Transcript Example 1, scaffold division

Multi-digit Numerical Long Division
© 2013 Meredith S. Moody
1


Divide numbers with 2 or more digits using a
variety of methods for long division
Divide numbers with 2 or more digits using
the standard algorithm for long division
© 2013 Meredith S. Moody
2



Division is determining how many groups of
one number can be made out of another
number
For example, I have the number 15 and I want
to make 3 groups; how many will be in each
group? The answer would be 5
That is the same as dividing 15 by 3
© 2013 Meredith S. Moody
3





What if there are not a whole number of
groups?
Let’s say I have 15 cookies and I want to
make 4 bags (equal groups) of cookies.
If I divide 15 into 4 equal groups, I would
have 3 cookies in each bag, but I would have
3 cookies left over.
3 cookies would ‘remain’
In other words, 3 is my remainder if I want to
divide 15 by 4
© 2013 Meredith S. Moody
4





Mathematical operations come in pairs
Which operations do you think are pairs?
Addition and subtraction are a pair
Multiplication and division are a pair
In order to divide, you have to understand
multiplication
© 2013 Meredith S. Moody
5

Multiplication is repeated addition
◦ 3 x 5 = 15
◦ 3 + 3 + 3 + 3 + 3 = 15

Division is repeated subtraction
◦ 15 ÷ 5 = 3 (3 groups of 5, none left over)
◦ 15 – 5 – 5 – 5 = 0
© 2013 Meredith S. Moody
6






If I have 15 cookies and want to make 5 equal
bags of cookies, there must be 3 cookies in
each bag
I can make 5 bags of 3 cookies.
5 x 3 = 15
15 ÷ 3 = 5
15 ÷ 5 = 3
Division and multiplication are inverse
operations
© 2013 Meredith S. Moody
7





What happens if the numbers are too large to
divide mentally?
What if I want to divide 487 by 32?
How could I do that?
I could use a calculator, yes, but what if I
don’t have one?
Let’s look at three different methods of
dividing by hand
◦ Repeated subtraction
◦ Standard algorithm
◦ Scaffold division
© 2013 Meredith S. Moody
8






Division is actually repeated subtraction
How many times can I subtract 32 from 487?
487–32=455–32=423–32=391–32=359
359-32=327-32=295-32=263-32=321
321-32=199-32=167-32=135-32=103
103-32=71-32=39-32=7
How many times did we subtract 32? 15
How many is left over? 7
Wow! That took a long time. Is there another
way?
© 2013 Meredith S. Moody
9




An “algorithm” is a step-by-step procedure
for calculations
We can use a division algorithm for multidigit division
In this method, there are specific parts with
universal names
Knowing these names are important so
everyone can discuss division without
becoming confused
© 2013 Meredith S. Moody
10

The division
bracket is the
“box” into which
we put the
dividend
© 2013 Meredith S. Moody
11




487 is the dividend, it goes in
the “box”
32 is the divisor, it goes outside
the “box”
The answer is called the
“quotient”
The left over amount is called
the “remainder”
© 2013 Meredith S. Moody
12




The most efficient way to
divide multi-digit numbers by
hand is called ‘long division’
How many groups of 32 are in
the number 4? 0. 32x0=0.
subtract 4-0=4. ‘Bring down’
the next digit (8)
How many groups of 32 are in
the number 48? 1. 32x1=32.
subtract 48-32=16. ‘Bring
down’ the next digit (7)
How many groups of 32 are in
the number 167? 32x5=160.
subtract 167-160=7
© 2013 Meredith S. Moody
13





Wow, that standard algorithm doesn’t make
sense to me
Is there another way?
Yes
Instead of trying to divide 487 by 32, we can
break up our steps into smaller chunks
This is called scaffold division
© 2013 Meredith S. Moody
14





We can break up large numbers using the
place value system
487 becomes 400+80+7
How many groups of 32 can I make out of
400? Well, I know 3x4=12; I should be able
to make about 12 groups of 32 out of 400
Well, if I make 12 groups of 32, how much of
the 400 have I ‘used’? 12x32=384
How much of the 400 do I still have to ‘use’?
400-384=16; I have 16 ‘left over’
© 2013 Meredith S. Moody
15





Now I work with the number 80
How many groups of 32 can I make out of the
number 80?
I know 3x3=9, but 90 is too much; I should
be able to make 2 groups of 32 out of 80
If I make 2 groups of 32, how much of the 80
have I ‘used’? 32x2=64
How much do I have left to ‘use’? 80-64=16;
I have 16 ‘left over’
© 2013 Meredith S. Moody
16








