Math 208, Section 7.2 Solutions: Understanding Long Division
1. a.) Calculate 4215 ÷ 6 and 62,635 ÷ 32 in two ways: with the standard division method
and with scaffold method.
Standard:
Scaffold:
2
100
100
500
702 rem 3
6
4215
42
6
15
12
3
Answer: 702
rem: 3
4215
3000
1215
600
615
600
15
12
3
7
50
400
500
1000
1957 rem 11
32 62635
32
306
288
183
160
235
224
11
32
Answer: 1957
rem 11
62635
32000
30635
16000
14635
12800
1835
1600
235
224
11
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b.) Compare the standard division method and the scaffold method. How are the methods
alike? How are the methods different? What are advantages and disadvantages of each
method?
Both the standard and the scaffold division methods find the correct answer using several
intermediate steps of multiplication and subtraction. The scaffold method works with entire
numbers (the entire dividend), rather than just portions of numbers. When you use the scaffold
method, you have to keep track of place value. The standard method is really just an abbreviated
version of the scaffold method. The scaffold method allows users to work the problem in small
steps, which is less efficient than the standard algorithm, but may be clearer for early learners.
Furthermore, in the Scaffold method, we do not have to erase and try to guess a closer answer as
long as we guess under the quotient.
2. a.) Use the scaffold method to calculate 793 ÷ 4.
8
40
50
100
4
793
400
393
200
193
160
33
32
1
4. a.) Use long division to determine the decimal number to answer to 2893 ÷ 6 to the
hundredths place.
482.166...
6
So, we would round to 482.17
2893.00
24
49
48
13
12
10
6
40
36
40
36
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5. Tamarin calculates 834 ÷ 25 in the following way: I know that 4 25s make 100, so I
counted 4 for each of the 8 hundreds. This gives me 32. Then there is one more 25 in 34, but
there will be 9 left. So the answer is 33, remainder 9.
a.) Explain Tamarin’s method in detail, and explain why her method is legitimate. (Do not
just state that she gets the correct answer; explain why her method gives the correct
answer.) Include equations as part of your explanation.
We want to divide 834 by 25. We are thinking of adding copies of 25: how many do we have to
add to get close to 834? If I add 4 copies of 25 together I get 100; so if I do that eight times (4
copies each repeated 8 times) I have added together 8×4 copies = 32 copies. Now I have gotten
to 800 (that is, 8 × 4 × 25) so I have 32 left. I can only fit in one more copy of 25, leaving 9. So
the answer is 32 + 1= 33 and the remainder is 9.
As an equation: 832 = 8 × (4 × 25) + 1 × 25 + 9 = (32 × 25) + (1 × 25) + 9
= 33 × 25 + 9.
b.) Use Tamarin’s method to calculate 781 ÷ 25.
Let’s see. 4 copies of 25 is 100 and I need to get to 700 so that’s 7 × 4 = 28. Now that gets me
700 so I have 81 to go. 3 more copies of 25 is 75 out of the 81; so that’s 28 + 3 copies of 25...31
copies of 25 and I have 81 – 75 = 6 left over. So the answer is 31, remainder 6.
8. Meili calculates 1200 ÷ 45 in the following way: (Given in the book)
a.) Explain why Meili’s strategy makes sense. It may help you to work with a story problem
for 1200 ÷ 45.
Meili is just doing a sort of scaffold method without putting in the scaffold framework. She is
merely putting down various things to multiply 45 by and keeping track of how many copies of
45 she uses. Let’s put this all into a story problem. Meili needs to distribute 1200 candies
equally (or as equally as possible) among the 45 students in first grade. She lines up the kids and
gives each kid 10 candies. That uses up 450 of the candies. So she does it again. Each kid now
has 20 candies and she’s used 450 + 450 = 900 candies.
She gives every kid 2 candies; that’s 90 more candies given out. Now they each have 22
candies and she’s given out 900 + 90 = 990 candies. Still a lot to go. She gives out 90 more
candies. Each kid has 24 and she’s given out 990 + 90 = 1080 candies .
One more time she gives out 2 candies to each kid; they have 26 candies and she’s given out
1080 + 90 = 1170. There are (1200 – 1170) = 30 candies left . That’s not enough to give one
to every kid. So they each get 26 candies and there are 30 left over. Whew!
b.) Write equations that correspond to Meili’s work and that demonstrate that 1200÷
÷ 45 = 26
remainder 30.
1200 = (10 × 45) + (10 × 45) + (2 × 45) + (2 × 45) + (2 × 45) + 30
= ( 10 + 10 + 2 + 2 + 2 ) × 45 + 30
= 26 × 45 + 30.
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