Level III Whole Number
Operations Bridging Unit
(Pilot Materials)
NSSAL
(Draft)
C. David Pilmer
2013
(Last Updated: October, 2013)
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This resource is the intellectual property of the Adult Education Division of the Nova Scotia
Department of Labour and Advanced Education.
The following are permitted to use and reproduce this resource for classroom purposes.
 Nova Scotia instructors delivering the Nova Scotia Adult Learning Program
 Canadian public school teachers delivering public school curriculum
 Canadian non-profit tuition-free adult basic education programs
The following are not permitted to use or reproduce this resource without the written
authorization of the Adult Education Division of the Nova Scotia Department of Labour and
Advanced Education.
 Upgrading programs at post-secondary institutions
 Core programs at post-secondary institutions
 Public or private schools outside of Canada
 Basic adult education programs outside of Canada
Individuals, not including teachers or instructors, are permitted to use this resource for their own
learning. They are not permitted to make multiple copies of the resource for distribution. Nor
are they permitted to use this resource under the direction of a teacher or instructor at a learning
institution.
Acknowledgments
The Adult Education Division would like to thank Dr. Genevieve Boulet (MSVU) for reviewing
this resource and providing valuable feedback.
The Adult Education Division would also like to thank the following ALP instructors for piloting
this resource and offering suggestions during its development.
Eileen Burchill (IT Campus)
Elliott Churchill (Waterfront Campus)
Lynn Cuzner (Marconi Campus)
Carissa Dulong (Truro Campus)
Krys Galvin (Truro Campus)
Barbara Gillis (Burridge Campus)
Nancy Harvey (Akerley Campus)
Barbara Leck (Pictou Campus)
Suzette Lowe (Lunenburg Campus)
Alice Veenema (Kingstec Campus)
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Table of Contents
Introduction (for Instructors) ………………………………………………………………
iii
Introduction to Whole Number ……………………………………………………………
Math Facts: Addition and Multiplication ………………………………………………….
Fact Families ……………………………………………………………………………….
Order of Operations ………………………………………………………………………..
Multiples of Ten, One Hundred, and One Thousand ………………………………………
Divisibility and Prime ………………………………………………………………………
Estimating ………………………………………………………………………………….
Adding Multi-Digit Numbers ………………………………………………………………
Subtracting Multi-Digit Numbers ………………………………………………………….
Multiplying Multi-Digit Numbers ………………………………………………………….
Dividing Multi-Digit Numbers ……………………………………………………………..
1
7
17
22
26
33
35
38
41
45
53
Appendix …………………………………………………………………………………..
Connect Four Games ……………………………………………………………………
Flash Cards ……………………………………………………………………………..
Answers …………………………………………………………………………………
61
62
71
85
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Introduction (for Instructors)
This unit is concerned with having learners:
1. Develop strong mental math skills as they pertain to the addition, subtraction, multiplication,
and division facts associated with whole numbers. This means learners should develop a
high level of automaticity associated these math facts.
2. Add, subtract, multiply, and divide multi-digit whole numbers, not limited to the traditional
algorithms. They are also expected to use their estimation skills to judge the reasonableness
of their final answer.
Let's take a few minutes to discuss the importance of mental math. Mental math is the process of
calculating the exact numerical answer without the aid of any external calculating or recording
device. Research shows that as adults over 80% of the mathematics we encounter in our daily
lives involves the mental manipulation of numerical quantities rather than the traditional paper
and pencil math so often stressed in schools. Most learners feel that mental math is important
however, they mistakenly believe that written math is learned in school while mental math is
learned outside of school. Although learners may value mental math, they may not be able to
perform even the most straightforward calculations mentally. Consider that on the Third
National Mathematics Assessment, only 45% of 17 year olds were able to multiple 90 and 70
mentally.
Research states that mental math activities should be integrated into daily classroom practices,
rather than being taught as an isolated unit. The exposure should be gradual and continuous. For
the Level III Math program, we recommend that such activities serve as a 4 to 6 minute warm-up
at the beginning of a session for the first few weeks. There are three reasons for this approach.
The first is that mental math is pervasive in the real-world and the curriculum. The continued
classroom exposure to mental math is meant to be reflective of the math that learners encounter
in the real-world and in the curriculum. The second reason deals with the memory demands of
mental math. Mental math tends to be easier for individuals whose addition and multiplication
facts are firmly entrenched in their long-term memory. If this is so, the working memory is
available to work flexibility with number and operations. For learners who have not retained the
addition and multiplication facts in their long-term memory, mental math can be very
challenging. In these cases, their working memory is consumed with determining the fact, and
they have little time or space left to address the question that was asked. When such memory
deficits occur, we, as instructors, need to present the learners with a series of strategies that allow
quick retrieval of pertinent facts. This goes beyond asking learners to go back and memorize
addition and multiplication charts. Many adult learners have been unsuccessful with this
approach, so more innovative strategies are required; strategies that are less taxing on the longterm memory. Consider the following examples.
(1) Flip Flop (actually the commutative property)
Many learners struggle with 9  3 because they attempt to figure out 9 sets of 3, however,
if they know 9  3  3  9 , then the question is more accessible to some. Using the
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commutative property cuts the long-term memory requirements for the addition and
multiplication tables almost in half.
(2) Next Even/Next Odd
When you add 2 to a whole number you just have to find the next even or odd number
depending on the number you start with. If you start with an even number and add 2,
then the sum is the next even number. If you start with an odd number and add 2, then
the sum is the next odd number.
(3) Add Ten, Then Subtract One
This strategy is used when adding 9 to a number (e.g. 9  6 ). This strategy relies on the
fact that most people are comfortable adding 10 to a number (e.g. 10  3  13 ,
7  10  17 ). So if we want to add 9 and 4 (i.e. 9  4 ), we will add 10 and 4, then
compensate by subtracting 1 (i.e. 10  4  1).
(4) Double Double
Multiplying a number by 4 is the same as doubling the number and then doubling that
new answer. If you have 4  6 , then you double 6 and then double that answer.
(5) Snake Legs
People often confuse the rule for multiplying by 0 with the rule for adding 0. For the
question 5 0 or 0  5 , learners need to remember that these mean 5 sets of 0, or 0 sets
of 5. In either case, the answer is 0. The easiest way to remember this is to use the snake
leg strategy. How many legs does one snake have? (Answer: 0) How many legs do five
snakes? (Again the answer is 0). Therefore, we can conclude that 5  0  0 .
Although these strategies, and the others that you will be exposed to in second section of this
resource, reduce the demands on the long-term memory, many learners will still require time and
practice to solidify the strategies in their long-term memory. If this is accomplished then
learners can develop a higher level of automaticity.
It should be mentioned that not all Level III learners will need to learn these strategies. A
significant group of these individuals will recall these facts with a quick review/refresher of the
addition and multiplication tables. However, there are other individuals who have repeatedly
struggled to retain these facts through memorization; these strategies are designed for those
individuals. Although all of these strategies only appear in one section of this resource, do not
feel that most learners will master all their facts using all these strategies in a few lessons. It will
likely take the learner several weeks to retain the strategies and the accompanying facts. As
stressed earlier in this introduction, the exposure to these strategies should be both gradual and
continuous. For further information on these strategies, and others, you may wish to access the
Mental Math resource (found on SharePoint).
As mentioned at the beginning of this introduction, we also want learners to be able to add,
subtract, multiply, and divide multi-digit whole numbers, but not necessarily limiting them to the
traditional algorithms. For example, there are learners who struggle to remember all the steps for
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multiplying multi-digit whole numbers. For these individuals, they may prefer multiplying such
numbers by multiplying them in their expanded forms. An example of this is shown below.
e.g. 84  57
Answer:
80  4
 50  7
2
5 6
2 0
4 0 0
8
0
0
0
The 7 must be multiplied by both the 80 and the 4.
The 50 must be multiplied by both the 80 and the 4.
4 7 8 8
Does that mean that everyone must learn this alternate method? No; only those who are not best
served by the traditional algorithm.
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Introduction to Whole Numbers
The set of whole numbers is 0, 1, 2, 3, 4, 5,…
(This set goes on indefinitely.)
Whole numbers do not include:
2
9
and
, (which are both between the whole numbers 0 and 1).
5
10

fractions like

mixed numbers like 2

decimals like 0.09 (which is between the whole numbers 0 and 1) and 4.7 (which is
between the whole numbers 4 and 5).

negative numbers like -2 or -9.2 (which are both less than the whole number 0).
1
5
(which is between the whole numbers 2 and 3) and 7 (which
6
16
is between the whole numbers 7 and 8).
For large whole numbers, we separate the digits into groups of three, called periods. Each
period has a name: ones, thousands, millions, billions, and trillions.
3 480 120 000
ones
thousands
millions
billions
In Canada, we use spaces to separate the periods. In the United States, they use commas to
separate the periods. We choose to use spaces because in Europe, those countries use the comma
in the same way we and the US use a decimal point.
Canada
2 649
705 830
3 834 000
United States
2,649
705,830
3,834,000
For this course, we are not really concerned with whether you use spaces or commas; just
recognize that all Canadian math print resources will use spaces.
Writing Numbers Using Words
Many two digit numbers, like 46 and 73, are written using a hyphen. For example, the number
38 is written as thirty-eight. Similarly, the number 95 is written as ninety-five.
The three-digit number 627 is written as six hundred twenty-seven. The most common mistake is
writing it as six hundred and twenty-seven. The word "and" is not used when writing whole
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numbers using words. Rather, it is used with decimals and mixed numbers. For example, the
1
decimal 2.1 or the mixed number 2
is correctly written as two and one-tenth.
10
When writing larger whole numbers (i.e. four or more digits), it is important to know and
understand the chart below.
Example 1
Write each number in words.
(a) 23 735
(c) 6 207 500
(e) 3 190 700 000
Ones
Tens
Ones
Period
Hundreds
Thousands
Ten Thousands
Thousands
Period
Hundred Thousands
Millions
Ten Millions
Hundred Millions
Billions
Ten Billions
Hundred Billions
Billions
Period
Trillions
Ten Trillions
Hundred Trillions
Trillions
Period
Place Value Chart
Millions
Period
(b) 270 516
(d) 15 049 670
(f) 56 000 987 000
Answers:
When writing out your answers using words, separate each period using commas. Ironically
we do not do this in Canada when writing the number in numerical form.
(a) 23 735
twenty-three thousand, seven hundred thirty-five
(b) 270 516
two hundred seventy thousand, five hundred sixteen
(c) 6 207 500
six million, two hundred seven thousand, five hundred
(d) 15 049 670
fifteen million, forty-nine thousand, six hundred seventy
(e) 3 192 700 000
(f) 56 000 987 000
three billion, one hundred ninety-two million, seven hundred thousand
fifty-six billion, nine hundred eighty-seven thousand
Example 2
Give the place value of each underlined digit in each whole number.
(a) 5 976 050
(b) 7 398 050 000
Answers:
(a) 5 976 050
(b) 7 398 050 000
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The 7 is in the ten thousands place.
The 3 is in the hundred millions place.
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Example 3
Write each number in expanded form.
(a) 359
(c) 65 078
(e) 304 609
(b) 2 895
(d) 157 800
(f) 4 081 540
Answers:
(a) 359  300  50  9
(b) 2 895  2 000  800  90  5
(c) 65 078  60 000  5 000  70  8
(d) 157 800  100 000  50 000  7 000  800
(e) 304 609  300 000  4 000  600  9
(f) 4 081 540  4 000 000  80 000  1 000  500  40
Questions
1. Write each number in words
(a) 986
(b) 412
(c) 2 397
(d) 4 609
(e) 12 750
(f) 347 052
(g) 506 900
(h) 2 870 040
(i) 350 026 000
(j) 17 509 100
(k) 7 360 800 000
(l) 11 094 300 002
(m) 6 532 000 000 080
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2. Write each number using digits
(a) six thousand, seven hundred forty-two
(b) three million, one hundred twenty-six thousand, five hundred eleven
(c) eight hundred nine thousand, three hundred twenty-seven
(d) thirteen million, fifty-two thousand, seventy-one
(e) sixty-three thousand, one hundred forty-seven
(f) four hundred eight million, seven hundred thousand, two hundred sixty
(g) three hundred forty-nine thousand, eight
(h) two billion, nine hundred ten million, seven hundred six thousand, fifty
(i) one hundred sixty-two billion, eighty-three million, four hundred thousand
(j) eighty billion, three hundred seven million, five thousand, two hundred
3. Give the place value of each underlined digit in each whole number.
(a) 42 937 050
(b) 152 856 200
(c) 894 751
(d) 23 872 000 000
(e) 6 930 850 000
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4. Write each number in expanded form.
(a) 4 296
(b) 136 942
(c) 6 056 730
(d) 45 760 500
(e) 9 602 000 000
5. Order the numbers from smallest to largest.
(a) 546, 93, 3, 800, 52, 9
(b) 8 700, 27, 529, 29, 8 698, 546
(c) 796, 34 000, 9 360, 790, 33 870, 9 502
(d) 56 943, 38 500, 38 099, 102 000, 56 899, 76 000
6. Matching Question
You do not have to look up any of this information; you will likely be able to match most of
these correctly using existing knowledge (and some common sense).
(a)
Average Canadian Mortgage in 2013
$500 000
(b)
Province of Nova Scotia's Debt in 2012
$2200
(c)
Approximate Cost of a New Family Sedan
$14 000 000 000
(d)
Approximate Cost of a 42 inch Flat Screen TV
$40 000
(e)
Overall Cost of 3 Year Phone Plan with Data Package
$245 000
(f)
Average Individual Income of Nova Scotian in 2010
$600 000 000 000
(g)
Canada's Federal Debt in 2012
$159 000 000
(h)
Cost of Purchasing a Tim Hortons' Franchise in 2012
$500
(i)
Estimated Cost of New Halifax Convention Center
$26 000
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7. Determine the number(s). Four examples have been provided.
Number
e.g. What whole number is before 23 120?
23 119
e.g. What whole number is between 1 456 and 1458?
1 457
e.g. What whole number is after 59999?
60000
e.g. What whole numbers are between 498 and 502?
(a)
What whole number is after 325?
(b)
What whole number is before 6 421?
(c)
What whole number is between 45 188 and 45 190?
(d)
What whole numbers are between 763 and 768?
(e)
What whole number is after 7 239?
(f)
What whole number is between 9 398 and 9400?
(g)
What whole number is before 82 600?
(h)
What whole number is between 62 985 and 62 987?
(i)
What whole numbers are between 69 997 and 70 000?
(j)
What whole number is before 120 000?
(k)
What whole number is before 7 300 000?
(l)
What whole numbers are between 27 029 and 27 032?
499, 500, 501
(m) What whole number is after 15 999?
(n)
What whole number is between 386 638 and 386 640?
(o)
What whole numbers are between 9 997 and 10 001?
(p)
What whole number is before 430 000 000?
(q)
What whole number is after 549 999?
(r)
What whole number is between 31 652 and 31 654?
(s)
What whole numbers are between 72 999 and 73 002?
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Math Facts: Addition and Multiplication
Calculators, especially graphing calculators, and calculator applications for smartphones or
tablets have revolutionized the mathematics one can access and learn. Teacher, instructors and
professors recognize that these devices have a place in today's classrooms. However, calculators
are not always to most efficient way to complete a mathematical task, and they don't replace the
thinking and problem solving necessary in mathematics.
In terms of efficiency, does it make sense to grab a calculator to complete simple calculations
like 7  4 , 8  3 , 9  6 , and 24  3 ? If you know the math facts, then these types of questions
can be completely mentally in a split second. In these cases, it takes longer to obtain the answer
using a calculator.
In terms of thinking and problem solving, consider the following tasks.