Now I have to look at the number 7
How many groups of 32 can I make out of 7?
None
Let’s use our ‘leftovers’
I had 16 left over from the 400, 16 left over from
the 80, and 7 left over from my original work
16 + 16 + 7 = 39
How many groups of 32 can I make out of 39?
I can make 1 group of 32 out of 39, with 7 left
over
© 2013 Meredith S. Moody
17








Now I just add my groups together:
I had 12 groups in the 400
I had 2 groups in the 80
I had 1 group in the ‘leftovers’
12+2+1=15
I have 7 ‘left over’ now, so the answer to my
problem: what is 487÷32, is 15 remainder 7
That was a little hard to follow; is there an
easier way to write this?
Yes
© 2013 Meredith S. Moody
18



Let’s put our scaffold
method into an easyto-read structure:
487 = 400 + 80 + 7
400÷32 = 12
◦ 32 x 12 = 384
◦ 400-384 = 16

80÷32 = 2



7÷32 = 0
16+16+7 = 39
39÷32 = 1
◦ 32 x 1 + 32
◦ 39-32 = 7


12+2+1 = 15
487÷32 = 15 r7
◦ 32 x 2 = 64
◦ 80-64 = 16
© 2013 Meredith S. Moody
19



The scaffold method took quite a while, too
Is there a more efficient way to scaffold?
Yes
© 2013 Meredith S. Moody
20




Let’s try another together
Two individuals are to equally share an
inheritance of $860. How much should each
receive?
To solve the problem, we want to divide 860
by 2
Let’s look at the three ways we could solve
(no calculators!)
© 2013 Meredith S. Moody
21


Trying to repeatedly subtract 2 from 860
would take a LONG time
It makes sense to use a faster method
© 2013 Meredith S. Moody
22






Let’s use the extended
scaffold division method
First, we break up 860
using place values:
800 + 60 = 860
We can easily divide 800
by 2. 800÷2=400.
Each person would get
$400 so far
400+400=800. Since we
have ‘used’ $800, we
subtract 860-800 = 60.
We still have $60 to
share.
© 2013 Meredith S. Moody
23





Next, we share the $60.
Dividing $60 by 2 is
easy. Each person would
get $30.
We need to add another
30 to our quotient.
Notice we place the 30 in
the proper place value
above the 400. We have
‘used’ the last $60,
60-60 = 0.
We have no money left to
share.
© 2013 Meredith S. Moody
24


The last step is to sum the two partial
quotients to obtain the final quotient
400+30=$430
Each person would each receive $430
© 2013 Meredith S. Moody
25









Let’s use the standard
algorithm
The dividend is 860
The divisor is 2
There are 4 groups of
2 in 8
‘bring down’ the 6
There are 3 groups of
2 in 6
‘bring down’ the 0
There are 0 groups of
2 in 0
The quotient is 430
© 2013 Meredith S. Moody
26




What if we had three people and they needed
to split $986 evenly among them?
Repeated subtraction would take too long
The extended scaffold division method would
take a long time, too
Let’s start with the efficient scaffold division
method
© 2013 Meredith S. Moody
27



How much money
would each person
receive if 3 people had
to split $986 evenly?
Each person would
receive $328
There would be $2 left
over
© 2013 Meredith S. Moody
28

Let’s use the
standard algorithm
to solve
© 2013 Meredith S. Moody
29




If 14 children had 239 cookies, what is the
highest number of cookies each child could
receive if each one had to have the same
number?
Repeated subtraction would take too long.
The extended scaffold method would take too
long
Let’s start with the efficient scaffold method
and then try the standard algorithm
© 2013 Meredith S. Moody
30
239 ÷ 14 = 17 r 1
Each child would receive 17
cookies
There would be 1 cookie left over
© 2013 Meredith S. Moody
31
© 2013 Meredith S. Moody
32
 Use
either the traditional scaffold
division, efficient scaffold
division, or standard algorithm
method to solve: 236÷4
© 2013 Meredith S. Moody
33

© 2013 Meredith S. Moody
34

© 2013 Meredith S. Moody
35

© 2013 Meredith S. Moody
36




Use any long division method (repeated
subtraction, standard, scaffold, or extended
scaffold) to solve
193 ÷ 11
Repeated subtraction solution:
193-11=182-11=171-11=160-11=149-11=138
138-11=127-11=116-11=105-11=94-11=83-11=72
72-11=61-11=50-11=39-11=28-11=17-11=6
17 remainder 6
© 2013 Meredith S. Moody
37

© 2013 Meredith S. Moody
38

© 2013 Meredith S. Moody
39

© 2013 Meredith S. Moody
40