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Q&A #19612


Showing a student how to do division in the simpliest way

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From: Gail (for Teacher2Teacher Service)
Date: May 28, 2008 at 18:49:13
Subject: Re: Showing a student how to do division in the simpliest way

Hi Annette,  

I have used a different algorithm with students in the past that seems to
help them get a "grip" on long division.

Here is how it works:

This is an alternate algorithm that helps students who are having difficulty
with the traditional version.  It helps students gain an understanding of
what is happening during the division process, and 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. Here is how it works...

   ____
7 /478     7 X 10     the student can select any factor they wish to use to
                     find a product to subtract from the original amount
  - 70             (the dividend) -- usually a student selects an 
 _____               easy-to-use factor like  1, 2, 5, or 10
   408     7 X 10
   -70
 _____
   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      7 X 2
  -14
______
   30       7 X 2
 - 14
______
  16       7 X 2    When you have found all the "groups" of 7 that can be
                      subtracted out, you are finished.
 -14               Now add all the factors you multiplied by 7, 
_____               and that will give you the quotient.
   2

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.

Hope this gives you something to think about...   :-)

 -Gail, for the T2T service

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