Find the least common multiple of 6 and 8. (Answer: 24)
Find the greatest common factor of 35 and 49. (Answer: 7)
Complete the prime factorization of the number 90. (Answer: 90  2  3  3  5 )
Find the two numbers that multiply to give 30 and add to give 17. (Answer: 2 and 15)
The calculator doesn't possess the mental flexibility (i.e. the thinking skills) necessary to
complete the questions above. However, you, armed with the math facts, are more than capable
of solving these types of questions. Taking the time to learn the math facts will ultimately be
beneficially to your learning. If you choose not to do so, you will likely reach an impasse in this
course or another.
Addition Facts
+
0
1
2
3
4
5
6
7
8
9
10
0
0
1
2
3
4
5
6
7
8
9
10
1
1
2
3
4
5
6
7
8
9
10
11
2
2
3
4
5
6
7
8
9
10
11
12
3
3
4
5
6
7
8
9
10
11
12
13
4
4
5
6
7
8
9
10
11
12
13
14
5
5
6
7
8
9
10
11
12
13
14
15
6
6
7
8
9
10
11
12
13
14
15
16
7
7
8
9
10
11
12
13
14
15
16
17
8
8
9
10
11
12
13
14
15
16
17
18
9
9
10
11
12
13
14
15
16
17
18
19
10
10
11
12
13
14
15
16
17
18
19
20
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On the previous page, you were given the addition facts table, which you are expected to know
for this course. Is it really necessary to memorize all 121 entries in this table? No; there are
ways to reduce the need for memorization. That is, there are strategies that can be used to help
us recall these facts. You do not need to know these strategies; we are only providing them for
learners who have had repeated difficulties recalling the addition facts.
Addition Fact Strategies (Optional)
1. Flip Flop
Notice that 6  8  14 and 8  6  14 . Similarly 3  9  12 and 9  3  12 . What we see is
that changing the order in which two numbers are added has no effect on the sum. Another
way of expressing this is a  b  b  a . So when we "flip flop" the numbers, the answer
remains the same. (This is referred to as the commutative property.) So why is this
important? It means that almost half of the 121 entries in our addition facts table can be
viewed as a duplication of another fact.
 If you know that 6  4  10 , then you can conclude that 4  6  10 .
 If you know that 9  7  16 , then you can conclude that 7  9  16 .
2. Doubles
When we add same numbers together, it is the same as doubling that number. For example,
3  3 is the same as doubling 3. Many people are familiar with the concept of doubling, so it
becomes an easy way to remember what you get when you add the same numbers together.
0  0  0 (double 0)
1  1  2 (double 1)
2  2  4 (double 2)
3  3  6 (double 3)
4  4  8 (double 4)
5  5  10 (double 5)
6  6  12 (double 6)
7  7  14 (double 7)
8  8  16 (double 8)
9  9  18 (double 9)
10  10  20 (double 10)
3. Nothing Changes
If you have some quantity and add nothing to it, you end up with the quantity you started
with (i.e. nothing changes). That is what happens when 0 is added to a number (e.g.
7  0  7 , 4  0  4 , 0  9  9 ). The "nothing changes" strategy allows one to quickly
understand what happens when 0 is added to a number.
4. Next Number
When you add 1 to a whole number, it means that you just find to next whole number.
1 2  3
1 3  4
1 4  5
and so on…
2 1  3
3 1  4
4 1  5
5. Next Even/Next Odd
When you add 2 to a whole number you just have to find the next even or odd number
depending on the number you start with. If you start with an even number and add 2, then
the sum is the next even number. If you start with an odd number and add 2, then the sum is
the next odd number.
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(Even)
24  6
42  6
(Odd)
25 7
52  7
26 8
2  8  10
62 8
8  2  10
27 9
2  9  11
729
9  2  11
and so on…
and so on….
6. Doubles Plus One
This strategy combines the Double strategy (#2) and the Next Number strategy (#4). The
Doubles Plus One strategy is used with questions like 2  3 , where the two numbers being
added together differ by 1. For this example, the learner should mentally change 2  3 into
2  2  1. By doing so, the learner just needs to double 2 and then go to the next number.
2  3  2  2 1
 4 1
5
4  5  4  4 1
 8 1
9
7  8  7  7 1
 14  1
 15
7. Add Ten, Then Subtract One
This strategy is used when adding 9 to a number (e.g. 9  6 ). This strategy relies on the fact
that most people are comfortable adding 10 to a number (e.g. 10  3  13 , 7  10  17 ). So if
we want to add 9 and 4 (i.e. 9  4 ), we will add 10 and 4, then compensate by subtracting 1
(i.e. 10  4  1).
9  4  10  4  1
 14  1
 13
9  6  10  6  1
 16  1
 15
9  7  10  7  1
 17  1
 16
8. Add Ten, Then Subtract Two
This strategy is very similar to the last strategy except it is used when adding 8 to a number
(e.g. 8  6 ). Again, this strategy relies on the fact that most people are comfortable adding
10 to a number (e.g. 10  3  13 , 7  10  17 ). So if we want to add 8 and 4 (i.e. 8  4 ), we
will add 10 and 4, then compensate by subtracting 2 (i.e. 10  4  2 ).
8  4  10  4  2
 14  2
 12
8  6  10  6  2
 16  2
 14
8  7  10  7  2
 17  2
 15
If you use all of these strategies, there are only six addition facts that you are forced to
memorize.
53  8
63 9
6  4  10
7  3  10
7  4  11
7  5  12
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Overview of the Addition Fact Strategies:
1. Flip Flop: The order when doing addition makes no difference 3  5  5  3  8
2. Doubles: Adding the same two numbers together is the same as doubling the number.
3. Nothing Changes: Nothing changes when you add 0.
4. Next Number: When adding 1, just count up to the next number.
5. Next Even/Next Odd: Used when adding 2.
6. Doubles Plus One: Used when adding 3.
7. Add Ten, Then Subtract One: Used when adding 9.
8. Add Ten, Then Subtract Two: Used when adding 8.
Multiplication Facts

