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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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