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Section 4.1 Mental Math

The availability and widespread use of calculators and computers have permanently changed the way we compute. Consequently, there is an increasing need to develop students’ skill in estimating answers when checking the reasonableness of results obtained electronically.

Computational estimation, in turn, requires a good working knowledge of mental math. Thus this section begins with several techniques for doing mental calculations.

NTCM Standard

Grade 4 and 5: Students select appropriate methods and apply them accurately to estimate products and calculate them mentally, depending on the context and numbers involved.


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Compatible Numbers (friendly numbers)

are numbers whose sums, differences, products, or quotients are easy to calculate mentally.

  • Examples

  • 3 and 7 are compatible under addition because 3 + 7 = 10

  • 86 and 14 are compatible under addition because 86 + 14 = 100

  • 25 and 4 are compatible under multiplication because 25 × 4 = 100

  • 480 and 24 are compatible under division because 480 ÷ 24 = 20


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Compensation

is a process that reformulates a sum, difference, product, or quotient to another one that is easier to compute.

Additive compensation

98 + 56 = (98 +2) + (56 –2) = 100 + 54 = 154

Subtractive compensation (equal additions method)

47 – 29 = (47 +1) – (29 +1) = 48 – 30 = 18

Multiplicative compensation

16 × 35 = (16 ÷2) × (2× 35) = 8 × 70 = 560

Division compensation

180 ÷ 15 = (180 ×2) ÷ (15 ×2) = 360 ÷ 30 = 12

270 ÷18 = (270 ÷9) ÷ (18 ÷9) = 30 ÷ 2 = 15


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Left-to-Right Methods

478

– 263

342

+ 136

2

1

5

4

7

8


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Multiplying Powers of 10

Examples

  • 45 × 1000 = 45000 i.e. we simply attach 3 zeros to the end of 45.

  • 20 × 300 = 6000 i.e. we multiply 2 by 3 and then attach 3 zeros to the end of our answer.

Remark:

It is not correct to say “adding 3 zeros to our answer”, because adding 0’s is the same as adding nothing. The most precise way to say is “concatenating” 3 zeros to the end of our answer.


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Multiplying by Special Factors

Numbers such as 5, 25, 11, and 99 are considered as special factors.

  • 628 × 5 = (628 × 10) ÷ 2 = 6280 ÷ 2 = 3140

  • 47 × 5 = 470 ÷ 2 = 235

  • 46 × 99 = 46 × (100 – 1) = 4600 – 46 = 4554

  • 231 × 11 = 231 × 10 + 231 = 2310+ 231 = 2541


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Special squares

The squares of 15, 25, 35, … , 95 can be computed mentally, such as

152 = 225, 452 = 2025, …

Secret

The squares of these numbers always end in 25, and to compute the first 2 digits, we do the following

852 = 7225 where the 72 comes from 8×8+8


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Special Products (I)

43 × 47 = 2021 can be performed mentally by 1. 4×4 + 4 = 20

2. 3×7 = 21 3. Linking 20 to 21 and we get 2021

  • Remark:

  • This method works only for multiplying two-digit numbers, and with the extra conditions that

  • The 1st digits must be equal,

  • The 2nd digits add up to 10.

More examples:

24×26, 62×68, 73×77, 41×49.


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Special Products (II)

63 × 43 = 2709 can be performed mentally by 1. 6×4 + 3 = 27

2. 3×3 = 9 3. Linking 27 to 09 and we get 2709

  • Remark:

  • This method works only for multiplying two-digit numbers, and with the extra conditions that

  • The 2nd digits must be equal,

  • The 1st digits add up to 10.

More examples:

24×84, 96×16, 37×77, 54×54.


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Section 4.2

Written Algorithms for

whole number operations




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Chip Abacus Model for base five

Compute 123five + 34five

123five

34five

Now we have more than four cubes, so we need to join them end to end and form a long. (click to see)


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Chip Abacus Model for base five

Compute 123five + 34five

123five

34five


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five

Now we have more than four longs, so we need to glue them together and form a flat. (click to see)


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five


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Chip Abacus Model

Chip Abacus Model for base five

Compute 123five + 34five

123five

34five

The answer is therefore 212five.




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Subtraction in base five

Compute 243five – 112five

243five

take away 112five

Do we have enough chip to take away in each column?