0
1
2
3
4
5
6
7
8
9
10
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
2
3
4
5
6
7
8
9
10
2
0
2
4
6
8
10
12
14
16
18
20
3
0
3
6
9
12
15
18
21
24
27
30
4
0
4
8
12
16
20
24
28
32
36
40
5
0
5
10
15
20
25
30
35
40
45
50
6
0
6
12
18
24
30
36
42
48
54
60
7
0
7
14
21
28
35
42
49
56
63
70
8
0
8
16
24
32
40
48
56
64
72
80
9
0
9
18
27
36
45
54
63
72
81
90
10
0
10
20
30
40
50
60
70
80
90
100
Again, is it really necessary to memorize all 121 entries in this table? No; there are ways to
reduce the need for memorization. That is, there are strategies that can be used to help us recall
these facts. You do not need to know these strategies; we are only providing them for learners
who have had repeated difficulties recalling the multiplication facts.
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Multiplication Fact Strategies (Optional)
1. Flip Flop
Notice that 5  8  40 and 8  5  40 . Similarly 6  9  54 and 9  6  54 . What we see is
that changing the order in which two numbers are multiplied has no effect on the product.
Another way of expressing this is a  b  b  a . So when we "flip flop" the numbers, the
answer remains the same. (This is referred to as the commutative property.) So why is this
important? It means that almost half of the 121 entries in our multiplication facts table can
be viewed as a duplication of another fact.
 If you know that 6  4  24 , then you can conclude that 4  6  24 .
 If you know that 7  9  63 , then you can conclude that 9  7  63 .
2. Doubles
Multiplying a number by 2 is the same as doubling the number.
2  4  8 (double 4)
2  6  12 (double 6)
2  9  18 (double 9)
3. The Nine Pattern
Look at the multiplication facts for nine listed below. There are two patterns here we can
exploit.
9  9  81
8  9  72
7  9  63
6  9  54
5  9  45
4  9  36
Notice that the product’s tens digit is one less than the first factor.
e.g. 4  9  36
e.g. 7  9  63
e.g. 8  9  72
Also notice that the sum of the digits for any of these products is 9.
e.g. 4  9  36
3 6  9
3  9  27
2  9  18
e.g. 8  9  72
729
e.g. 9  9  81
8 1  9
4. The Five Chant
Most people remember that when multiplying a whole number by 5, the last digit in the
resulting product ends in 5 or 0 (e.g. 7  5  35 , e.g. 8  5  40 ). However, it's the five chant
that most people use to recall the multiplication facts for 5.
"five, ten, fifteen, twenty, twenty-five, thirty, thirty-five, forty, forty-five, fifty"
5. Snake Legs
People often confuse the rule for multiplying by 0 with the rule for adding 0. For the
question 5 0 or 0  5 , learners need to remember that these mean 5 sets of 0, or 0 sets of 5.
In either case, the answer is 0. The easiest way to remember this is to use the snake leg
strategy. How many legs does one snake have? (Answer: 0) How many legs do five
snakes? (Again the answer is 0). Therefore, we can conclude that 5  0  0 .
3 0  0
6 0  0
8 0  0
(3 snakes have 0 legs)
(6 snakes have 0 legs)
(8 snakes have 0 legs)
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6. No Change
The product of 1 and another number will be the other number. There is no change. For the
question 4  1 or 1 4 , learners should think of it as 4 sets of 1, or 1 set of 4. In either case,
the answer is 4. We started with 4 and ended with 4; there was no change.
1 7  7
1 4  4
1 9  9
7. Tic Tac Toe Threes
This technique allows one to quickly generate a 3 by 3 grid of the multiplication facts for 3.
Step 1: Draw a tic tac toe grid
Step 2: Starting in the lower left hand corner, moving up and then over to the next
column, fill in the numbers 1 to 9.
3
6
9
2
5
8
1
4
7
Step 3: Take all the numbers in the middle row and give them a tens digit of 1. Take all
the numbers in the bottom row and give them a tens digit of 2
3
6
9
12
15
18
21
24
27
If you look at the grid, you will notice that you have all the multiplication facts for 3.
31 = 3
3  2 =6
33 = 9
3  4 = 12
etc.
8. Double Double
Multiplying a number by 4 is the same as doubling the number and then doubling that new
answer. If you have 4  6 , then you double 6 and then double that answer.
4  6  22  6
 212
 24
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Double 6
Double 12
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4  3  2  2  3
4  4  2  2  4
4 7  2 2  7
 2  6
 2 8
 2 14 
 12
 16
 28
This strategy works nicely for questions like 4  4 , 4  6 , and 4  7 . It is, however, more
challenging for 4  8 because many people find it difficult to mentally double 16.
If you use all of these strategies, there are only six multiplication facts that you are forced to
memorize.
6  6  36
6  7  42
6  8  48
7  7  49
7  8  56
8  8  64
Overview of the Addition Fact Strategies:
1. Flip Flop: The order when doing multiplication makes no difference 3  5  5  3  15 .
2. Doubling: Multiplying a number by 2 is the same as double the number.
3. The Nine Patterns: Two patterns can be exploited to remember the multiplication facts for 9.
4. The Five Chant: "five, ten, fifteen, twenty, twenty-five,…"
5. Snake Legs: Used when multiplying by 0.
6. No Change: Used when multiplying by 1.
7. Tic Tac Toe Threes: Used when multiplying by 3
8. Double Double: Used when multiplying by 4.
Questions
You are not permitted to use a calculator, addition tables or multiplication tables to complete
these questions.
1. Fill in the blanks.
(a) 7  3  _____
(b) 5  1  _____
(c) 6  6  _____
(d) 0  8  _____
(e) 4  3  _____
(f) 1  8  _____
(g) 6  2  _____
(h) 8  8  _____
(i) 3  6  _____
(j) 5  9  _____
(k) 7  2  _____
(l) 5  6  _____
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(m) 9  1  _____
(n) 4  8  _____
(o) 2  0  _____
(p) 7  5  _____
(q) 9  4  _____
(r) 7  4  _____
(s) 2  9  _____
(t) 3  2  _____
(u) 6  7  _____
(v) 8  5  _____
(w) 5  4  _____
(x) 3  1  _____
(y) 4  4  _____
(z) 3  5  _____
2. Fill in the blanks.
(a) 4  2  _____
(b) 5  7  _____
(c) 1 8  _____
(d) 6  0  _____
(e) 3  9  _____
(f) 7  4  _____
(g) 4  3  _____
(h) 7  7  _____
(i) 0  9  _____
(j) 2  6  _____
(k) 5 1  _____
(l) 9  8  _____
(m) 6  8  _____
(n) 3  8  _____
(o) 6  9  _____
(p) 4  5  _____
(q) 3  7  _____
(r) 8  8  _____
(s) 8  2  _____
(t) 6  3  _____
(u) 5  6  _____
(v) 4  9  _____
(w) 9  7  _____
(x) 3  6  _____
(y) 9  5  _____
(z) 8  7  _____
3. Indicate whether the following phrase implies the operation of addition or multiplication.
(a) "5 is increased by 7"
operation: _______________________
(b) "8 plus 9"
operation: _______________________
(c) "4 times 5"
operation: _______________________
(d) "the sum of 3 and 5"
operation: _______________________
(e) "7 is increased by a factor of 3"
operation: _______________________
(f) "the product of 4 and 9"
operation: _______________________
(g) "combine 3 and 9"
operation: _______________________
(h) "triple the number 8"
operation: _______________________
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4. Find two numbers that:
(a) multiply to give 6, and
add to give 5.
(b)
Answer: ____ & ____
(d)
multiply to give 16,
and add to give 10.
Answer: ____ & ____
multiply to give 35,
and add to give 12.
(c)
Answer: ____ & ____
(e)
multiply to give 30,
and add to give 13.
multiply to give 3, and
add to give 4.
Answer: ____ & ____
(f)
Answer: ____ & ____
multiply to give 24,
and add to give 11.
Answer: ____ & ____
5. KenKen puzzles were invented in 2004 by Japanese math teacher Tetsuya Miyamoto. The
goal is to fill a grid with the indicated numbers such that:
 no number is repeated in the same row or column, and
 the numbers in the cages produce the target number using the indicated operation
(e.g. 6  : find the numbers that multiplied give you 6)
(e.g. 5+: find the numbers that added give you 5)
Complete the following KenKen Puzzles using the indicated numbers. The first four puzzles
are 3 by 3 puzzles because they are comprised of three rows and three columns. The last two
are 4 by 4 puzzles.
(a) 2, 3, 4 Puzzle
6
6+
5+
(b) 4, 5, 6 Puzzle
24 
16+
11+
10+
20 
(c) 5, 6, 7 Puzzle
(d) 7, 8, 9 Puzzle
30 
18+
56 
42 
63 
35 
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24+
72 
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C. D. Pilmer
(e) 2, 3, 4, 5 Puzzle
(f) 6, 7, 8, 9 Puzzle
12 
8+
8
48 
10 
6+
16+
9+
15 
6
63 
42 
8
25+
13+
54 
6. In the appendix of this resource, you will find Addition and Multiplication Connect Four
Games. Play two rounds of an Addition Game and two rounds of a Multiplication Game
with a classmate, instructor and/or family member. Record in the chart below whom you
played and who won.
Opponent
Winner
Addition Game #1
Addition Game #2
Multiplication Game #1
Multiplication Game #2
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Fact Families
In the last section we looked at the addition facts and multiplication facts. That leaves us with
the subtraction and division facts. Are these another two sets of facts that we have to memorize?
No; fact families allow us to connect our addition facts to the subtraction facts, and similarly, the
multiplication facts to the division facts.
 For example, if you know 7  3  10 , then you also know that:
3  7  10
10  7  3
10  3  7
These four facts, two addition and two subtraction facts, are referred to as a fact family.
 For example, if you know 6  5  30 , then you also know that:
5  6  30
30  5  6
30  6  5
These four facts, two multiplication and two division facts, are referred to as a fact family.
Questions
You are not permitted to use a calculator, addition tables or multiplication tables to complete
these questions.
1. Write the three other members of the corresponding fact family for each.
(a) 9  5  14
(b) 4  7  28
2. Complete each of the following operations (i.e. fill in the blank) and then write the three
other members of corresponding fact family.
(a) 4  8  _____
(b) 7  9  _____
(c) 6  7  _____
(d) 8  3  _____
(e) 15  6  _____
(f) 10  2  _____
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3. Fill in the blanks.
(a) 9  1  _____
(b) 7  2  _____
(c) 6  0  _____
(d) 5  4  _____
(e) 8  3  _____
(f) 11  4  _____
(g) 7  7  _____
(h) 14  7  _____
(i) 15  5  _____
(j) 11  6  _____
(k) 10  3  _____
(l) 8  4  _____
(m) 12  4  _____
(n) 12  9  _____
(o) 13  6  _____
(p) 9  9  _____
(q) 8  0  _____
(r) 18  9  _____
(s) 6  1  _____
(t) 11  3  _____
(u) 7  4  _____
(v) 7  5  _____
(w) 14  6  _____
(x) 12  5  _____
(y) 10  7  _____
(z) 11  9  _____
4. Fill in the blanks.
(a) 14  2  _____
(b) 18  6  _____
(c) 30  5  _____
(d) 8 1  _____
(e) 7  7  _____
(f) 10  2  _____
(g) 16  8  _____
(h) 30  6  _____
(i) 27  9  _____
(j) 12  3  _____
(k) 28  4  _____
(l) 32  8  _____
(m) 45  9  _____
(n) 81  9  _____
(o) 6  3  _____
(p) 40  5  _____
(q) 42  7  _____
(r) 48  6  _____
(s) 20  4  _____
(t) 36  6  _____
(u) 63  9  _____
(v) 54  6  _____
(w) 21  7  _____
(x) 9  9  _____
(y) 72  9  _____
(z) 24  4  _____
5. Fill in the blanks.
(a) 9  4  _____
(b) 63  7  _____
(c) 3  5  _____
(d) 7  7  _____
(e) 8  3  _____
(f) 12  5  _____
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(g) 2  7  _____
(h) 36  4  _____
(i) 2  8  _____
(j) 35  5  _____
(k) 8  5  _____
(l) 15  6  _____
(m) 3  3  _____
(n) 54  9  _____
(o) 14  8  _____
(p) 4  6  _____
(q) 4  8  _____
(r) 3  4  _____
(s) 6  5  _____
(t) 9  5  _____
(u) 6 1  _____
(v) 3  0  _____
(w) 6  6  _____
(x) 5  5  _____
6. Find two numbers that:
(a) multiply to give 15,
and add to give 8.
(b)
Answer: ____ & ____
(d)
multiply to give 12,
and add to give 13.
multiply to give 16,
and add to give 8.
(e)
multiply to give 12,
and add to give 8.
(h)
(k)
multiply to give 9, and
differ by 8.
(b)
multiply to give 24,
and differ by 5.
Answer: ____ & ____
NSSAL
©2013
(i)
multiply to give 10,
and add to give 11.
multiply to give 10,
and differ by 3.
(e)
multiply to give 16,
and differ by 0.
(l)
multiply to give 0, and
differ by 2.
Answer: ____ & ____
19
multiply to give 25,
and add to give 10.
Answer: ____ & ____
(c)
multiply to give 12,
and differ by 1.
Answer: ____ & ____
(f)
Answer: ____ & ____
(h)
multiply to give 36,
and add to give 13.
Answer: ____ & ____
Answer: ____ & ____
Answer: ____ & ____
(g)
multiply to give 24,
and add to give 10.
multiply to give 40,
and add to give 13.
Answer: ____ & ____
Answer: ____ & ____
Answer: ____ & ____
(d)
(f)
Answer: ____ & ____
Answer: ____ & ____
7. Find two numbers that:
(a) multiply to give 35,
and differ by 2.
multiply to give 18,
and add to give 9.
multiply to give 28,
and add to give 11.
Answer: ____ & ____
Answer: ____ & ____
Answer: ____ & ____
(j)
(c)
Answer: ____ & ____
Answer: ____ & ____
(g)
multiply to give 20,
and add to give 12.
multiply to give 18,
and differ by 7.
Answer: ____ & ____
(i)
multiply to give 20,
and differ by 8.
Answer: ____ & ____
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C. D. Pilmer
8. For each number, express it as:
 At least two number sentences involving addition,
 At least two number sentences involving subtraction,
 At least two number sentences involving multiplication, and
 At least two number sentences involving division.
(Please note that answers will vary from learner to learner.)
Example: Number 10
 Two number sentences involving addition: 6  4  10 , 5  5  10
 Two number sentences involving subtraction: 12  2  10 , 29  19  10
 Two number sentences involving multiplication: 1 10  10 , 2  5  10
 Two number sentences involving division: 30  3  10 , 80  8  10
(a) Number 6
(b) Number 8
9. Use the numbers in the charts below to complete the following RAD puzzles.
(a)
(b)


6
=
=

-


13
-
+
=
2

-

=


3
=
30
=

=
=
=
=
10
=
=
=
24
+
1
6
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©2013
1
7
5
32
=
=
2
8
=
3
9
=

