Download 5. factors and multiples

5. FACTORS AND MULTIPLES
5 - 1 Multiplying and dividing by whole numbers
2
5 - 2 Factor bingo
5
5 - 3 Common factors
6
5 - 4 Common multiples
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5 - 5 Divisibility tests
9
5 - 6 Prime and composite numbers
14
5 - 7 Divisibility hex
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5 - 8 Prime factorisation
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5 - 9 Using prime factors
18
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Activity 5 - 1
Multiplying and dividing by whole numbers
Multiples of cash.
The money in Alan’s hand must be a number of fifties eg. $900 (which is
18 fifties).
A guess of $1520 would be wrong because it is not a number of fifties.
The first multiple of 50 is 50 because 50 x 1 = 50.
The next 5 multiples of 50 are 100, 150, 200, 250, 300.
There are an infinite number of multiples of 50.
The first 12 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36.
You find multiples of 3 in the 3 times table.
You find multiples of 12 in the 12 times table.
A box in a maths classroom contains the cubes pictured below.
From this picture, you know that
the number of cubes in the box is:
- a multiple of 5 because it is made up
of columns of 5 blocks.
- a multiple of 7 because it is made up
of rows of 7 blocks.
- a multiple of 35 because it is made up
of slices of 35 blocks.
The number of cubes could also be a
multiple of 1, 11, 55, 77 and 385.
The number of cubes= 5 x 7 x 11
= 385
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Job dividends
SCOUT
JOBS PAYMENT
The recorded number of jobs done is
incorrect for Tim.
Steve
10
$190
Mal
5
$ 95
Roger
5
$ 95
Geoff
8
$152
Frank
9
$171
Ken
3
$ 57
Now that the error is corrected, each
payment is divisible by 19.
Alan
6
$114
Each payment is also a multiple of 19.
Tim
4 3
$ 57
Andy
4
$ 76
Bill
10
$190
Ben
5
$ 95
Mike
5
$ 95
To find out how much money the scouts
made altogether, the payments were
added together. The total amount of
money should be divisible by the total
number of jobs because the same
payment was made for each job.
Paul
8
$152
Lauren
3
$ 57
Abdul
3
$ 57
Israel
5
$ 95
Suraj
6
$114
Pradeep
10
$190
Sailosi
7
$133
Kyron
10
$190
Greg
7
$133
Phil
8
$152
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You could tell there was an error here
because 57 is not a multiple of 4.
The total should also be divisible by 19.
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Knitting arrangements
Listing all the possible rectangular arrangements for the knitted squares,
starting with the narrowest and longest arrangement (1 x 60):
1 x 60
2 x 30
3 x 20
4 x 15
5 x 12
6 x 10
60 is a multiple of 1 because 1 x 60 = 60
60 is a multiple of 2 because 2 x 30 = 60
60 is also a multiple of 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
The numbers that 60 is divisible by are 1, 2, 3, 4, 5, 6, 10,
12, 15, 20, 30, 60.
Mrs Smart suggested that there was a faster way to find them.
Because 2 is a factor of 60, she knew that 30 must also be a factor
because 60 = 2 x 30.
If 3 is a factor of 60, 20 must also be a factor of 60.
She continued finding pairs of factors this way.
When she tested 7, she found it was not a factor.
When she came to test 8, she said she didn’t need to test any more
numbers because 60  8 = 7.5 and she has already found all the whole
numbers smaller than this.
The factors of each of the following numbers are:
100: 1, 2, 4, 5, 10, 20, 25, 50, 100
144: 1, 2, 3, 4, 6, 8, 12, 18, 24, 36, 72, 144
400: 1, 2, 4, 5, 8, 10, 16, 20, 25, 10, 50, 80, 100, 200, 400
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Activity 5 - 2
Factor Bingo
1) Choose the numbers that have the most factors. Larger numbers
generally have more factors than smaller numbers.
2) Another strategy is to write the number with the most factors in the
middle square and those with the next highest numbers of factors in
the corner squares.
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Activity 5 - 3
Common factors
Sandbags
Each sack, when filled with sand would be 0.5 m long, so one metre
would require 2 sacks.
To calculate the number of sacks they required altogether, Greg pulled
out his calculator and entered:
(131 x 2 x 6) + (89 x 2 x 6) + (97 x 2 x 6) + (83 x 2 x 6) = 4800
Each term in Greg’s expression has 3 factors.