YES


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Subtraction in base five

Compute 243five – 112five

243five

take away 112five


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Subtraction in base five

Compute 243five – 112five

take away 112five


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Subtraction in base five

Compute 243five – 112five

take away 112five


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Subtraction in base five

Compute 243five – 112five

take away 112five


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Subtraction in base five

Compute 243five – 112five

take away 112five


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Subtraction in base five

Compute 243five – 112five

take away 112five

131five

The answer is therefore:



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Subtraction in base five

Compute 31five – 12five

31five

take away 12five

No

Do we have enough chips to take away in each column?


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Subtraction again

Compute 31five – 12five

31five

take away 12five

What can we do?

Trade one long for five small cubes.


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Subtraction again

Compute 31five – 12five

31five

take away 12five


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Subtraction again

Compute 31five – 12five

31five

take away 12five


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Subtraction again

Compute 31five – 12five

31five

take away 12five

Now we can start to take chips away.


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Subtraction again

Compute 31five – 12five

31five

take away 12five

Now we can start to take chips away.


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Subtraction again

Compute 31five – 12five

take away 12five

Now we can start to take chips away.


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Subtraction again

Compute 31five – 12five

take away 12five

14five

The answer is therefore:


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One more subtraction problem

Compute 102five – 23five

102five

take away 23five

Do we have enough chips to take away?

NO


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One more subtraction problem

Compute 102five – 23five

102five

take away 23five

What can we do?

Trade a flat for five longs etc.


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One more subtraction problem

Compute 102five – 23five

102five

take away 23five


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One more subtraction problem

Compute 102five – 23five

102five

take away 23five

Next trade a long for five small cubes.


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One more subtraction problem

Compute 102five – 23five

102five

take away 23five


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One more subtraction problem

Compute 102five – 23five

102five

take away 23five


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One more subtraction problem

Compute 102five – 23five

take away 23five

Now we can beginning taking chips away


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One more subtraction problem

Compute 102five – 23five

take away 23five

The answer is therefore:

24five


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Equal Addend Subtraction Algorithm

The term “borrow” was dropped from elementary math books about fifteen years ago because we are not really borrowing, we regroup or trade. The object that we took out for trading will never be returned. Hence the new terminology “regrouping” is introduced.

However, this term “borrow”, no matter how incorrect it is, is stuck to everyone’s mind. Even if the teachers are not using it, the parents will still do.

It is therefore desirable to create an algorithm that allows us to borrow and return.

This is called the

Equal Addend (Subtraction) Algorithm


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The standard algorithm for subtraction can become quite complicated when there are several zeros in the minuend (the 1st number in the subtraction problem).

For instance, in the problem

9

9

3

10

10

16

4 0 0 6

  • 1 3 2 8

    ¯¯¯¯¯¯¯¯

2

6

7

8

we don’t have anything in the tens column hence we need to go to the thousands column to regroup and bring something back to the ones column. (click to see animation)

Most students will not be able to memorize the long process correctly.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 1

When we try to subtract 3 by 5, we can easily see that there is not enough to do so, hence we borrow 10 (perhaps using a credit card) and we change the 3 to13.(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 2

After we borrow, our debt will increase. We then cross out the 7 and change it to 8.(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

8

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 3

Perform the subtraction 13 – 5. (click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

8

8

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 4

In the ten’s column, 4 is not big enough to be subtracted by 8, so we borrow again.(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

1

8

8

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 5

Increase the debt by crossing out the 6 and change it to a 7.(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

1

7

8

8

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 6

Perform the subtraction 14 – 8.

(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

1

7

8

6

8

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 7

In the hundred’s column, 0 is not big enough to be subtracted by 7, so we borrow again and change the 0 to 10.

(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

1

1

7

8

6

8

Click to see the next step.


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Method: complicated when there are several zeros in the minuend (the 1

The major feature of this method is that we increase the subtrahend whenever we borrow, hence whatever we borrowed will be returned in the next step. This is sometimes nicknamed as the “credit card method”.

Step 8

Increase the debt by adding a 1 under the 2.(click to see animation)

Example:

2 0 4 3

  • 6 7 5

    ¯¯¯¯¯¯¯¯

1

1

1

1

7

8

1

3

6

8

Click to see the next step.


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Another example complicated when there are several zeros in the minuend (the 1

Let’s first review the standard method.

Now we look at the

Equal Addend method

Standard Method

9

9

10

10

16

3

10

10

16

4 0 0 6

  • 1 3 2 8

    ¯¯¯¯¯¯¯¯

4 0 0 6

  • 1 3 2 8

    ¯¯¯¯¯¯¯¯

3

2

4

2

6

7

8

2

6

7

8


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Conclusions: complicated when there are several zeros in the minuend (the 1

Just like learning anything new, the beginning is difficult.

