Section 4.1 Mental Math.
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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.
Compatible Numbers (friendly numbers)
are numbers whose sums, differences, products, or quotients are easy to calculate mentally.
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
Examples
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.
Multiplying by Special Factors
Numbers such as 5, 25, 11, and 99 are considered as special factors.
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
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
More examples:
24×26, 62×68, 73×77, 41×49.
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
More examples:
24×84, 96×16, 37×77, 54×54.
Intermediate Algorithms for Addition
134 + 325 = ?
Intermediate Algorithms for Addition
37 + 46 = ?
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)
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)
Chip Abacus Model for base five
Compute 123five + 34five
123five
34five
The answer is therefore 212five.
Intermediate Algorithms for Subtraction
357 – 123 = ?
Compute 243five – 112five
243five
take away 112five
Do we have enough chip to take away in each column?
YES
Compute 243five – 112five
take away 112five
131five
The answer is therefore:
Intermediate Algorithms for Subtraction
423 – 157 = ?
Compute 31five – 12five
31five
take away 12five
No
Do we have enough chips to take away in each column?
Compute 31five – 12five
31five
take away 12five
What can we do?
Trade one long for five small cubes.
Compute 31five – 12five
31five
take away 12five
Now we can start to take chips away.
Compute 31five – 12five
31five
take away 12five
Now we can start to take chips away.
Compute 102five – 23five
102five
take away 23five
Do we have enough chips to take away?
NO
Compute 102five – 23five
102five
take away 23five
What can we do?
Trade a flat for five longs etc.
Compute 102five – 23five
102five
take away 23five
Next trade a long for five small cubes.
Compute 102five – 23five
take away 23five
Now we can beginning taking chips away
Compute 102five – 23five
take away 23five
The answer is therefore:
24five
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
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
¯¯¯¯¯¯¯¯
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.
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
¯¯¯¯¯¯¯¯
1
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
8
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
1
8
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
1
7
8
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
1
7
8
6
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
1
1
7
8
6
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
1
1
1
1
7
8
1
3
6
8
Click to see the next step.
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
¯¯¯¯¯¯¯¯
4 0 0 6
¯¯¯¯¯¯¯¯
3
2
4
2
6
7
8
2
6
7
8
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.
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?
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
The key idea to the “adding the complement” method is to split up the original subtraction into two easier subtractions:
Algorithms for Multiplication split up the original subtraction into two easier subtractions:
The first thing students should memorize is the multiplication table.
The purpose of learning multiplication algorithm(s) is to perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Multiplication of 2digit numbers with base ten blocks
34 × 12 = ?
32 × 14 = ?
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 2digit by 2digit multiplication
2 3
5
6
Intermediate Algorithm for Multiplication perform multiplications of multidigit 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:
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
2 4
× 6 8
3
1
9
2
2 4
× 6 8
3
1
9
2
2
1
4
4
0
2 4
× 6 8
3
1
9
2
2
1
4
4
0
+
1,632
7 perform multiplications of multidigit numbers without memorizing a huge multiplication table.
23 ) 1738
Division AlgorithmsDivision 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.
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
In this PowerPoint presentation, we will introduce two
steps that will help learning the long division algorithm.
(I) A handson algorithm with the base ten blocks.
(II) An intermediate written algorithm called the Scaffold Method
Perform 447 ÷ 3 by base ten blocks perform multiplications of multidigit 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
Let’s find out which way is more efficient.
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Now we need to split the flat and possibly the long (click)
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
We now give 3 more longs to each group. (click)
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Next we split the longs into cubes. (click)
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Next we split the longs into cubes. (click)
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Next we split the longs into cubes. (click)
Method I: separate the cubes first. perform multiplications of multidigit numbers without memorizing a huge multiplication table.
Finally we distribute the cubes.
Method I: separate the cubes first. perform multiplications of multidigit 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.
Let’s try again with Method II: separating the flats first.
Wethen break the flat into 10 longs (click).
Let’s try again with Method II: separating the flats first.
Wethen break the flat into 10 longs (click).
Let’s try again with Method II: separating the flats first.
We then distribute the longs into 3 groups. (click)
Let’s try again with Method II: separating the flats first.
Now we break the longs into cubes. (click)
Let’s try again with Method II: separating the flats first.
Finally we distribute the cubes evenly into three sets. (click)
Let’s try again with Method II: separating the flats first.
Finally we distribute the cubes evenly into three sets. (click)
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.
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.
Addition table for base Five first.
Multiplication table for base Five
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
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 “DivideMultiplySubtractBring downrepeat”
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)
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
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.
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.
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.
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
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
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.
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
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
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
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.
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)
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.
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)
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)
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.
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)
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)
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)
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)
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
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
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.