The factors 2 and 6 are in every term.
(131 + 89 + 97 + 83) x 2 x 6 = 4800
Yes. You would you expect Greg and Phil to get the same answer.
Another common factor of the terms in Greg’s calculation is 12.
You know that 12 is a common factor because 2 x 6 = 12.
Buying drinks
Total cost = 3 x 293 + 6 x 293 + 9 x 293 + 4 x 336 + 2 x 336
= 879 + 1758 + 2637 + 1344 + 672
The first 3 terms in the calculation have 293 as a common factor.
Another common factor of the first 3 terms is 3 because 3, 6 and 9 are
all divisible by 3.
The HCF factor of the last 2 terms: HCF= 2 x 336 = 672
The calculation as 2 terms, with the highest common factor of each term
in front of a set of brackets:
Total cost = 879(1 + 2 + 3) + 672(2 + 1)
= 879 x 6 + 672 x 3
= 7290 cents
= $ 72.90
Lists of factors:
60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
24: 1, 2, 3, 4, 6, 8, 12, 24
The common factors of these 3 numbers: 1, 2, 3, 4, 6, 12
The highest common factor of these 3 numbers is 12.
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Activity 5 - 4
Common multiples
Sand timers
1
7
13
19
25
31
37
43
49
55
61
67
73
79
85
91
2
8
14
20
26
32
38
44
50
56
62
68
74
80
86
92
3
9
15
21
27
33
39
45
51
57
63
69
75
81
87
93
4
10
16
22
28
34
40
46
52
58
64
70
76
82
88
94
5
11
17
23
29
35
41
47
53
59
65
71
77
83
89
95
6
12
18
24
30
36
42
48
54
60
66
72
78
84
90
96
If the time taken to turn over a sand timer
doesn’t count, time intervals of 35 minutes and
70 minutes can be measured by either the 5
minute timer or the 7 minute timer.
You could you get your answer from looking at
the number chart because 35 and 70 are the
numbers coloured twice.
You could get your answer from the numbers 5
and 7 because 35 and 70 are multiples of both
5 and 7.
Three other common multiples of 5 and 7 are
(for example) 105, 140 and 175.
There are an infinite number of common
multiples of 5 and 7.
Model trains
The trains pass each other at the station after 180 seconds because 180
is divisible by both 12 and 15.
Lists of the first 15 multiples:
12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180
15: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225
The LCM of 12 and 15 is 60.
Flynn will have to wait 1 minute (60 seconds) until he sees the two trains
passing each other at the station again.
Lists of multiples:
1) 8 and 20
8: 8, 16, 24, 32, 40
20: 20, 40
LCM = 40
2) 15 and 20 15: 15, 30, 45, 60
20: 20, 40, 60
LCM = 60
3) 30 and 45 30: 30, 60, 90
45: 45, 90
LCM = 90
The LCM of two numbers is always larger than their HCF.
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Dancers
Lists of the first 10 multiples:
8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80
6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
The LCM of 8, 6 and 4 is 24.
The girls re-choreographed the dance so that it no longer needed groups
of 4. It only needed groups of 8 or 6.
No. this does not change the minimum number of dancers they require
because the LCM of 8 and 6 is 24.
(This happens because 8 is a multiple of 4. Any number that is a multiple
of 8 is also a multiple of 4.)
1) 3, 6 and 5
LCM = 30
2) 3, 14 and 7
LCM = 42
3) 2, 3, 6 and 13.
LCM = 78
4) 2, 3, 11 and 66.
LCM = 66
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Activity 5 - 5
Divisibility Tests
Divisibility by factors of 10
Without making the division, you know that the number 131 880 is
divisible by 10 because it ends in a 0 digit.
Each of the expressions below is equal to 131 880:
13 188 x 10
13 188 x 2 x 5
13 188 x 5 x 2
Without making the division, you know that any number that is divisible
by 10 is also divisible by 5 and by 2 because 5 and 2 are factors of 10.
131 885 can be written as two terms: 131 885 = 131 880 + 5.
5 is a common factor of these terms: 131 885 = (13 188 x 2 x 5) + (1 x 5)
= (13 188 x 2 + 1) x 5
So 131 885 is divisible by 5.
Any number ending in 5 is divisible by 5 because it can be split into a
number ending in a 0 and a number ending in a 5, and both these
numbers are divisible by 5.