You may criticize this algorithm just because it is different from the one that you are familiar with. We hope you can be more open minded and give it a chance. Compare it with the standard algorithm fairly.

Even if you really don’t like it, you should still thoroughly understand it just in case someday, one of your students likes it.


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Subtraction by adding the Complement complicated when there are several zeros in the minuend (the 1

Complement of a number

Do you see a pattern?


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Add the complement complicated when there are several zeros in the minuend (the 1

314

+ 814

1128

Original problem

314

– 185

Final adjustment

314

+ 814

1128

+1

129

Compute the complement of 185, which is

999 – 185 = 814

Subtraction by adding the Complement


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The key idea to the “adding the complement” method is to split up the original subtraction into two easier subtractions:

  • finding the complement.

  • deleting the leading 1 from the result of addition.


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Algorithms for Multiplication split up the original subtraction into two easier subtractions:

The first thing students should memorize is the multiplication table.


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The purpose of learning multiplication algorithm(s) is to perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Multiplication of 2-digit numbers with base ten blocks

34 × 12 = ?


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Multiplication of 2-digit numbers with Grid Paper perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

32 × 14 = ?


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Multiplication Algorithms perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

The Lattice Method

was introduced to Europe in 1202 in Fibonacci's Liber Abaci. It requires the preparation of a lattice (a grid drawn on paper) which guides the calculation. Its advantage is in separating all the multiplications from the additions.

Example: To compute 23 × 56, we build a 2 by 2 lattice because it is a 2-digit by 2-digit multiplication

2 3

5

6


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Intermediate Algorithm for Multiplication perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

This method is a bit longer than the standard one but it shows the place value of each digit very clearly.

Example: 2 3

× 5 6

1 8 because 3 × 6 = 18

1 2 0 b/c 20 × 6 = 120

1 5 0 b/c 3 × 50 = 150

1 0 0 0 b/c 20 × 50 = 1000

1 2 8 8

final answer:


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Standard Algorithm for Multiplication perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Example: Find 68 × 24

Step 2

Step 1

Step 3

2

3

3

2

3

24

× 68

24

× 68

24

× 68

192

192

192

24×8

1440

+ 1440

24×60

1632


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Variation of the Standard Algorithm perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

2 4

× 6 8

3

1

9

2


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Variation of the Standard Algorithm perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

2 4

× 6 8

3

1

9

2

2

1

4

4

0


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Variation of the Standard Algorithm perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

2 4

× 6 8

3

1

9

2

2

1

4

4

0

+

1,632


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7 perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

23 ) 1738

Division Algorithms

Division is by all means the most feared operation even for adults.

The standard algorithm requires lots of mental estimation.

For example, in the following problem

how do we know we should put down a 7 on top but not an 8 or 6?

It’s mainly because we mentally estimated the product 23 × 7 and knew that it is smaller than 173, and mentally estimated 23 × 8 is too big, etc.


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Division Algorithms perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

It is true that we no longer need to perform long division by hand in this computer age. However, we still need to learnand teach the basic principle of long division because it enables us to

  • better understand the meaning of division so that we can properly apply it in problem solving.

  • catch significant errors by estimation in the event that wrong data were fed to the calculator or the calculator malfunctioned.


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Division Algorithms perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

In this PowerPoint presentation, we will introduce two

steps that will help learning the long division algorithm.

(I) A hands-on algorithm with the base ten blocks.

(II) An intermediate written algorithm called the Scaffold Method


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Perform 447 ÷ 3 by base ten blocks perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Dividing 447 by 3 means separating

into 3 equal groups (possibly with a remainder).

There are two ways to do this

  • Separate the cubes first, then the longs and flats,

  • Separate the flats first, then the longs and the cubes.

    Let’s find out which way is more efficient.


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Now we need to split the flat and possibly the long (click)


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

We now give 3 more longs to each group. (click)


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Next we split the longs into cubes. (click)


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Next we split the longs into cubes. (click)


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Next we split the longs into cubes. (click)


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Finally we distribute the cubes.


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Method I: separate the cubes first. perform multiplications of multi-digit numbers without memorizing a huge multiplication table.

Finally we distribute the cubes.

You can see that even though this method works,

it is inefficient because we have to distribute the cubes twice, and it is more difficult to see that the answer is 149.


