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### 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
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.

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/
```
Associated Topics:
Elementary Division
Elementary Subtraction

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