131 882 can be written as two terms: 131 882 = 131 880 + 2.
2 is a common factor of these terms: 131 882 = (13 188 x 5 x 2) + (1 x 2)
= (13 188 x 5 + 1) x 2
So 131 882 is divisible by 2.
Any number ending in 2 is divisible by 2 because it can be split into a
number ending in 0 and a 2, and both these numbers are divisible by 2.
131 884 = 131 880 + 4
= (65 940 x 2) + (2 x 2)
= 65 942 x 2
Other digits that numbers can end with, and be divisible by 2 are 6 and
8.
Numbers that are divisible by 2 are called even numbers.
Summary:
 a number is divisible by 10 if the last digit is 0.
 a number is divisible by 5 if the last digit is 0 or 5.
 a number is divisible by 2 if the last digit is 0, 2, 4, 6, or 8.
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Divisibility by factors of 100
Without making the division, you know that the number 131 800 is
divisible by 100 because it ends in 00.
Each of the expressions below is equal to 131 800:
1318 x 100
1318 x 2 x 50
1318 x 5 x 20
1318 x 10 x 10
1318 x 20 x 5
1318 x 50 x 2
1318 x 25 x 4
1318 x 4 x 25
So, any number ending with 00 is divisible by 100, 50, 20, 10, 5, 2, 4 and
25.
If a number is divisible
by 20, its last 2 digits are 00, 20, 40, 60, or 80.
by 25, its last 2 digits are 00, 25, 50 or 75.
by 4, its last 2 digits are 00, 04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44,
48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92 or 96.
Therefore, to find out whether a number is divisible by 4, just find out
whether the last 2 digits are divisible by 4.
Divisibility by factors of 1000
Without making the division, you know that the number 131 000 is
divisible by 1000 because it ends in 000.
Each of the expressions below is equal to 131 000:
131 x 1000
131 x 2 x 500
131 x 5 x 200
131 x 10 x 100
131 x 20 x 50
131 x 50 x 20
131 x 100 x 10
131 x 200 x 5
131 x 500 x 2
131 x 25 x 40
131 x 4 x 250
131 x 250 x 4
131 x 40 x 25
131 x 8 x 125
131 x 125 x 8
So, any number ending with 000 is divisible by 1000, 500, 200, 100, 50,
20, 10, 5, 2, 40, 250, 4, 25, 125 or 8,
If a number is divisible
by 250, its last 3 digits are 000, 250, 500 or 750
by 125 its last 3 digits are 000, 125, 250, 375, 500, 625, 750 or 875
Some 3 digit endings that mean that a number is divisible by 8 are:
008, 016, 024, 032, 040, 048, 056, 064, 072, 080, 088, 096, 104 etc.
Summary:
 a number is divisible by 4 if the last 2 digits of the number are
divisible by 4.
 a number is divisible by 8 if the last 3 digits of the number are
divisible by 8.
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Divisibility by 3
Altogether, there are 1110 small cubes in the blocks below.
1000
+
100
+
10
If each of the terms above are divided into groups of 3 small cubes,
1 cube will be left over from the 1000 cubes.
1 cube will be left over from the 100 cubes.
1 cube will be left over from the 10 cubes.
Altogether the number of left-over cubes will be 1 + 1 + 1 = 3
If these three lots of left-over cubes are put in groups of 3, there will be 0
cubes left over.
Yes. The total number is divisible by 3.
Altogether, there are 1312 small cubes in the blocks below.
1000
+
300
+
10
+
2
If each of the terms above are divided into groups of 3 small cubes,
The number of left-over cubes will be 1 + 3 + 1 + 2 = 7
If these four lots of left-over cubes are put in groups of 3, there will then
be 1 left over.
No. The total number is not divisible by 3.
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Divisibility by 9
1110 = 1000 + 100 + 10
If each of the terms above are divided by 9,
There will be 1 left over from 1000.
There will be 1 left over from 100.
There will be 1 left over from 10.
Altogether the number left over is 1 + 1 + 1 = 3
No. 1110 is not divisible by 9.
Yes. The number 1116 is divisible by 9 because the numbers left over
are 1 + 1 + 1 + 6 = 9. Another group of 9 is formed from the left-over
cubes so there are no more left over.
No. All numbers that are divisible by 3, are not also divisible by 9
because if 3 or 6 are left over they form groups of 3 but not a group of 9.