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Let’s try again with Method II: separating the flats first.

Wethen break the flat into 10 longs (click).


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Let’s try again with Method II: separating the flats first.

Wethen break the flat into 10 longs (click).


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Let’s try again with Method II: separating the flats first.

We then distribute the longs into 3 groups. (click)


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Let’s try again with Method II: separating the flats first.

Now we break the longs into cubes. (click)




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Let’s try again with Method II: separating the flats first.

Finally we distribute the cubes evenly into three sets. (click)


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Let’s try again with Method II: separating the flats first.

Finally we distribute the cubes evenly into three sets. (click)


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Let’s try again with Method II: separating the flats first.

Now it is easy to see that the answer is 149 and that this method is more efficient.

Conclusion: we should divide the “bigger” objects first.


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Intermediate Algorithms first.for Division

It is impractical to use manipulatives for large numbers, hence we have to learn some written algorithms.

Let us begin with some intermediate ones that are easy to learn but may be slower than the standard one.


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Addition table for base Five first.

Multiplication table for base Five


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Division by repeated subtraction first.

211Five ÷ 12Five = ?

211Five – 12Five = 144Five


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Division by a single-digit number first.

In this case, there is no need to introduce any intermediate algorithm, but a grid paper may help lining up the digits.

1

Example:

6 7 3 9

3

-6

1


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Division by a single-digit number first.

In this case, there is no need to introduce any intermediate algorithm, but a grid paper may help lining up the digits.

1

2

3

r 1

Example:

6 7 3 9

9

-6

3

1

-12

1

-1 8

1

The 4 steps to remember “Divide-Multiply-Subtract-Bring down-repeat”


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Intermediate Algorithms first.for Division by a 2-digit number

The Scaffold method

A scaffold is a temporary or movable platform for workers to stand or sit on when working at a height above the floor or ground level.

(Click to see a picture)



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Intermediate Algorithms first.for Division

The Scaffold method

A scaffold is a temporary or movable platform for workers to stand or sit on when working at a height above the floor or ground level.

We use this word for our division algorithm because

  • it is a temporary method to help us learn the standard algorithm.

  • it has structures that look like a real scaffold.


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23 first.) 1247

Guess and Check

Example:

Find the quotient and remainder of the division problem 1247 ÷ 23

Let’s start building a scaffold.

(23) × 100 = 2300

too big!

Since 2300 is bigger than 1247, we know that 23 cannot go into 1247 a hundred times. We have to scratch this out and try 23×10.


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23 first.) 1247

Guess and Check

Example:

Find the quotient and remainder of the division problem 1247 ÷ 23

Let’s start building a scaffold.

(23) × 100 = 2300

too big!

- 230

(23) × 10 = 230

1017

It looks like that we still have a lot left, so we can be more aggressive.


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23 first.) 1247

Guess and Check

Example:

Find the quotient and remainder of the division problem 1247 ÷ 23

Let’s start building a scaffold.

(23) × 100 = 2300

too big!

- 230

(23) × 10 = 230

1017

- 460

(23) × 20 = 460

557

Clearly we can take away another 20 copies of 23.


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23 first.) 1247

Guess and Check

Example:

Find the quotient and remainder of the division problem 1247 ÷ 23

Let’s start building a scaffold.

(23) × 100 = 2300

too big!

- 230

(23) × 10 = 230

1017

- 460

(23) × 20 = 460

557

- 460

(23) × 20 = 460

97


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Example: first.

Find the quotient and remainder of the division problem 1247 ÷ 23

Let’s start building a scaffold.

23 ) 1247

Guess and Check

(23) × 100 = 2300

too big!

- 230

(23) × 10 = 230

1017

- 460

(23) × 20 = 460

557

- 460

(23) × 20 = 460

97


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Example: first.

Find the quotient and remainder of the division problem 1247 ÷ 23

Let’s start building a scaffold.

23 ) 1247

Guess and Check

(23) × 100 = 2300

too big!

- 230

(23) × 10 = 230

1017

- 460

(23) × 20 = 460

557

- 460

(23) × 20 = 460

97

- 46

(23) × 2 = 46

51

- 46

(23) × 2 = 46

5

Therefore the quotient is 10+20+20+2 +2 = 54,

and the remainder is 5.


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23 first.) 1247

Guess and Check

Remarks:

we suggest using simple multiples of 23 in the calculations because we want the student to focus on the subtraction process rather than the tedious multiplication.

For example, 23 × 27 is not a good choice because it cannot be done mentally by an average student.

