Multiple Subtraction by Missing Addend Method

Date: 09/26/2002 at 00:35:19
From: Ivan Isop
Subject: Multiple subtraction by missing addend method

      11 111
    -  2 345
    -  1 638
    -  3 172
    -  1 932
   ----------
         ?              

To get the difference by the missing addend method we are not supposed 
to change the given problem into subtraction of one subtrahend, but to 
work out the problem as it is. We are not even to use regrouping. I 
have heard of this method but my teachers are 
unable to describe it to me.

- II

Date: 09/27/2002 at 10:52:31
From: Doctor Rick
Subject: Re: Multiple subtraction by missing addend method

Hi, Ivan.

I haven't heard of such a method, but I can make one up, using the 
name you gave it as a clue.

The name "missing addend" suggests that we treat the problem as an 
addition problem with one of the addends missing. Your problem is 
equivalent to the following:

       2 345
    +  1 638
    +  3 172
    +  1 932
    +      ?
   ----------
      11 111

Let's see how it works. I start adding in the rightmost column: 
5+8+2+2 = 17. I need a 1 in the ones column, so the sum must be 21 in 
order to be greater than 17. I need 4 more to reach 21, so I write 
down the 4; now the complete sum in this column is 21, which means I 
must carry the 2.

          2
       2 345
    +  1 638
    +  3 172
    +  1 932
    +     ?4
   ----------
      11 111

Now I add the tens column: 2+4+3+7+3 = 19; again I need 21 so I put 
down 2 (21-19), and carry the 2 (from 21).

         22
       2 345
    +  1 638
    +  3 172
    +  1 932
    +    ?24
   ----------
      11 111

In the hundreds column, I get 2+3+6+1+9 = 21; that's just what I need, 
so I put down 0. I must again carry the 2 from 21.

       2 22
       2 345
    +  1 638
    +  3 172
    +  1 932
    +  ? 024
   ----------
      11 111

In the thousands column, I get 2+2+1+3+1 = 9; I need 11, so I put 
down 2, and since 11 is just what I've got in the total, I'm done. 
The answer is 2024.

We can do the same operations in the oiginal subtraction 
configuration. You might find it easiest to add UP the columns so 
you're looking at the "total" (the top line) when you're done - don't 
forget to add in the carry. (I might circle the top number so I don't 
forget and keep adding.) Here is the result:

      12 22
      11 111         2+2+8+5=17, add 4 to get 21: 4, carry 2.
    -  2 345         3+7+3+4+2=19, add 2 to get 21: 2, carry 2.
    -  1 638         9+1+6+3+2=21, add 0 to get 21: 0, carry 2.
    -  3 172         1+3+1+2+2=9, add 2 to get 11: 2, carry 1.
    -  1 932         1(carry), add 0 to get 1:we're done.
  ----------
       2 024

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

Date: 09/29/2002
From: Ivan Isop
Subject: Multiple subtraction by missing addend method

Thank you for a sensible and informative reply to my question. 

This way is easy to do for any subtraction, which makes me wonder why 
it is not known on this continent. Even you say that you have not 
heard of it. It seems to me that this method eliminates altogether our 
way of dealing with multiple subtraction, where we need to rewrite all 
subtrahends separately to obtain one subtrahend, then rewrite the 
problem as subtraction with one subtrahend, then do our regrouping, 
and only then are we able to do our main task. Isn't this awkward? 

Can we assume this method could work even division? What division is 
all about is actually another form of multiple subtraction. I tried to 
use this method and here is the result:    

   11 111  :  1 638  = 6 1 283 

I would like to describe my thinking process.

Traditional estimation does take place. With estimation of 6 I am 
multiplying the 8 of ones. This gives 48. To my 1 I have to  "fill" to 
51 with 3, which I write as my first number of difference (missing 
addend). I have 5 to carry over (I don't write carry overs at all). 
Next, 6 times 3 is 18, plus 5 makes 23, missing is 8 to 31, and so on. 

I think this procedure is neater than the present teaching of 
division. What do you think?

Date: 09/30/2002 at 11:24:46
From: Doctor Rick
Subject: Re: Multiple subtraction by missing addend method

Hi, Ivan.

I think you have a good idea for reducing the amount you need to 
write down while minimizing what you need to remember - a single 
borrowed digit. I would teach long division to start with, to give an 
understanding of what is going on, but your method makes a good 
shortcut to be taught later. 

Your example problem done with long division looks like

             6.7
       ---------
  1638 ) 11111.0
          9828
         -----
          1283 0
          1146 6
          ------
           136 4

Your shortcut involves doing the multiplication and subtraction (e.g. 
11111 - 6 * 1638) in one step, so we only write down the partial 
remainder (1283):

  6*8 = 48;           51 - 48 = 3; write the 3, remember the 5
  6*3 = 18, + 5 = 23; 31 - 23 = 8; write the 8, remember the 3
  6*6 = 36, + 3 = 39; 41 - 39 = 2; write the 2, remember the 4
  6*1 =  6, + 4 = 10; 11 - 10 = 1; write the 1, remember the 1
                       1 -  1 = 0, we're done.

For the second digit, we do as follows:

  7*8 = 56;           60 - 56 = 4; write the 4, remember the 6.
  7*3 = 21, + 6 = 27; 33 - 27 = 6; write the 6, remember the 3.
  7*6 = 42, + 3 = 45; 48 - 45 = 3; write the 3, remember the 4.
  7*1 =  7, + 4 = 11; 12 - 11 = 1; write the 1, remember the 1.
                       1 -  1 = 0, we're done.

This is what I'd write:

             6.7
       ---------
  1638 ) 11111.0
          1283
           136 4

I don't know if everyone would find the missing-addend method easier 
for simple two-number subtraction, but I do like how you've applied 
it to division. With this method, you don't need to deal with both 
the carries in multiplication and the borrows in subtraction.

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