Egyptian Division

Date: 06/23/98 at 13:28:29
From: Anne Fogg
Subject: Egyptian Division

My fourth graders are studying ancient Egypt. We have tried working 
with Egyptian multiplication but haven't found any reference to 
division. What format did the ancient Egyptians use for dividing? Can 
you help?

Date: 06/24/98 at 16:57:52
From: Doctor Mateo
Subject: Re: Egyptian Division

Hello Anne,

Egyptian division is basically Egyptian multiplication in reverse.
The divisor is repeatedly doubled to give the dividend.

For example, 153 divided by 9.

        powers of two       divisor and successive doubling     
         (doubling)

           2^0 = 1                   9
           2^1 = 2                  18
           2^2 = 4                  36
           2^3 = 8                  72
           2^4 = 16                144 
           2^5 = 32                288 
        
        288>153 so you can stop.
   
Look for the combination of numbers that add up to 153 in the divisor 
column.

This can be like a puzzle for the students and an excellent way to 
teach the problem-solving method known as guess and check. 
         
The combination that works here is 144 + 9 since 144 + 9 = 153.

To determine the divisor, look at the corresponding column of powers 
of two. Here we have:

      2^4 corresponding with 144 and 
      2^0 corresponding with 9.

So the divisor is 2^4 + 2^0 = 16 + 1 = 17.  

The complication with Eqyptian division comes with remainders.
        
For example,  17 divided by 3.

        Powers of two column        divisor doubling column
               
            2^0 = 1                       3
            2^1 = 2                       6
            2^2 = 4                      12
            2^3 = 8                      24
        24>17 so we can stop.

Looking at the combinations of 3, 6, and 12 we see that 12 + 3 = 15 is 
the closest we can seem to get without going over 17.

So what did the Eqyptians do to take care of the remainder? 
They found 2/3 of the divisor and then took one-half of that result. 
We choose 2/3 in this problem because 3 is the divisor. And since we 
know that 3/3 = 1, we can go ahead and determine the value associated 
with 1/3 too. When 3 is the divisor the possible remainders are 0/3, 
1/3, and 2/3.

So now we have:
      
        Powers of two column        divisor doubling column
               
            2^0 = 1                       3
            2^1 = 2                       6
            2^2 = 4                      12
                2/3                       2     (2/3 of 3 = 2)
                1/3                       1     (1/2 of 2/3 of 3 = 1)

So now in the divisor column what sums up to 17?
          
          3 + 12 + 2 = 17
         
and this corresponds to what in the powers of two column.

          2^0 + 2^2 + 2/3 =
              1 + 4 + 2/3 = 5 2/3

If you are looking for resources on Eqyptian division I would consider 
checking out some books on the history of mathematics or methods of 
teaching elementary mathematics.

Hope that this helps.  Thank you for writing to Ask Dr. Math.

- Doctor Mateo, The Math Forum
  http://mathforum.org/dr.math/