Usually we will choose simple numbers that the multiplication does not involve a “carry”, but sometimes doubling a larger number is also acceptable.

- 230

(23) × 10 = 230

1017

- 460

(23) × 20 = 460

557

- 460

(23) × 20 = 460

97

- 46

(23) × 2 = 46

51

- 46

(23) × 2 = 46

5


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Division in Base Five first.

24Five ) 1430Five

-240Five 10Five × 24Five = 240Five

1140Five

-240Five 10Five × 24Five = 240Five

400Five

-240Five 10Five × 24Five = 240Five

110Five

- 24Five 1Five × 24Five = 24Five

31Five

- 24Five 1Five × 24Five = 24Five

2Five

Therefore quotient = 10Five + 10Five + 10Five + 1Five + 1Five = 32Five

Remainder = 2Five


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Division in Base Five first.(example 2)

42Five ) 24232Five

- 4200Five 100Five × 42Five = 4200Five

20032Five

- 4200Five 100Five × 42Five = 4200Five

10332Five

- 4200Five 100Five × 42Five = 4200Five

1132Five

- 420Five 10Five × 42Five = 420Five

212Five

- 42Five 1Five × 42Five = 42Five

120Five

Therefore quotient = 100Five + 100Five + 100Five + 10Five + 1Five +1Five = 312Five

Remainder = 23Five


Advantages of the scaffold method l.jpg
Advantages of the Scaffold Method first.

There are many advantages of using scaffolding:

1. It's fun and it makes sense.

2. It develops estimation skills.

3. Students are engaged in mental arithmetic – they are thinking throughout the process, not just following an algorithm.

4. Students develop number sense.

5. The more number sense that students possess, the more efficient the process.

6. There are many correct ways to arrive at a solution.

7. There are fewer opportunities for error than with long division.

8. Students who practice scaffolding are better able to divide mentally.


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Intermediate algorithm first.

This algorithm uses appropriate rounding to make estimation easier.

Question: What is the quotient and remainder of 5667 ÷ 37?

100

37 5667

3700

Think: How many times can 37 go into 5600?

Answer: 100.

We then put 100 on top of 667 and then put 3700 under 5667. (click)

The next step is to subtract. (click)


Slide115 l.jpg

Intermediate algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

100

37 5667

3700

1967

Think: How many times can 37 go into 5600?

Answer: 100.


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Intermediate algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

40

100

37 5667

3700

1967

Think: How many times can 37 go into 1967?

1480

Answer: Here we should estimate. How many times can 40 go into 1900? It should be 40.

We then put 40 on top of 100 and then multiply 37×40 = 1480. (click)


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Intermediate algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

12

40

100

37 5667

3700

1967

Think: How many times can 37 go into 487?

1480

Answer: Here we should estimate. How many times can 40 go into 480? It should be 12.

487

444

We then put 12 on top of 40 and then multiply 37×12 = 444. (click)


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Intermediate algorithm first.

1

12

40

100

37 5667

3700

1967

Think: How many times can 37 go into 43?

1480

Answer: 1

487

444

43

37

6

The final answer is then 100 + 40 + 12 + 1 = 153, with remainder 6.


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

Think: How many times can 37 go into 5?

37 5667

Answer: none.

We then move the purple rectangle one position to the right. (click)


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

1

37 5667

-37

Think: How many times can 37 go into 56?

Answer: 1.

We then put 1 on top of 6 and then put 37 under 56. (click)

The next step is to subtract. (click)


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

1

37 5667

-37

1

9

Think: How many times can 37 go into 56?

Answer: 1.

Now we move the purple rectangle one more step to the right. (click)


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

1

37 5667

-37

19

Now we move the purple rectangle one more step to the right. (click)


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

1

37 5667

6

-37

Think: How many times can 37 go into 196?

19

Answer: 5

Bring down the 6.

We then put a 5 on top and mentally calculate 37×5 = 185


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

1

5

37 5667

-37

Think: How many times can 37 go into 196?

19

6

-

185

Answer: 5

11

We then put a 5 on top and mentally calculate 37×5 = 185


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Standard algorithm first.

Question: What is the quotient and remainder of 5667 ÷ 37?

1

5

3

37 5667

7

-37

Think: How many times can 37 go into 117?

19

6

-

185

11

Answer: 3

- 111

6

We then put a 3 next to the 5 in the quotient, and subtract 3 × 37 (= 111).

The final answer is 153 with remainder 6.


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