=
=
=
4
=
=

=
+

3 3 3 5
10 14 15 31
0
6
20
3
=
3
-

=

=
40

+
=
1
8
=
6

-

=
=
1
8
=
1
8

+
1
8
2
9

=
-
2
9
3 3 4
12 14 18
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10. Indicate whether the following phrase implies the operation of addition, subtraction,
multiplication, or division.
(a) "how many times does 6 go into 18?"
operation: _______________________
(b) "take 9 and make it 5 times larger"
operation: _______________________
(c) "12 is decreased by 7"
operation: _______________________
(d) "the total of 6 and 12"
operation: _______________________
(e) "5 is removed from 9"
operation: _______________________
(f) "break 12 into 4 equal parts"
operation: _______________________
11. In the appendix of this resource, you will find Subtraction and Division Connect Four
Games. Play two rounds of a Subtraction Game and two rounds of the Division Game with a
classmate, instructor and/or family member. Record in the chart below whom you played
and who won.
Opponent
Winner
Subtraction Game #1
Subtraction Game #2
Division Game #1
Division Game #2
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Order of Operations
Exponents
Before we talk about order of operations, we need to talk about exponents, specifically the
exponents of 2 and 3.
When a number is raised to an exponent of 2 (i.e. the number is being squared), you are
multiplying the number by itself.
e.g. 32  3  3  9
e.g. 62  6  6  36
e.g. 72  7  7  49
When a number is raised to an exponent of 3 (i.e. the number is being cubed), you are
multiplying the number by itself three times.
e.g. 23  2  2  2  8
e.g. 33  3  3  3  27
e.g. 53  5  5  5  125
Order of Operations
Below, the same question was done by four different learners. The problem was that everyone
ended up with different answers.
Kimi's Answer:
Ryan's Answer
Paulette's Answer
Ajay's Answer
4  5  2
4  5  2
4  5  2
4  5  2
2
2
2
2
20  22
20  4
16
4  32
49
36
45  4
4 1
4
122
144
Kimi started with the
multiplication,
followed by the
squaring, and then
did the subtraction.
Ryan started with the
operation in the
brackets, followed by
the squaring, and
then did the
multiplication.
Paulette started with
the squaring,
followed by the
subtraction, and then
did the
multiplication.
Ajay started with the
operation in the
brackets, followed by
the multiplication,
and then did the
squaring.
4  32
All of these learners started with the same question, but ended up with very different answers
based on the order they chose to do the operations. Only one of the learners is correct. Do you
know which one? The correct answer is 36. Ryan did the question correctly because he knew
the order of operations, the rules used to clarify which mathematical operations are done first in a
mathematical expression. The proper order can be remembered using the acronym BEDMAS.
B
E
DM
AS
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- brackets first
- then exponents (e.g. squaring, cubing)
- followed by division and multiplication in the order they appear (i.e. from left to right)
- followed by addition and subtraction in the order they appear (i.e. from left to right)
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Example 1
Evaluate each of the following.
(a) 5  7  3
(b) 10  3  8  4
3
3
(d) 29  3  24  6
(e)  7  5  3  7
Answers:
(a) 5  7  3
5  21
26
(c) 5  32  2
2
(f) 2  7  1  1  33
The mathematical expression 5  7  3 only involves the
operations of addition and multiplication. According to
BEDMAS, we do multiplication before addition.
(b)
10  3  8  4
10  3  2
72
9
The mathematical expression 10  3  8  4 involves the
operations of subtraction, addition and multiplication.
According to BEDMAS, multiplication is done before
subtraction or addition. Once this is done we have to decide
between the subtraction and addition. When these
operations occur in the same question, we always work
from left to right. That means we will do the subtraction
before the addition.
(c)
5  32  2
5  9 2
5  18
23
The mathematical expression 5  32  2 involves the
operations of addition, squaring, and multiplication.
According to BEDMAS, we would do the squaring (i.e.
exponents) first, followed by multiplication, and finally the
addition.
(d)
29  33  24  6
29  27  24  6
29  27  4
24
6
The mathematical expression 29  33  24  6 involves the
operations of subtraction, cubing, addition, and division.
According to BEDMAS, we would do the cubing (i.e.
exponents) first. Next we would do the division. Once this
is done we have to decide between the subtraction and
addition. When these operations occur in the same
question, we always work from left to right. That means we
will do the subtraction before the addition.
(e)
 7  5
With the mathematical expression  7  5  3  7 we have
subtraction embedded within a set of brackets, cubing,
addition, and multiplication. According to BEDMAS, we
start with the operations in the brackets. This will be
followed by the cubing (i.e. exponents). We will then do
the multiplication, and then finish up with the addition.
3
 3 7
23  3  7
8  3 7
8  21
29
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(f)
2  7  1  1  33
2
2  7  22  33
2  7  4  27
2  28  27
30  27
3
With the mathematical expression 2  7  1  1  33 we
will start with the addition that is embedded within a set of
brackets. Next we will work with the exponents (i.e. the
squaring and the cubing). Following this we do the
multiplication. That leaves us with the addition and
multiplication. When these operations occur in the same
question, we always work from left to right. That means we
will do the addition before the subtraction.
2
Questions
1. Evaluate each of the following. Show your work and do not use a calculator or math fact
tables.
(a) 10  3  2
(b) 18  10  2
(c) 9  2  3
(d) 12  6  5
(e) 3  2  6  1
(f) 8  12  4  3
(g) 12  4  3  9
(h) 7  3 16  2
(i) 5  4  2  7  2
(j) 24  42  8
(k) 25  23  3
(l) 4  2  32
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(m) 2  32  5  2
(n) 10  4  32  2
(o) 5  23  10  22
(p) 62  4  2  52
(q) 10   5  1
(r)
(s) 10  25   7  2 
(t)
(v) 3   7  5  5
(w) 7   3  3  9
2
2
 7  20  33
2
 3  6  23
(u) 32   5  3
3
(x) 5   7  4   1  1
3
2. Create your own expression that when correctly evaluated produces an answer of 5. Your
expression must include brackets, exponents, and at least two different operations (i.e.
addition, subtraction, multiplication, division).
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Multiples of Ten, One Hundred, and One Thousand
Numbers that are multiples of ten are divisible by 10 (i.e. ten divides evenly into them). The
numbers are easy to recognize because they end with a zero.
Multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130,…
Numbers that are multiples of one hundred are divisible by 100 (i.e. one hundred divides
evenly into them). The numbers are easy to recognize because they end with two zeros.
Multiples of 100: 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100, 1200,…
Numbers that are multiples of one thousand are divisible by 1000 (i.e. one thousand divides
evenly into them). The numbers are easy to recognize because they end with three zeros.
Multiples of 1000: 1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10000, 11000,…
So obviously, a multiple of 1000, like 7000, is also a multiple of 100 and 10. Similarly, a
multiple of 100, like 500, is also a multiple of 10.
In this section, we will mentally add, subtract, multiply and divide whole numbers that are
multiples of ten, one hundred, and one thousand.
e.g. 40  50  90
e.g. 600  200  400
e.g. 40  600  12 000
e.g. 150  50  3
Addition
We will only be adding "like multiples" in this section. That is, we will be adding multiples of
ten with multiples of ten, adding multiples of one hundred with multiples of one hundred, and so
on.
e.g. 40  30
If 4  3  7 , then 40  30 must be equal to 70.
e.g. 80  50
If 8  5  13 , then 80  50 must be equal to 130.
e.g. 600  200
If 6  2  8 , then 600  200 must be equal to 800.
e.g. 400  700
If 4  7  11 , then 400  700 must be equal to 1 100.
e.g. 1 000  8 000
If 1  8  9 , then 1 000  8 000 must be equal to 9 000.
e.g. 9 000  5 000
If 9  5  14 , then 9 000  5 000 must be equal to 14 000.
Subtraction
We will only be subtracting "like multiples" in this section. That is, we will be subtracting
multiples of ten from multiples of ten, subtracting multiples of one hundred from multiples of
one hundred, and so on.
e.g. 90  70
If 9  7  2 , then 90  70 must be equal to 20.
e.g. 140  80
If 14  8  6 , then 140  80 must be equal to 60.
e.g. 600  200
If 6  2  4 , then 600  200 must be equal to 400.
e.g. 1 500  700
If 15  7  8 , then 1 500  700 must be equal to 800.
e.g. 8 000  5 000
If 8  5  3 , then 8 000  5 000 must be equal to 3 000.
e.g. 14 000  5 000
If 14  5  9 , then 14 000  5 000 must be equal to 9 000.
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Multiplication
Many of you are likely familiar with the three step process for completing the type of
multiplication shown below. You are not expected to show this process; rather complete the
work in your head.
e.g. Evaluate 80  6 .
Answer:
To work out 80  6 , it is a three step process.
(i) Omitting the zeros, multiply the two numbers ( 8  6  48 )
(ii) Next count the number of zeros in the original question (There is 1; one from the
number 80 and none from the 6)
(iii)Take the product from step (a) and tack on the number of zeros from step (b).
Therefore: 80  6 = 480
e.g. Evaluate 90  50 .
Answer:
To work out 90  50 , it is a three step process
(i) Omitting the zeros, multiply the two numbers ( 9  5  45 )
(ii) Next count the number of zeros in the original question (There are 2; one from the
number 90 and one from the 50)
(iii)Take the product from step (a) and tack on the number of zeros from step (b).
Therefore: 90  50 = 4 500
e.g. Evaluate 400  70 .
Answer:
To work out 400  70 , it is a three step process
(i) Omitting the zeros, multiply the two numbers ( 4  7  28 )
(ii) Next count the number of zeros in the original question (There are 3; two from the
number 400 and one from the 70)
(iii)Take the product from step (a) and tack on the number of zeros from step (b).
Therefore: 400  70 = 28 000
Division
With these questions, we are going to show you two approaches; you choose the approach that
you prefer. Like multiplication, we expect that all this work will be done mentally, as opposed to
writing everything out using pencil and paper.
e.g. 160  4
Answer:
Method 1: Use Related Multiplication Fact
Based on our work with fact families in the previous section, we know that
operations of multiplication and division are connected to each other. So when we
are asked, "What is 160 divided by 4?" (i.e. 160  4 ), we know that this can be
restated as, "What multiplied by 4, gives us 160?" (i.e. ?  4  160 ). Many realize
that 40 multiplied by 4 gives 160. Therefore 160  4  40
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Method 2: Number of Sets
Knowing that 16  4  4 can help one answer 160  4 . The question that remains is
whether 160  4 is equal to 4, 40, or 400? The way to decide is to determine how
many sets of 4 are in 160. Logically, there are 40 sets of 4 in 160. Therefore
160  4  40 .
e.g. 560  70
Answer:
Method 1: Use Related Multiplication Fact
"What is 560 divided by 70?" (i.e. 560  70 ), can be restated as, "What multiplied
by 70, gives us 560?" (i.e. ?  70  560 ). Many realize that 8 multiplied by 70
gives 560. Therefore 560  70  8
Method 2: Number of Sets
Knowing that 56  7  8 can help one answer 560  70 . The question that remains
is whether 560  70 is equal to 8, 80, or 800? The way to decide is to determine
how many sets of 70 are in 560. Logically, there are 8 sets of 70 in 560. Therefore
560  70  8 .
e.g. 4 500  5
Answer:
Method 1: Use Related Multiplication Fact
"What is 4500 divided by 5?" (i.e. 4 500  5 ), can be restated as, "What multiplied
by 5, gives us 4500?" (i.e. ?  5  4 500 ). Many realize that 900 multiplied by 5
gives 4 500. Therefore 4 500  5  900
Method 2: Number of Sets
Knowing that 45  5  9 can help one answer 4 500  5 . The question that remains
is whether 4 500  5 is equal to 9, 90, or 900? The way to decide is to determine
how many sets of 5 are in 4 500. Logically, there are 900 sets of 5 in 4 500.
Therefore 4 500  5  900 .
e.g. 2 800  40
Answer:
Method 1: Use Related Multiplication Fact
The question can restated as, "What multiplied by 40, gives us 2 800?"
(i.e. ?  40  2 800 ). Many realize that 70 multiplied by 40 gives 2 800.
Therefore 2 800  40  70
Method 2: Number of Sets
Knowing that 28  4  7 can help one answer 2 800  40 . The question that
remains is whether 2 800  40 is equal to 7, 70, or 700? The way to decide is to
determine how many sets of 400 are in 2 800. Logically, there are 60 sets of 40 in
2 800. Therefore 2 800  40  70 .
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e.g. 2 700  900
Answer:
Method 1: Use Related Multiplication Fact
The question can be restated as, "What multiplied by 900, gives us 2 700?" (i.e.
?  900  2 700 ). Many realize that 3 multiplied by 900 gives 2 700. Therefore
2 700  900  3
Method 2: Number of Sets
Knowing that 27  9  3 can help one answer 2 700  900 . The question that
remains is whether 2 700  900 is equal to 3, 30, or 300? The way to decide is to
determine how many sets of 900 are in 2 700. Logically, there are 3 sets of 900 in
2 700. Therefore 2700  900  3 .
Questions
You are not permitted to use a calculator, addition tables or multiplication tables to complete
these questions.
1. (a) 20  60  _______
(b) 500  400  _______
(c) 60  60  _______
(d) 5000  3000  _______ (e) 600  300  _______
(f) 1000  8000  _______
(g) 90  60  _______
(h) 700  400  _______
(i) 40  40  _______
(j) 800  600  _______
(k) 5000  6000  _______ (l) 900  700  _______
2. (a) 70  40  _______
(b) 500  100  _______
(c) 6 000  6 000  _______
(d) 800  200  _______
(e) 7 000  3 000  _______ (f) 90  70  _______
(g) 1 200  500  _______
(h) 170  90  _______
(i) 13 000  4 000  _______
(j) 150  80  _______
(k) 1 100  300  _______
(l) 14 000  7 000  _______
(b) 30  50  _______
(c) 2000  8  _______
(d) 90  400  _______
(e) 60  80  _______
(f) 50  5  _______
(g) 600  40  _______
(h) 90  90  _______
(i) 8  300  _______
(j) 5  70  _______
(k) 6  7000  _______
(l) 80 100  _______
3. (a) 7  30  _______
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4. (a) 320  8  _______
(b) 280  70  _______
(c) 420  60  _______
(d) 4 500  50  _______
(e) 5 400  600  _______
(f) 140  2  _______
(g) 1 500  5  _______
(h) 480  60  _______
(i) 1 800  900  _______
(j) 6400  80  _______
(k) 490  7  _______
(l) 3500  7  _______
(m) 7 200  9  _______
(n) 160  80  _______
(o) 1 200  20  _______
(p) 210  3  _______
(q) 2 500  500  _______
(r) 3600  6  _______
5. (a) 70  20  _______
(b) 400  8  _______
(c) 1 400  900  _______
(d) 80  30  _______
(e) 90  80  _______
(f) 3  90  _______
(g) 4200  7  _______
(h) 800  500  _______
(i) 12 000  7 000  _______
(j) 40  40  _______
(k) 2400  800  _______
(l) 8 000  6 000  _______
(m) 700  600  _______
(n) 800  400  _______
(o) 2800  40  _______
(p) 70  10  _______
(q) 700  80  _______
(r) 130  90  _______
(s) 4 200  60  _______
(t) 40  80  _______
(u) 270  30  _______
6. Find two digit numbers, which are also multiples of ten, that:
(a) multiply to give 800,
(b) multiply to give 1200, (c)
and add to give 60.
and add to give 70.
Answer:
_______ & _______
(d)
multiply to give 8100,
and add to give 180.
Answer:
_______ & _______
(e)
Answer:
_______ & _______
(g)
multiply to give 1800,
and differ by 30.
Answer:
_______ & _______
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multiply to give 5400,
and add to give 150.
Answer:
_______ & _______
(f)
Answer:
_______ & _______
(h)
multiply to give 4200,
and differ by 10.
Answer:
_______ & _______
30
multiply to give 3500,
and add to give 120.
multiply to give 1600,
and differ by 60.
Answer:
_______ & _______
(i)
multiply to give 3600,
and differ by 50.
Answer:
_______ & _______
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7. With KenKen puzzles, the goal is to fill a grid with the indicated numbers such that:
 no number is repeated in the same row or column, and
 the numbers in the cages produce the target number using the indicated operation
(e.g. 600  : find the numbers that multiplied give you 600)
(e.g. 50+: find the numbers that added give you 50)
Complete the following KenKen Puzzles using the indicated numbers.
(a) 10, 20, 30 Puzzle
600 
(b) 20, 30, 40 Puzzle
300 
50+
50+
70+
1200 
800 
40+
20
(c) 30, 40, 50 Puzzle
1500 
(d) 40, 50, 60 Puzzle
120+
90+
150+
1200 
3000 
(e) 50, 60, 70 Puzzle
4200 
110+
2400 
90+
(f) 70, 80, 90 Puzzle
3000 
170+
6300 
190+
70
5600 
7200 
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8. Use the numbers in the charts below to complete the following RAD puzzles.
(a)
3


=
-
=
40
+
-
+
-
=
=
=
27
=
=

=
3
=
=
60
=
+
=
=
20
=
240

+
0
30

-
2
30
=
4
30
5
30


=
6
40

9
50
20
60
21
90
=
200
+
24
180
(b)

-
=
+
-


=
=
=
20
=
30
=

=
1200
=
2

=
=
=
=
=
+

+
-

30

0
40
2
60
4
70
=
300
=
5
100
7
200
10
600

10
900
20
1200
40
1400
9. There are flashcards in the appendix of this resource. Over the next few days, regularly use
these cards and see how fast you can answer 40 randomly selected cards. Record your best
three times.
_________
_________
_________
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Divisibility and Prime
Divisibility:
When one number can be divided by another and the result is an exact whole
number (i.e. no remainder). For example, 18 is divisible by 3, because 18  3  6
exactly. However, 11 is not divisible by 5 because 11  5  2 with a remainder
of 1.
Divisibility Rules:

Numbers that are divisible by 2 are even numbers.
e.g. 6, 34, 798, 2050, and 3942 are all divisible by 2

If the sum of the digits of a multi-digit number produces a number that is divisible by three,
then the original multi-digit number is divisible by 3.
e.g. 72 is divisible by 3 because 7 + 2 = 9 and 9 is divisible by 3.
e.g. 561 is divisible by 3 because 5 + 6 + 1 = 12 and 12 is divisible by 3.

If the ones digit is a 0 or a 5, then the number is divisible by 5.
e.g. 75, 120, 375, and 2960 are all divisible by 5.

If a number is divisible by 2 (i.e. an even number) and divisible by 3 (i.e. sum of the digits
is a number divisible by 3), then the number is divisible by 6.
e.g. 48 is divisible by both 2 and 3, and therefore divisible by 6.
e.g. 132 is divisible by both 2 and 3, and therefore divisible by 6.

