giving 5:

   (5) 5  7  4  2

Then you add 4*1, giving 9, which you write down (that's the leftmost 
digit of your product):

    9  5  7  4  2

So your answer is all of the digits you've written down, namely 95742.

If you have any questions about this or need more help, please write 
back and show me what you have been able to do, and I will try to 
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Elementary Multiplication

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