Teacher2Teacher Q&A #4130

European long division

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From: Gail (for Teacher2Teacher Service)
Date: Jun 17, 2000 at 07:16:19
Subject: Re: European long division

I have used the following method with some students who were having trouble with the "traditional American" Algorithm. It seemed to be easier for them to catch on to... I think it helps students gain a better understanding of what is happening during the division process, since it deals with the value of the total number, not small chunks of it (for example, if the dividend is 345, that is what the student works with, not 3, then 32, then what ever is left with 5 tacked onto the end). It also assists those who do not have a good recall of all the multiplication facts, because they don't have to come up with the ONE correct digit for each step. Instead, they can use a succession of factors to get to the final quotient. It is very similar to one of the first methods Pat posted in his responses. Here is how it works... ____ 7 /478 7 X 10 the student can select any factor they wish to use - 70 to find a product to subtract from the original _____ amount (the dividend)-- usually a student 408 7 X 10 selects an easy to use factor like -70 1, 2, 5, or 10 ____ 338 7 X 10 -70 As students become more familiar with the ____ process they will select larger factors 268 7 X 10 -70 ____ 198 7 X 10 -70 ____ 128 7 X 10 -70 ___ 58 7 X 2 -14 ___ 44 7X 2 -14 ___ 30 7 X 2 -14 ____ 16 7X 2 -14 ___ 2 When no more subtractions can take place, all the factors along the left side are added together, like this: 10 + 10 + 10 + 10 + 10 + 10 + 2 + 2 + 2 + 2= 68 See, this student was able to finish the problem without using any of the facts except 7 X 10 and 7 X 2. This could be of great assistance to someone who understood why we needed to divide, but didn't have the facts mastered yet. This is easier for struggling students because they don't have to have the perfect factor each time, just one that will work. They continue finding factors and dividing until they have a remainder that is smaller than the divisor (in this case, 7). When they have found all the factors, they just add them up (not the 7, but the other one each time). That is their quotient, and what is left is the remainder, which can be put in fraction form (in this case, 2 out of 7, or 2/7) , or they can place a decimal, and add zeroes to continue dividing. Besides that help, this process treats the dividend (in this case, 478) as a whole amount, instead of looking at 4 hundreds, then 48 tens, as the traditional algorithm does. For some students, that is a new way to look at the procedure. They have been used to looking at the amounts in terms of single digits... "How many 7's are in 4?" then "How many sevens are in 48?" No one has helped them see the big picture, that the 4 is really 4 hundreds, and the 48, 48 tens... I try to move my students to try larger "chunks" once we have that first step mastered... ____ 7 /478 7 X 30 -210 _____ 268 7 X 30 -210 ____ 58 7 X 5 -35 ____ 23 7 X 3 -21 ____ 2 30 + 30 + 5 + 3 = 68 Then we have a challenge to try to find the one "best chunk" for each place value. In the next example, there is one "chunk" for the tens, and one for the ones. (There is no hundreds chunk, because 7 X 100 would be more than 478). _____ 7 /478 7 X 60 -420 _____ 58 7 X 8 -56 ____ 2 60 + 8 = 68 You will probably notice that that is VERY similar to the traditional algorithm. It is just a small nudge away for most students, at that point:-) -Gail, for the Teacher2Teacher service

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