If a number is only divisible by itself and 1, then the number is prime.
e.g. 2, 3, 17, 41, and 89 are all prime numbers
Questions
1. Beside each number you will find a box corresponding to a number that might divide evenly
into the original number. Check off the appropriate boxes to identify whether the original
number is divisible by 2, 3, 5 and/or 6, or a prime number (P). Do not use a calculator.
e.g.
78
(a)
2
3


5
6
P
2

e.g.
345
22
(b)
55
(c)
13
(d)
72
(e)
75
(f)
29
(g)
90
(h)
63
(i)
61
(j)
86
(k)
261
(l)
440
(m)
428
(n)
546
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5


6
P
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2
3
5
6
P
2
(o)
417
(p)
948
(q)
510
(r)
179
(s)
8001
(t)
4315
(u)
3218
(v)
5202
(w)
2870
(x)
1320
(y)
2853
(z)
7302
3
5
6
P
2. For this question, do not use any of the numbers that you encountered in question 1.
(a) Create a three digit number that is divisible by 2, but not divisible by 3, 5, or 6. _______
(b) Create a three digit number that is divisible by 3, but not divisible by 2, 5, or 6. _______
(c) Create a three digit number that is divisible by 5, but not divisible by 2, 3, or 6. _______
(d) Create a three digit number that is divisible by 2 and 5, but not divisible by 3.
_______
(e) Create a three digit number that is divisible by 3 and 5, but not divisible by 2.
_______
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Estimating
In the sections following this one, we will learning how to add subtract, multiply, and divide
multi-digit numbers. When learners do these types, they sometimes do not take the time to judge
the reasonableness of their final answer. Often they are just pleased to get through the question
and do not spend additional time to know if their answer is "in-the-ballpark." In reality, it only
takes a few seconds to determine whether your answer is reasonable; it all comes down to using
estimation skills.
Another issue is that many individuals have become so dependent on calculators that they are
unaware when the device outputs an unreasonable answer, often because the user has pressed the
wrong key or omitted a decimal point or digit. Using your estimation skills can prevent one from
blindly accepting an unreasonable output from a calculator.
Example 1
Hannah completed the question 118  42 and obtained an answer of 4956. Is her answer
reasonable?
Answer:
Please note that this question is not asking whether Hannah's answer is correct; rather, it is
asking whether it is reasonable. If you were checking the answer, you would work it out in
full using paper and pencil, or check it with a calculator. However, to judge the
reasonableness of Hannah's answer, we can use a variety of estimation strategies. We have
shown several acceptable ways of handling this question below.
Estimate 1:
120  40  4800
Estimate 2:
100  50  5000
Estimate 3:
110  40  4400
110  50  5500
Between 4400 and 5500
Change 118 to 120, and change 42 to 40. By increasing one
number slightly and decreasing the other number slightly, the
estimate should be fairly close to the actual answer. This
technique is only useful if you can mentally multiply 12 by 4.
Change 118 to 100 and change 42 up to 50. We decreased one
number significantly and increased the other number significantly,
but the estimate is still fairly close to the actual answer.
Change 118 to 110 and 42 to 40, and then 50. In this case, they
calculate two estimates; one whose product is likely under the
actual answer, and one whose product is likely over. They
concluded that the real answer should be between 4400 and 5500.
Regardless of the estimation strategy used, all three techniques illustrate that Hannah's
answer of 4956 is reasonable.
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Example 2
Samir completed the question 356  248  109 and obtained an answer of 893. Is his answer
reasonable?
Answer:
Again, we have shown a variety of acceptable estimation strategies that can be used to handle
this question.
Estimate 1:
350  250  110  710
Estimate 2:
400  200  100  700
Estimate 3:
300  200  100  600
400  300  100  800
Between 600 and 800
Change 356 to 350, change 248 to 250, and change 109 to 110.
This technique is only useful if you are comfortable mentally
adding 350 and 250. This estimate should be fairly close to the
actual answer.
Change 356 to 400, change 248 to 200, and change 109 to 100.
The first number, 356, was increased by almost 50, and the second
number, 248, was decreased by almost 50. Increasing one number
significantly and decreasing the second number by roughly the
same amount will not greatly affect the sum. This estimate should
be fairly close to the actual answer.
In this case, they calculate two estimates; one whose sum is likely
under the actual answer, and one whose sum is likely over. They
concluded that the real answer should be between 600 and 800.
Regardless of the estimation strategy used, all three techniques illustrate that Samir's answer
of 893 is unreasonable.
Questions
Do not use a calculator to complete any of these questions.
1. In each case, indicate whether the answer is reasonable or unreasonable. Show your work in
the space provided on the right (i.e. show the estimation technique that you employed).
(a) 1852 1179  3031
Reasonable
Unreasonable
(b) 61 89  5429
Reasonable
Unreasonable
(c) 1260  4  315
Reasonable
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Unreasonable
36
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C. D. Pilmer
(d) 488  307  685
Reasonable
Unreasonable
(e) 919  228  691
Reasonable
Unreasonable
(f) 395  79  50
Reasonable
Unreasonable
(g) 201 72  1512
Reasonable
Unreasonable
(h) 648  553  95
Reasonable
Unreasonable
2. In each case, find in the missing digit.
(a) If __845 6 is about 300, then what number should be filled in to
replace the missing digit?
______
(b) If __2 81 is close to 3400, then what number should be filled in to
replace the missing digit?
______
(c) If __49 - 754 is about 200, then what number should be filled in to
replace the missing digit?
______
(d) If 3945 + __078 is close to 6000, then what number should be filled in to
replace the missing digit?
______
(e) If __95 5 is about 80, then what number should be filled in to
replace the missing digit?
______
(f) If __93 + 389 is almost 900, then what number should be filled in to
replace the missing digit?
______
(g) If 71  __9 is close to 6300, then what number should be filled in to
replace the missing digit?
______
NSSAL
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C. D. Pilmer
Adding Multi-Digit Numbers
To add multi-digit whole numbers, start by stacking the numbers vertically such that
corresponding place values line up (e.g. units with units, tens with tens) and add from right to
left. If the sum in any corresponding place value is 10 or greater, we regroup (i.e. carry the
excess to the next larger place value).
e.g. 324 + 45
Answer:
Stack the numbers vertically (i.e. one on top of another) such corresponding place values
line up.
Add the Units
Add the Tens
Add the Hundreds

3 2 4

4 5

3 2 4

4 5

3 2 4

4 5
9
6 9
3 6 9
4 units plus 5 units is 9
units.
2 tens plus 4 tens is 6
tens.
3 hundreds plus 0
hundreds is 3 hundreds.
Does this answer of 369 look reasonable? The easiest way to check is to round the original
numbers to values that we can mentally add. We can round 324 to 320, and round 45 to 50.
When 320 is added to 50, we obtain 370, which is very close to the original answer of 369.
Our answer looks reasonable.
e.g. 158 + 265
Answer:
Add the Units
Add the Hundreds


1 5 8
 2 6 5
1 5 8
 2 6 5
1 5 8
 2 6 5
3
2 3
4 2 3
1
8 units plus 5 units is 13
units. Regroup the 13 to
1 ten and 3 units. Write
the 3 in the units place
and carry the 1 to the
next place value (tens).
NSSAL
©2013
Add the Tens

1
1
1
1
1 ten plus 5 tens plus 6
1 hundred plus 1
tens is 12 tens. Regroup hundred plus 2 hundreds
the 12 to 1 hundred and is 4 hundreds
2 tens. Write the 2 in
the tens place and carry
the 1 to the next place
value (hundreds).
38
Draft
C. D. Pilmer
e.g. 451 + 75 + 192
Answer:
Add the Units
Add the Tens

4 5 1
7 5
 1 9 2

4 5 1
7 5
 1 9 2
8
Add the Hundreds

2
2
4 5 1
7 5
 1 9 2
1 8
7 1 8
1 unit plus 5 units plus 2 5 tens plus 7 tens plus 9
units is 8 units.
tens is 21 tens. Regroup
the 21 to 2 hundreds and
1 ten. Write the 1 in the
tens place and carry the
2 to the next place value
(hundreds).
2 hundreds plus 4
hundreds plus 0
hundreds plus 1 hundred
is 7 hundreds.
Does this answer of 718 look reasonable? The easiest way to check is to round the original
numbers to values that we can mentally add. We can round 451 to 450, round 75 to 100,
and round 192 to 200. When we add 450, 100, and 200, we obtain 750. This estimate is
higher than the original answer of 718, but this is to be expected because we rounded two of
the numbers up significantly. Regardless of this, the answer of 718 seems reasonable.
e.g. 926 + 437
e.g. 2 943 + 4 864
Answer:
Answer:
1
1
1
9 2 6
 4 3 7
2 9 4 3
 4 8 6 4
1 3 6 3
7 8 0 7
Note:
Is the technique (i.e. algorithm), which we have been teaching, the only way to add multi-digit
whole numbers? No; however, it is the most commonly taught technique. Examples of other
techniques are shown below.
 Suppose a learner was required to add 36 and 57. They could say 30 + 50 = 80,
6 + 7 = 13, and 80 + 13 = 93. Therefore 36 + 57 = 93. These individuals did not use the
algorithm we have shown but obtained the correct solution using a perfectly valid method.
 Suppose a learner was required to add 88 and 54. They could increase the first number by
2 and decrease the second number by 2. That changes the question from 88 + 54 to
90 + 52, a question that many can solve in their head. 88 + 54 = 90 + 52 = 142
NSSAL
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C. D. Pilmer
Questions
Complete the following operations. Do not use a calculator or an addition facts table.
1.
2.
(a)
6 3
 2 5
(b)
3 8
 4 6
(c)
7 5
 6 1
(d)
2 4 6
 5 1 3
(e)
3 2 5
 3 4 7
(f)
5 0 8
 9 7 7
(g)
4 2 3
3 5 1

2 6
(h)
2 5 3
4 7
 6 3 4
(i)
3 6 2
7 8 4
 2 9 1
(j)
3 7 5 9
 1 4 2 6
(k)
6 1 5 9
 8 0 4 9
(l)
8 0 1 6
5 8 8
 7 1 5 6
(a)
78 + 94
(b)
367 + 243
(c)
48 + 313 + 925
3. There are 183 women and 68 men on the sales staff?
How many people are on the sales staff?
NSSAL
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C. D. Pilmer
Subtracting Multi-Digit Numbers
To subtract multi-digit whole numbers, start by stacking the numbers vertically such that
corresponding place values line up (e.g. units with units, tens with tens) and subtract from right
to left. If the digit being subtracted is larger than the digit from which it is being subtracted,
regroup (i.e. borrow) one from the digit in the next larger place value.
e.g. 597 - 62
Answer:
Stack the numbers vertically (i.e. one on top of another) such corresponding place values
line up.
Subtract the Units
Subtract the Tens
Subtract the Hundreds

5 9 7

6 2

5 9 7

6 2

5 9 7

6 2
5
3 5
5 3 5
7 units minus 2 units is
5 units.
9 tens minus 6 tens is 3
tens.
5 hundreds minus 0
hundreds is 5 hundreds.
To determine if our answer is reasonable, we can round 597 to 600, round 62 to 60, and
take the difference. Since 600  60  540 , it appears that our answer of 535 is reasonable.
e.g. 392 - 145
Answer:
Subtract the Units
3
 1


8
12
8
12
9
4
2
5
9
4
2
5
4
7
7
We cannot take 5 units
from 2 units. Therefore
we regroup (i.e. borrow)
1 from the tens, which
leaves us with 8 tens
and 12 units. 12 units
minus 5 units is 7 units.
NSSAL
©2013
Subtract the Tens
3
 1

8 tens minus 4 tens is 4
tens.
41
Subtract the Hundreds
8
12
9
4
2
5
2 4
7
3
 1
3 hundreds minus 1
hundred is 2 hundreds.
Draft
C. D. Pilmer
e.g. 647 - 391
Answer:
Subtract the Units

6 4 7
 3 9 1
Subtract the Tens



5
14
5
14
6
3
4
9
7
1
6
3
4
9
7
1
5
6
2
5
6
6
7 units minus 1 unit is 6
units.
Subtract the Hundreds

We cannot take 9 tens
3 hundreds minus 1
from 4 tens. Therefore
hundred is 2 hundreds.
we regroup (i.e. borrow)
1 from the hundreds,
which leaves us with 5
hundreds and 14 tens.
14 tens minus 9 tens is 5
tens.
e.g. 934 - 268
Answer:
Subtract the Units
Subtract the Tens


9
 4
12
12
2
14
8
2
14
8
2
14
3
6
4
8
9
4
3
6
4
8
9
4
3
6
4
8
6
6
4
6
6
6
We cannot take 8 units
from 4 units. Therefore
we regroup (i.e. borrow)
1 from the tens, which
leaves us with 2 tens
and 14 units. 14 units
minus 8 units is 6 units.
NSSAL
©2013
Subtract the Hundreds



We cannot take 6 tens
8 hundreds minus 4
from 2 tens. Therefore
hundreds is 4 hundreds.
we regroup (i.e. borrow)
1 from the hundreds,
which leaves us with 8
hundreds and 12 tens.
12 tens minus 6 tens is 6
tens.
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Draft
C. D. Pilmer
e.g. 803 - 288
Answer:
Subtract the Units
Subtract the Tens
Subtract the Hundreds


9

7
10
13
7
10
8
2
0
8
3
8
8
2
0
8

5
We cannot take 8 units
from 3 units. Therefore
we regroup (i.e.
borrow). However, we
have a zero in the tens
place. We need to
borrow 1 hundred from
the hundreds place, then
borrow 10 from the tens
place. That leaves us
with 7 hundreds, 9 tens,
and 13 units. 13 units
minus 8 units is 5 units.

