Multiple Subtraction by Missing Addend MethodDate: 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/ |
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