9
13
7
10
13
3
8
8
2
0
8
3
8
5
1
5
9

1 5
9 tens minus 8 tens is 1
ten.
e.g. 6 946 - 2 439
7 hundreds minus 2
hundreds is 5 hundreds.
e.g. 5 248 - 761
Answer:
Answer:
3
16
6 9
 2 4
4
3
6
9
4 5
0
7
11
4
1
14
5
2
7
4
6
4
4
8 7

8
1
Questions
Complete the following operations without using a calculator or an addition facts chart. Notice
that we have included a few addition questions.
1.
(a)
NSSAL
©2013
7 8
 5 3
(b)
8 3
 2 6
43
(c)
5 6
 9 7
Draft
C. D. Pilmer
2.
(d)
6 4 8
 5 2 1
(e)
5 6 7
 1 8 7
(f)
2 5 4

6 7
(g)
4 7 1
 5 8 3
(h)
9 8 4
 7 5 8
(i)
4 0 5
 2 7 9
(j)
4 2 9 7
 3 0 5 2
(k)
6 3 7 4
 2 5 1 4
(l)
5 4 6 3
 1 2 9 5
(m)
6 7 0 3

3 6 9
(n)
4 2 5 4
 7 1 8 3
(o)
9 0 2 8
 8 2 5 7
(a)
674 - 58
(b)
1 243 - 87
(c)
5 692 - 825
3. On payday Tyrus had $1250 in his chequing account.
After paying his bills, his balance in this account was
$367. How much did he spend on bills?
4. A power drill costs $89 (before taxes). A circular saw
costs $124 (before taxes). Find the total cost for the
two tools before taxes.
5. On Tuesday, 763 people attended the fair. On
Wednesday, 927 people attended the fair. How many
more people attended on Wednesday compared to
Tuesday?
NSSAL
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C. D. Pilmer
Multiplying Multi-Digit Numbers
In this section, we will show three ways to multiply multi-digit whole numbers: the traditional
algorithm, multiplying using the expanded form, and the lattice method. You choose the
technique you are most comfortable with.
Example 1
Complete the following operation. 67  49
Answer:
Method 1: Traditional Algorithm
6
6 7
 4 9
3
6 7
 4 9
6 0 3
0
2
6 7
 4 9
6 0 3
2 6 8 0
Step 1:
Multiply 9 by 7. The
resulting product is 63.
Write 3, and carry (i.e.
regroup) the 6.
6
6 7
 4 9
6 0 3
Step 3:
Place a 0 in the units place
as in our next step we will
be multiplying using the
second digit (tens)
Step 5:
Multiply 4 by 6. The
resulting product is 24.
Add the 2 that you carried
to the 24, and write down
the resulting sum.
2
6 7
 4 9
Step 2:
Multiply 9 by 6. The
resulting product is 54.
Add the 6 that you carried
to the 54, and write down
the resulting sum.
Step 4:
Multiply 4 by 7. The
resulting product is 28.
Write 8, and carry the 2.
6 0 3
8 0
6 7
 4 9
Step 6:
Add to get the desired
product.
6 0 3
2 6 8 0
3 2 8 3
Method 2: Multiplying Using the Expanded Form
We first need to express the two numbers in their expanded forms (67 = 60 + 7 and
49 = 40 + 9) and then set the numbers up so that we can do the multiplication.
60  7
 40  9
6
5 4
2 8
2 4 0
3
0
0
0
9  7 ; first set of multiplication
9  60 ; second set of multiplication
40  7 ; third set of multiplication
40  60 ; fourth set of multiplication
3 2 8 3
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C. D. Pilmer
Method 3: Lattice Method
In this case we are multiplying a two digit number by another two digit number, therefore we
create a 2 by 2 grid with diagonals drawn in each of the resulting squares.
6
7
4
6
2
4
2
4
9
Step 1:
We put the 67 along the top
of the chart, and the 49 along
the right side of the grid.
6
7
7
2
4
8
9
9
Step 2:
Multiply the 6 by the 4, and
place the two digits of the
product (24) in the two
spaces in the upper left hand
corner of the grid.
Step 3:
Multiply the 7 by the 4, and
place the two digits of the
product (28) in the two
spaces in the upper right
hand corner of the grid.
carry 1
6
2
4
5
4
7
6
2
8
4
9
2
4
5
7
2
8
6
4
3
4
4
3
9
2
2
carry 1
2
4
5
8
6
4
3
8
3
67  49 = 3283
Step 4:
Multiply the 6 by the 9, and
place the two digits of the
product (54) in the two
spaces in the lower left hand
corner of the grid.
NSSAL
©2013
Step 5:
Multiply the 7 by the 9, and
place the two digits of the
product (63) in the two
spaces in the lower right
hand corner of the grid.
46
Step 6:
Now ignore the 67 and 49
along the outside of our grid.
Starting at the bottom, add
the numbers along each
diagonal placing the answer
along the outer edge of the
chart. If a sum exceeds 9,
carry the tens digit up to the
next diagonal. The numbers
along the outside, starting at
the upper left, represent the
digits of your product.
Draft
C. D. Pilmer
Of the three methods shown, most learners find the second method the easiest to understand and
the easiest to complete successfully. The author of this resource generally recommends this
technique. Although, the lattice method has a "cool factor" to it, learners don't always
understand why this technique works. It is actually very similar to the expanded form technique;
notice that the diagonals in the lattice match the columns in the expanded form. We have a video
on YouTube that explains the lattice method. Google search "YouTube Lattice Method
nsccalpmath."
Example 2
Complete the following operation. 497  53
Answer:
Method 1: Traditional Algorithm
2
4 9 7

5 3
1
2
4 9 7

5 3
1 4 9 1
3
4 9 7

5 3
1 4 9 1
5 0
4
4 9 7

5 3
1 4 9 1
2 4 8 5 0
Step 1:
Multiply 3 by 7. The
resulting product is 21.
Write 1, and carry the
2.
Step 3:
Multiply 3 by 4. The
resulting product is 12.
Add the 2 that you
carried to the 12, and
obtain 14.
Step 5:
Multiply 5 by 7. The
resulting product is 35.
Write 5, and carry the
3.
Step 7:
Multiply 5 by 4. The
resulting product is 20.
Add the 4 that you
carried to the 20, and
obtain 24.
2
2
4 9 7

5 3
9 1
4 9 7

5 3
1 4 9 1
0
4
3
4 9 7

5 3
1 4 9 1
8 5 0
4 9 7

5 3
Step 2:
Multiply 3 by 9. The
resulting product is 27.
Add the 2, which you
carried, to the 27, and
obtain 29. Write down
9, and carry the 2.
Step 4:
Place a 0 in the units
place as in our next
step we will be
multiplying using the
second digit (tens)
Step 6:
Multiply 5 by 9. The
resulting product is 45.
Add the 3, which you
carried, to the 45, and
obtain 48. Write down
8, and carry the 4.
Step 8:
Add to get the desired
product.
1 4 9 1
2 4 8 5 0
2 6 3 4 1
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C. D. Pilmer
Method 2: Multiplying Using the Expanded Form
We first need to express the two numbers in their expanded forms (497 = 400 + 90 + 7 and
53 = 50 + 3) and then set the numbers up so that we can do the multiplication.
400  90  7

50  3
2
1 2
3
4 5
2 0 0
2
7
0
5
0
0
1
0
0
0
0
0
The 3 must be multiplied by the 400, 90, and 7.
The 50 must be multiplied by the 400, 90, and 7.
2 6 3 4 1
Method 3: Lattice Method
4
9
7
4
5
3
2
4
0
1
5
7
2
5
5
2
1
3
6
2
2
2
1
7
3
9
4
3
0
1
5
2
5
2
2
3
1
1
7
4
1
We put the 497 along the top Multiply the numbers. (e.g. 4 Now ignore the 497 and 53
of the grid, and the 53 along times 5 is 20, 9 times 5 is 45, along the outside of our grid.
the right side of the grid.
etc.)
Starting at the bottom, add
the numbers along each
diagonal placing the answer
along the outer edge of the
chart. If a sum exceeds 9,
carry the tens digit up to the
next diagonal. The numbers
along the outside, starting at
the upper left, represent the
digits of your product.
NSSAL
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C. D. Pilmer
Example 3
Complete the following operation. 534  728
Answer:
Method 1: Traditional Algorithm
Method 2: Multiplying Using Expanded
Form
500  30  4
 700  20  8
5 3 4
 7 2 8
4 2 7 2
1 0 6 8 0
3 7 3 8 0 0
3
4
0
8
0
0
0
0
0
2
4 0
3 8 8 7 5 2
6
0
8
0
0
1 0
2
2 1
3 5 0
2
0
0
0
0
0
0
0
0
3 8 8 7 5 2
Method 3: Lattice Method
5
3
2
2
3
5
1
1
4
3
8
2
8
2
8
8
8
6
2
0
7
0
0
0
1
4
3
4
2
3
5
NSSAL
©2013
0
1
0
5
1
0
8
3
4
0
49
8
6
2
4
7
2
1
2
2
Draft
C. D. Pilmer
Questions
Complete the following operations without using a calculator, an addition facts chart, or a
multiplication facts chart. Notice that we have included a few addition and subtraction
questions. Show your work and use the method(s) you prefer.
1. (a) 46  39
(b) 74  63
(c) 86  45
(d) 57  41
(e) 83  56
(f) 65  28
(g) 435 14
(h) 542  36
(i) 348  65
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(j) 368  92
(k) 647  84
(l) 915  83
(m) 142  263
(n) 852  273
(o) 621 364
(p) 748  459
(q) 937  291
(r) 602  237
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C. D. Pilmer
2. The banquet hall had 18 tables with 12 chairs at each.
How many people can be seated at one time?
3. A landscaping company needs 89 cubic yards of topsoil
at one site, and 134 cubic yards at another site. How
many cubic yards do they need in total for the two
sites?
4. A particular spreadsheet is comprised on 24 columns
and 16 rows. How many entries can be placed on this
spreadsheet?
5. The tank, when filled, held 530 litres of a particular
chemical. After a period of time, 183 litres remained.
How many litres of the chemical were used during that
period of time?
6. With a reforestation project, they plant 350 seedlings
every acre. If they wish to reforest 76 acres, how many
seedlings will be needed?
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Dividing Multi-Digit Numbers
In this section, we learn to divide multi-digit numbers by single digit numbers (e.g. 3759  6 ).
We show two approaches (i.e. algorithms) to complete these types of questions; you choose the
method you prefer. We will not be dividing a multi-digit number by another multi-digit number
(e.g. 5768  74 ). Our rationale is that these latter types of question are tedious and best solved
using a calculator.
Example 1
Solve 2856  8
Answer:
Method 1: Traditional Algorithm
This is the technique that many people have worked with in the past, however, many
individuals struggle to understand why this technique works. That is why we generally
suggest that you learn method 2, which generally makes more sense.
?
8 2856
3
8 2856
24
3
8 2856
Ask: "How many times does 8 divide into 2?"
Answer: It does not go into it, therefore we move to the next step
where we include the next place value.
Ask: "How many times does 8 divide into 28?"
Answer: We estimate 3 times.
Place 3 in the hundreds column, multiply 8 by 3, and place the
answer 24 below the 28.
Subtract the 24 from the 28. Following this, bring down the 5
from the tens position.
- 24 
45
35
8 2856
- 24
45
40
35
8 2856
Ask: "How many times does 8 divide into 45?"
Answer: We estimate 5 times.
Place 5 in the tens column, multiply 8 by 5, and place the answer
40 below the 45.
Subtract the 40 from the 45. Following this, bring down the 6
from the units position.
- 24
45
- 40 
56
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Draft
C. D. Pilmer
357
Ask: "How many times does 8 divide into 56?"
Answer: 7 times.
Place 7 in the units column, multiply 8 by 7, and place the
answer 56 below the 56.
8 2856
- 24
45
- 40
56
56
Subtract the 56 from the 56. There is no remainder, therefore we
know that 8 divides evenly into 2856.
357
8 2856
- 24
45
- 40
56
- 56
0
Method 2: Partial Quotient Method
8 2856
1600
8 2856
1600
1256
8 2856
1600
1256
800
8 2856
1600
1256
800
456
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©2013
200
How many times does 8 go into 1600? This learner goes with
200 because he/she knows that 8 times 200 is 1600. (The learner
could have gone with a larger number like 300, but it does not
matter with this method.)
Now he/she subtracted 1600 from 2856.
200
200
How many times does 8 go into 1256. This learner goes with
100 because he/she knows that 8 times 100 is 800.
100
Now he/she subtracted 800 from 1256.
200
100
54
Draft
C. D. Pilmer
8 2856
1600
1256
800
456
400
8 2856
1600
1256
800
456
400
56
8 2856
1600
1256
800
456
400
56
56
0
357
8 2856
1600
1256
800
456
400
56
56
0
NSSAL
©2013
200
How many times does 8 go into 456. This learner goes with 50
because he/she knows that 8 times 50 is 400.
100
50
Now he/she subtracted 400 from 456.
200
100
50
200
How many times does 8 go into 56. This learner goes with 7
because he/she knows that8 times 7 is 56. After we filled in the
56, we did the subtraction and found that we had a remainder of
0. That means 8 divides evenly into 2856.
100
50
7
In our last step the learner simply adds 200, 100, 50, and 7. That
means that the quotient is 357.
200
2856  8 = 357
100
50
7
55
Draft
C. D. Pilmer
Example 2
Solve 4785  7
Answer:
Method 1: Traditional Algorithm
683 R:4
7 4785
- 42
58
- 56
25
- 21
4
Method 2: Partial Quotient Method
We have shown you three solutions to this question. The first student was the most efficient
because he/she did the question in the fewest number of steps, however, all of the students
have correct answers. That is the great thing about the partial quotient method; there is more
than one way to do it right. In this case, 7 does not go evenly into 4785; we have a
remainder of 4 when we complete the division.
First Learner:
683 R: 4
7 4785
4200
585
560
25
21
4
NSSAL
©2013
600
80
3
Second Learner:
683 R: 4
7 4785
400
2800
1985
1400
585
560
25
21
4
200
80
3
56
Third Learner:
683 R: 4
7 4785
4200
600
585
350
50
235
210
30
25
21
4
3
Draft
C. D. Pilmer
Example 3
Solve 31 654  9
Answer:
Method 1: Traditional Algorithm
3517 R:1
9 31654
- 27
46
- 45
15
- 9
64
- 63
1
Method 2: Partial Quotient Method
We have again supplied multiple solutions; all of them are correct.
First Learner
3517 R: 1
9 31654
27000
4654
4500
154
90
64
63
1
3000
500
10
7
Second Learner:
3517 R: 1
9 31654
18000
2000
13654
9000
1000
4654
400
3600
1054
900
154
90
64
63
1
100
10
Third Learner:
3517 R: 1
9 31654
27000
3000
4654
4500
500
154
90
64
45
19
18
1
10
5
2
7
We have a video on YouTube that explains the partial quotient method. Google search
"YouTube Long Division Partial Quotient Method nsccalpmath."
NSSAL
©2013
57
Draft
C. D. Pilmer
Questions
Complete the following operations without using a calculator, an addition facts chart, or a
multiplication facts chart. Notice that we have included a few addition, subtraction, and
multiplication questions. Show your work and use the method(s) you prefer.
1. (a) 268  4
(d) 54  83
NSSAL
©2013
(b) 178  3
(c) 2538  6
(e) 3270  4
(f) 825  467
58
Draft
C. D. Pilmer
(g) 2919  5
(h) 5646  9
(i) 2474  653
(j) 6259  6
(k) 392  79
(l) 26194  7
NSSAL
©2013
59
Draft
C. D. Pilmer
2. The prize money of $4710 had to be shared equally by
the 6 ticket holders. How much did it person receive?
3. A particular car holds 38 litres of gasoline and travels
14 kilometres per litre. How far can the car travel on a
full tank of gasoline?
4. The land developer wants to take her 234 acre property
and break it into 9 lots of equal size. What will be the
acreage of these new lots?
5. Tylena earned $2860 this month and her expenses
during that same period were $2395. How much could
Tylena potentially save during this month?
6. You are selling lemonade at the fair. You large
dispensing container holds 192 ounces of lemonade. If
you are serving the drinks in 6 ounce cups, how many
cups can you potentially sell?
NSSAL
©2013
60
Draft
C. D. Pilmer
Appendix
NSSAL
©2013
61
Draft
C. D. Pilmer
Connect Four Whole Number Addition Game (A)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers on the Addend Strip whose sum is that desired square.
Once they have chosen the two numbers, they can capture one square with that appropriate
sum. They either mark the square with an X or place a colored counter on the square. There
may be other squares with that same sum but only one square can be captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips on the Addend Strip. They then mark the square with that sum using a O or a
different colored marker. If a player cannot move a single paperclip to capture a square, a
paperclip must still be moved on the addend strip in order to ensure that the game can
continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
10
12
7
13
8
11
9
11
10
12
6
13
6
14
9
11
10
9
8
11
12
7
14
11
13
10
8
6
9
10
9
7
14
10
12
8
6
7
Addend Strip:
3
NSSAL
©2013
4
5
62
Draft
C. D. Pilmer
Connect Four Whole Number Addition Game (B)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers on the Addend Strip whose sum is that desired square.
Once they have chosen the two numbers, they can capture one square with that appropriate
sum. They either mark the square with an X or place a colored counter on the square. There
may be other squares with that same sum but only one square can be captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips on the Addend Strip. They then mark the square with that sum using an O or a
different colored marker. If a player cannot move a single paperclip to capture a square, a
paperclip must still be moved on the addend strip in order to ensure that the game can
continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
12
11
18
14
16
14
15
14
13
16
12
13
16
12
15
10
14
17
15
17
14
18
15
13
13
10
13
16
11
18
15
12
11
17
14
10
Addend Strip:
5
NSSAL
©2013
6
7
8
9
63
Draft
C. D. Pilmer
Connect Four Whole Number Subtraction Game (A)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers; one from Value 1 and one from Value 2. Once they
have chosen the two numbers, they can capture one square with that appropriate difference
(i.e. Value 1 subtract Value 2). They either mark the square with an X or place a colored
counter on the square. There may be other squares with that same difference but only one
square can be captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips. They then mark the square with that difference using an O or a different colored
marker. If a player cannot move a single paperclip to capture a square, a paperclip must still
be moved in order to ensure that the game can continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
3
2
5
4
6
4
7
4
6
2
3
5
6
0
1
0
5
6
3
5
3
7
2
4
4
3
2
4
1
0
1
7
0
5
3
6
Value 1:
13
NSSAL
©2013
Value 2:
12
11
10
9
6
64
7
8
9
Draft
C. D. Pilmer
Connect Four Whole Number Subtraction Game (B)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers; one from Value 1 and one from Value 2. Once they
have chosen the two numbers, they can capture one square with that appropriate difference
(i.e. Value 1 subtract Value 2). They either mark the square with an X or place a colored
counter on the square. There may be other squares with that same difference but only one
square can be captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips. They then mark the square with that difference using an O or a different colored
marker. If a player cannot move a single paperclip to capture a square, a paperclip must still
be moved in order to ensure that the game can continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
3
5
4
5
6
5
6
8
7
9
10
7
9
7
6
3
5
9
7
10
5
7
8
4
8
6
9
4
9
3
4
7
8
10
5
6
Value 1:
15
NSSAL
©2013
Value 2:
14
13
12
5
65
6
7
8
9
Draft
C. D. Pilmer
Connect Four Whole Number Multiplication Game (A)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers on the Factor Strip whose product is that desired
square. Once they have chosen the two numbers, they can capture one square with that
appropriate product. They either mark the square with an X or place a colored counter on the
square. There may be other squares with that same product but only one square can be
captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips on the Factor Strip. They then mark the square with that product using an O or a
different colored marker. If a player cannot move a single paperclip to capture a square, a
paperclip must still be moved on the fraction strip in order to ensure that the game can
continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
6
45
27
5
45
8
10
0
36
18
20
15
36
8
12
4
0
36
2
18
45
27
6
12
20
4
15
0
10
9
27
12
3
6
36
20
Factor Strip:
0
NSSAL
©2013
1
2
3
4
66
5
9
Draft
C. D. Pilmer
Connect Four Whole Number Multiplication Game (B)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers on the Factor Strip whose product is that desired
square. Once they have chosen the two numbers, they can capture one square with that
appropriate product. They either mark the square with an X or place a colored counter on the
square. There may be other squares with that same product but only one square can be
captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips on the Factor Strip. They then mark the square with that product using an O or a
different colored marker. If a player cannot move a single paperclip to capture a square, a
paperclip must still be moved on the fraction strip in order to ensure that the game can
continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
18
2
30
8
12
24
9
54
12
18
10
6
24
5
8
6
54
20
10
30
18
5
24
3
24
4
20
12
2
18
12
54
9
30
5
8
Factor Strip:
1
NSSAL
©2013
2
3
4
5
67
6
9
Draft
C. D. Pilmer
Connect Four Whole Number Multiplication Game (C)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers on the Factor Strip whose product is that desired
square. Once they have chosen the two numbers, they can capture one square with that
appropriate product. They either mark the square with an X or place a colored counter on the
square. There may be other squares with that same product but only one square can be
captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips on the Factor Strip. They then mark the square with that product using an O or a
different colored marker. If a player cannot move a single paperclip to capture a square, a
paperclip must still be moved on the fraction strip in order to ensure that the game can
continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
14
63
6
28
15
30
42
12
30
63
14
10
8
21
54
18
54
21
35
15
8
28
42
12
18
54
14
63
6
35
10
28
42
12
21
18
Factor Strip:
2
NSSAL
©2013
3
4
5
6
68
7
9
Draft
C. D. Pilmer
Connect Four Whole Number Multiplication Game (D)
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers on the Factor Strip whose product is that desired
square. Once they have chosen the two numbers, they can capture one square with that
appropriate product. They either mark the square with an X or place a colored counter on the
square. There may be other squares with that same product but only one square can be
captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips on the Factor Strip. They then mark the square with that product using an O or a
different colored marker. If a player cannot move a single paperclip to capture a square, a
paperclip must still be moved on the fraction strip in order to ensure that the game can
continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
42
12
16
8
24
48
6
72
45
54
15
18
56
24
21
16
56
20
14
30
10
40
6
27
54
18
36
12
42
21
15
72
27
14
35
10
5
6
Factor Strip:
2
NSSAL
©2013
3
4
69
7
8
9
Draft
C. D. Pilmer
Connect Four Whole Number Division Game
Number of Players: Two
Objective: The winner is the first player to connect four of his/her pieces horizontally, vertically
or diagonally.
Instructions:
1. Roll a die to see which player will go first.
2. The first player looks at the board and decides which square he/she wishes to capture. They
place two paperclips on two numbers; one from Value 1 and one from Value 2. Once they
have chosen the two numbers, they can capture one square with that appropriate quotient (i.e.
Value 1 divided by Value 2). They either mark the square with an X or place a colored
counter on the square. There may be other squares with that same quotient but only one
square can be captured at a time.
3. Now the second player is ready to capture a square but he/she can only move one of the
paperclips. They then mark the square with that quotient using an O or a different colored
marker. If a player cannot move a single paperclip to capture a square, a paperclip must still
be moved in order to ensure that the game can continue.
4. Play alternates until one player connects four squares. Remember that only one paperclip is
moved at a time. If none of the players is able to connect four, then the winner is the
individual who has captured the most squares.
Game Board:
6
24
12
15
12
2
8
3
6
30
4
15
18
12
10
9
8
12
6
8
2
24
6
9
30
4
15
12
4
3
6
18
9
2
10
18
Value 1:
30
NSSAL
©2013
Value 2:
24
18
12
6
1
70
2
3
Draft
C. D. Pilmer
Flash Cards: Addition
Remove the following pages, cut out the flash cards, and regularly practice your math fact using
these cards.
NSSAL
©2013
30 + 20
400 + 300
2 000 + 5 000
40 + 50
100 + 600
3 000 + 6 000
60 + 70
500 + 800
8 000 + 4 000
90 + 30
700 + 400
1 000 + 2 000
10 + 80
200 + 900
9 000 + 5 000
70 + 70
800 + 800
6 000 + 6 000
60 + 40
300 + 700
2 000 + 8 000
71
Draft
C. D. Pilmer
NSSAL
©2013
7 000
700
50
9 000
700
90
12 000
1 300
130
3 000
1 100
120
14 000
1 100
90
12 000
1 600
140
10 000
1 000
100
72
Draft
C. D. Pilmer
NSSAL
©2013
20 + 20
900 + 400
2 000 + 3 000
80 + 70
700 + 800
9 000 + 9 000
30 + 50
300 + 100
4 000 + 3 000
90 + 80
200 + 400
5 000 + 8 000
10 + 60
500 + 700
6 000 + 2 000
20 + 40
100 + 900
7 000 + 7 000
60 + 60
600 + 800
8 000 + 6 000
73
Draft
C. D. Pilmer
NSSAL
©2013
5 000
1 300
40
18 000
1 500
150
7 000
400
80
13 000
600
170
8 000
1 200
70
14 000
1 000
60
14 000
1 400
120
74
Draft
C. D. Pilmer
Flash Cards: Subtraction
Remove the following pages, cut out the flash cards, and regularly practice your math fact using
these cards.
NSSAL
©2013
50 - 10
600 - 300
3 000 - 1 000
80 - 30
500 - 400
7 000 - 4 000
90 - 20
900 - 300
9 000 - 5 000
40 - 40
700 - 100
8 000 - 7 000
30 - 10
800 - 800
6 000 - 4 000
90 - 50
900 - 500
9 000 - 6 000
80 - 70
300 - 200
5 000 - 5 000
75
Draft
C. D. Pilmer
NSSAL
©2013
2 000
300
40
3 000
100
50
4 000
600
70
1 000
600
0
2 000
0
20
3 000
400
40
0
100
10
76
Draft
C. D. Pilmer
100 - 60
1 200 - 700
13 000 - 8 000
120 - 50
1 100 - 900
16 000 - 9 000
140 - 60
1 700 - 800
11 000 - 3 000
150 - 90
1 500 - 600
15 000 - 7 000
110 - 70
1 600 - 800
12 000 - 6 000
160 - 80
1 300 - 400
18 000 - 9 000
120 - 50
1 400 - 800
14 000 - 6 000
NSSAL
©2013
77
Draft
C. D. Pilmer
NSSAL
©2013
5 000
500
40
7 000
200
70
8 000
900
80
8 000
900
60
6 000
800
40
9 000
900
80
8 000
600
70
78
Draft
C. D. Pilmer
Flash Cards: Multiplication
Remove the following pages, cut out the flash cards, and regularly practice your math fact using
these cards.
NSSAL
©2013
2 × 40
70 × 30
60 × 200
30 × 5
4 × 600
1000 × 8
80 × 6
80 × 80
7 × 3000
7 × 70
50 × 10
400 × 30
40 × 7
200 × 9
2000 × 9
9 × 50
3 × 500
40 × 700
60 × 3
60 × 30
6 × 6000
79
Draft
C. D. Pilmer
NSSAL
©2013
12 000
2 100
80
8 000
2 400
150
21 000
6 400
480
12 000
500
490
18 000
1 800
280
28 000
1 500
450
36 000
1 800
180
80
Draft
C. D. Pilmer
NSSAL
©2013
8 × 60
70 × 70
30 × 900
70 × 5
300 × 8
4 × 8000
30 × 9
5 × 500
500 × 30
8 × 70
30 × 90
1000 × 7
90 × 4
60 × 80
800 × 60
3 × 40
900 × 9
90 × 700
20 × 7
7 × 600
20 × 800
81
Draft
C. D. Pilmer
NSSAL
©2013
27 000
4 900
480
32 000
2 400
350
15 000
2 500
270
7 000
2 700
560
48 000
4 800
360
63 000
8 100
120
16 000
4 200
140
82
Draft
C. D. Pilmer
Flash Cards: Division
Remove the following pages, cut out the flash cards, and regularly practice your math fact using
these cards.
NSSAL
©2013
80  2
90  30
2 100  7
50  5
180  90
1 600  40
160  8
270  30
2 400  300
120  4
320  80
5 400  9
350  7
420  70
4 900  70
480  6
250  50
3 200  400
630  9
720  90
8 100  900
83
Draft
C. D. Pilmer
NSSAL
©2013
300
3
40
40
2
10
6
9
20
600
4
30
70
6
50
8
5
80
9
8
70
84
Draft
C. D. Pilmer
Answers
Introduction to Whole Numbers (pages 1 to 6)
1. (a) nine hundred eighty-six
(b) four hundred twelve
(c) two thousand, three hundred ninety-seven
(d) four thousand, six hundred nine
(e) twelve thousand, seven hundred fifty
(f) three hundred forty-seven thousand, fifty-two
(g) five hundred six thousand, nine hundred
(h) two million, eight hundred seventy thousand, forty
(i) three hundred fifty million, twenty-six thousand
(j) seventeen million, five hundred nine thousand, one hundred
(k) seven billion, three hundred sixty million, eight hundred thousand
(l) eleven billion, ninety-four million, three hundred thousand, two
(m) six trillion, five hundred thirty-two billion, eighty
2. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
6 742
3 126 511
809 327
13 052 071
63 147
408 700 260
349 008
2 910 706 050
162 083 400 000
80 307 005 200
3. (a)
(b)
(c)
(d)
(e)
millions place
hundred millions place
ten thousands place
ten billions place
hundred thousands place
4. (a)
(b)
(c)
(d)
(e)
4 296  4000  200  90  6
136 942  100 000  30 000  6 000  900  40  2
6 056 730  6 000 000  50 000  6 000  700  30
45 760 500  40 000 000  5 000 000  700 000  60 000  500
9 602 000 000  9 000 000 000  600 000 000  2 000 000
5. (a) 3, 9, 52, 93, 546, 800
(b) 27, 29, 529, 546, 8 698, 8 700
(c) 790, 796, 9 360, 9 502, 33 870, 34 000
(d) 38 099, 38 500, 56 899, 56 943, 76 000, 102 000
NSSAL
©2013
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Draft
C. D. Pilmer
6.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Average Canadian Mortgage in 2013
Province of Nova Scotia's Debt in 2012
Approximate Cost of a New Family Sedan
Approximate Cost of a 42 inch Flat Screen TV
Overall Cost of 3 Year Phone Plan with Data Package
Average Individual Income of Nova Scotian in 2010
Canada's Federal Debt in 2012
Cost of Purchasing a Tim Hortons' Franchise in 2012
Estimated Cost of New Halifax Convention Center
7.
(h)
(e)
(b)
(f)
(a)
(g)
(i)
(d)
(c)
$500 000
$2200
$14 000 000 000
$40 000
$245 000
$600 000 000 000
$159 000 000
$500
$26 000
Number
(a)
What whole number is after 325?
326
(b)
What whole number is before 6 421?
6 420
(c)
What whole number is between 45 188 and 45 190?
45 189
(d)
What whole numbers are between 763 and 768?
(e)
What whole number is after 7 239?
7 240
(f)
What whole number is between 9 398 and 9400?
9 399
(g)
What whole number is before 82 600?
82 599
(h)
What whole number is between 62 985 and 62 987?
62 986
(i)
What whole numbers are between 69 997 and 70 000?
(j)
What whole number is before 120 000?
(k)
What whole number is before 7 300 000?
(l)
What whole numbers are between 27 029 and 27 032?
764, 765, 766, 767
69 998, 69 999
119 999
7 299 999
27 030, 27 031
(m) What whole number is after 15 999?
16 000
(n)
What whole number is between 386 638 and 386 640?
386 639
(o)
What whole numbers are between 9 997 and 10 001?
9 998, 9 999, 10 000
(p)
What whole number is before 430 000 000?
(q)
What whole number is after 549 999?
550 000
(r)
What whole number is between 31 652 and 31 654?
31 653
(s)
What whole numbers are between 72 999 and 73 002?
NSSAL
©2013
86
429 999 999
73 000, 73 001
Draft
C. D. Pilmer
Math Facts: Addition and Multiplication (pages 7 to 16)
1. (a) 10
(d) 8
(g) 8
(j) 14
(m) 10
(p) 12
(s) 11
(v) 13
(y) 8
(b) 6
(e) 7
(h) 16
(k) 9
(n) 12
(q) 13
(t) 5
(w) 9
(z) 8
(c)
(f)
(i)
(l)
(o)
(r)
(u)
(x)
12
9
9
11
2
11
13
4
2. (a) 8
(d) 0
(g) 12
(j) 12
(m) 48
(p) 20
(s) 16
(v) 36
(y) 45
(b) 35
(e) 27
(h) 49
(k) 5
(n) 24
(q) 21
(t) 18
(w) 63
(z) 56
(c)
(f)
(i)
(l)
(o)
(r)
(u)
(x)
8
28
0
72
54
64
30
18
(b) 5 and 7
(e) 3 and 10
(c) 1 and 3
(f) 3 and 8
3. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
addition
addition
multiplication
addition
multiplication
multiplication
addition
multiplication
4. (a) 2 and 3
(d) 2 and 8
5. (a)
NSSAL
©2013
(b)
4
2
3
5
6
4
3
4
2
4
5
6
2
3
4
6
4
5
87
Draft
C. D. Pilmer
(c)
(d)
5
7
6
8
7
9
7
6
5
9
8
7
6
5
7
7
9
8
(e)
(f)
3
5
4
2
8
6
7
9
5
2
3
4
9
7
6
8
2
4
5
3
7
8
9
6
4
3
2
5
6
9
8
7
Fact Families (pages 17 to 21)
1. (a) 5  9  14
14  9  5
14  5  9
(b) 7  4  28
28  7  4
28  4  7
2. (a) 4  8  12
8  4  12
12  4  8
12  8  4
(b) 7  9  63
9  7  63
63  9  7
63  7  9
(e) 6  7  13
7  6  13
13  7  6
13  6  7
(d) 8  3  24
3  8  24
24  8  3
24  3  8
NSSAL
©2013
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Draft
C. D. Pilmer
(f) 15  6  9
15  9  6
6  9  15
9  6  15
(g) 10  2  5
10  5  2
2  5  10
5  2  10
3. (a) 8
(d) 1
(g) 0
(j) 5
(m) 8
(p) 0
(s) 5
(v) 2
(y) 3
(b) 5
(e) 5
(h) 7
(k) 7
(n) 3
(q) 8
(t) 8
(w) 8
(z) 2
(c)
(f)
(i)
(l)
(o)
(r)
(u)
(x)
6
7
10
4
7
9
3
7
4. (a) 7
(d) 8
(g) 2
(j) 4
(m) 5
(p) 8
(s) 5
(v) 9
(y) 8
(b) 3
(e) 1
(h) 5
(k) 7
(n) 9
(q) 6
(t) 6
(w) 3
(z) 6
(c)
(f)
(i)
(l)
(o)
(r)
(u)
(x)
6
5
3
4
2
8
7
1
5. (a) 5
(d) 14
(g) 9
(j) 7
(m) 9
(p) 10
(s) 1
(v) 0
(b) 9
(e) 24
(h) 9
(k) 13
(n) 6
(q) 32
(t) 14
(w) 0
(c)
(f)
(i)
(l)
(o)
(r)
(u)
(x)
15
7
16
9
6
7
6
1
6. (a)
(d)
(g)
(j)
(b)
(e)
(h)
(k)
(c)
(f)
(i)
(l)
4&7
5&8
9&4
5&5
3&5
12 & 1
4&4
2&6
7. (a) 5 & 7
(d) 1 & 9
(g) 8 & 3
NSSAL
©2013
2 & 10
6&3
6&4
10 & 1
(b) 2 & 5
(e) 4 & 4
(h) 0 & 2
(c) 3 & 4
(f) 2 & 9
(i) 2 & 10
89
Draft
C. D. Pilmer
9. (a)
(b)

6
6
=
1
=
14
-
13
18

9
=
2
=

2
1

-


-
+
-



5
3
10
7
8
14
1
2
6
3
=
=
=
=
=
=
=
=
=
=
5
32

30
3
=
10
=
2


8
=
4
=
12

3
=
=
=
=
=
=
=
=
=
=
31
24
1
6
15
40
0
3
3
9
-

+


-
+
+


3
8
1
10. (a)
(b)
(c)
(d)
(e)
(f)
+
8
=
9
=
3


8
=
(c)
(f)
(i)
(l)
(o)
(r)
(u)
(x)
15
8
8
17
0
72
4
29
1
=
4
-
3
division
multiplication
subtraction
addition
subtraction
division
Order of Operations (pages 22 to 25)
1. (a) 4
(d) 11
(g) 30
(j) 26
(m) 28
(p) 59
(s) 7
(v) 23
(b) 23
(e) 14
(h) 13
(k) 1
(n) 44
(q) 26
(t) 1
(w) 3
2. Answers will vary.
Multiples of Ten, One Hundred and One Thousand (pages 26 to 32)
1. (a)
(d)
(g)
(j)
80
8000
150
1400
NSSAL
©2013
(b)
(e)
(h)
(k)
900
900
1100
11000
(c)
(f)
(i)
(l)
90
120
9000
80
1600
Draft
C. D. Pilmer
2. (a)
(d)
(g)
(j)
30
600
700
70
(b)
(e)
(h)
(k)
400
4000
80
800
(c)
(f)
(i)
(l)
0
20
9000
7000
3. (a)
(d)
(g)
(j)
210
36000
24000
350
(b)
(e)
(h)
(k)
1500
4800
8100
42000
(c)
(f)
(i)
(l)
16000
250
2400
8000
4. (a) 40
(d) 90
(g) 300
(j) 80
(m) 800
(p) 70
(b)
(e)
(h)
(k)
(n)
(q)
4
9
8
70
2
5
(c)
(f)
(i)
(l)
(o)
(r)
7
70
2
500
60
600
5. (a) 1400
(d) 110
(g) 600
(j) 1600
(m) 1300
(p) 80
(s) 70
(b)
(e)
(h)
(k)
(n)
(q)
(t)
50
10
1300
3
400
56000
3200
(c)
(f)
(i)
(l)
(o)
(r)
(u)
500
270
5000
14000
70
40
9
6. (a) 20 & 40
(d) 90 & 90
(g) 30 & 60
(b) 30 & 40
(e) 60 & 90
(h) 60 & 70
7. (a) 10, 20, 30 Puzzle
NSSAL
©2013
(c) 50 & 70
(f) 20 & 80
(i) 40 & 90
(b) 20, 30, 40 Puzzle
20
10
30
20
30
40
30
20
10
40
20
30
10
30
20
30
40
20
91
Draft
C. D. Pilmer
(c) 30, 40, 50 Puzzle
(d) 40, 50, 60 Puzzle
50
40
30
50
40
60
30
50
40
60
50
40
40
30
50
40
60
50
(e) 50, 60, 70 Puzzle
(f) 70, 80, 90 Puzzle
70
60
50
80
90
70
50
70
60
90
70
80
60
50
70
70
80
90
30
=
8. (a)
3

90
=
40
+
50

-
-
+
-
60
27
30
0
30
=
=
=
=
=
180

3
=
60
=
40
+
20
=
=
=
=
=
20
24
2
240
4

-



9
NSSAL
©2013
+
21
=
30
=
6
92

5
Draft
C. D. Pilmer
(b)

70
20
=
1400
=
200
+
1200
-
+
-


40
20
200
100
2
=
=
=
=
=

30
40
=
1200
=

2
600
=
=
=
=
=
0
4
900
7
10
+

+
-


30
10
=
300
=

5
60
Divisibility and Prime (pages 33 and 34)
1.
2
(a)
22
(c)
13
(e)
75
(g)
90
(i)
61
(k)
261
(m)
428
(o)
417
(q)
510
(s)
8001
(u)
3218
(w)
2870
(y)
2853
3
5
6
P
2





















(b)
55
(d)
72
(f)
29
(h)
63
(j)
86
(l)
440
(n)
546
(p)
948
(r)
179
(t)
4315
(v)
5202
(x)
1320
(z)
7302
3
5
6
P



























2. Answers will vary.
NSSAL
©2013
93
Draft
C. D. Pilmer
Estimating (pages 35 to 37)
1. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
unreasonable
reasonable
reasonable
unreasonable
reasonable
unreasonable
unreasonable
reasonable
2. (a)
(b)
(c)
(d)
(e)
(f)
(g)
1
4
9
2
3
4
8
Adding Multi-Digit Numbers (pages 38 to 40)
1. (a)
(d)
(g)
(j)
88
759
800
5 185
2. (a) 172
(b)
(e)
(h)
(k)
84
672
934
14 208
(c)
(f)
(i)
(l)
(b) 610
136
1 485
1 437
15 760
(c) 1286
3. 251 people
Subtracting Multi-Digit Numbers (pages 41 to 44)
1. (a) 25
(d) 127
(g) 1 054
(j) 1 245
(m) 6 334
(b)
(e)
(h)
(k)
(n)
57
380
226
3 860
11 437
2. (a) 616
(b) 1 156
(c)
(f)
(i)
(l)
(o)
153
187
126
4 168
771
(c) 4 867
3. $883
4. $213
NSSAL
©2013
94
Draft
C. D. Pilmer
5. 164 people
Multiplying Multi-Digit Numbers (pages 45 to 52)
1. (a) 1 794
(d) 2 337
(g) 6 090
(j) 33 856
(m) 37 346
(p) 343 332
(b)
(e)
(h)
(k)
(n)
(q)
4 662
27
19 512
731
1 125
272 667
(c)
(f)
(i)
(l)
(o)
(r)
131
1 820
283
75 945
226 044
365
(c)
(f)
(i)
(l)
423
358
3 127
3742
2. 216 people
3. 223 cubic yards
4. 384 entries
5. 347 litres
6. 26 600 seedlings
Dividing Multi-Digit Numbers (pages 53 to 60)
1. (a)
(d)
(g)
(j)
67
4482
583 R: 4
1 043 R: 1
(b)
(e)
(h)
(k)
59 R: 1
817 R: 2
627 R: 3
30 968
2. $785
3. 532 kilometres
4. 26 acres
5. $465
6. 32 cups
NSSAL
©2013
95
Draft
C. D. Pilmer