Yes. All numbers that are divisible by 9, are also divisible by 3 because
every group of 9 can be made into 3 groups of 3.
Divisibility by 6 and 12
7120 1116 36123 2220 2200 13189
Numbers divisible by 2: 7120, 1116, 2220, 2200.
Numbers divisible by 3: 1116, 36123, 2220.
Numbers divisible by 6: 1116, 2220.
Yes. All numbers that are divisible by 6, are also divisible by 2 because
every group of 6 can be made into 3 groups of 2.
No. All numbers that are divisible by 2, are not also divisible by 6
because, for example, 2 and 4 are divisible by 2 but not by 6.
Yes. All numbers that are divisible by 2 and 3, are also divisible by 6
because the factors 2 and 3 when multiplied together give 6.
If a number is divisible by 3 and by 4, it is divisible by all the factors of 12
(because 3 x 4 = 12) i.e. it is also divisible by 1, 2, 6 and 12.
Summary:
 a number is divisible by 3 if the sum of its digits is divisible by 3.
 a number is divisible by 9 if the sum of its digits is divisible by 9.
 a number is divisible by 6 if it is divisible by 2 and by 3.
 a number is divisible by 12 if it is divisible by 3 and by 4.
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Divisibility by 11
Look at the pattern that occurs when dividing by 11, then continue it:
10 = 11 - 1
100 = (9 x 11) + 1
1 000 = (91 x 11) - 1
10 000 = (909 x 11) + 1
100 000 = (9091 x 11) - 1
1 000 000 = (90909 x 11) + 1
10 000 000 = (909091 x 11) - 1
100 000 000 = (9090909 x 11) + 1
1 000 000 000 = (90909091 x 11) -1
To test 131 888, 8 - 8 + 8 - 1 + 3 - 1 = 9
No. 131 888 is not divisible by 11.
2
28080
55440
15836
73210
65093
14641
3
4
5
6
7
8
9
10
11
12
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   
        







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Activity 5 - 6
Prime Numbers
2 is prime (ie. a prime number) because its only factors are 1 and 2.
3 is prime because its only factors are 1 and 3.
The number 1 is not a prime number or a composite number because it
has only one factor (not 2 or more than 2).
Multiples of 2 (larger than 2) are composite numbers because they have
at least 3 factors (1, 2 and the number itself).
X1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100 101 102 103 104 105 106 107 108 109 110
111 112 113 114 115 116 117 118 119 120 121
There are 12 prime numbers between 0 and 40.
There are 11 prime numbers between 40 and 80.
There are 9 prime numbers between 80 and 120.
As the numbers get larger, there are fewer prime numbers because
larger numbers have more numbers smaller than them and hence more
possible factors
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The only prime number that is even is 2.
There can be no other even prime number because every even number
is divisible by 2, 1 and itself.
3 and 5 are prime numbers which differ by 2.
Other prime numbers on the chart that differ by 2 are 5 and 7, 11 and 13,
17 and 19, 29 and 31, 41 and 43, 43 and 47, 59 and 61, 71 and 73, 77
and 79, 101 and 103, 107 and 109.
7 and 11 are prime numbers which differ by 4.
Other prime numbers on the chart that differ by 4 are 13 and 17,19 and
23 37 and 41, 67 and 71, 79 and 83, 97 and 101,103 and 107,109 and
113
2 and 5 are prime numbers which differ by 3.
There are no other prime numbers that differ by 3.
This is because if numbers differ by 3, one must be odd and the other
must be even. Since 2 is the only even prime number, this cannot
happen.
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Activity 5 - 7
Divisibility Hex
You need to develop your own strategies for this game.
It helps to:
1) know the divisibility tests (Activity 5 -4).
2) know the prime numbers in the chart in Activity 5 -6 and use them
as much as possible.
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Activity 5 - 8
Prime factorisation
If a branch cannot be divided, the number at the end must be a prime
number.
Division by primes
1)
3)189
3) 63
3) 21
7) 7
1
189 = 33 x 7
© 2008, McMaster & Mitchelmore
2)
2)484
2)242
11)121
11) 11
1
484 = 22 x 112
17
3)
5)1225
5) 245
7) 49
7)
7
1
1225 = 52 x 72
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Activity 5 - 9
Using prime factors
Finding the HCF and the LCM of two large numbers
2)756
2)378
3)189
3) 63
3) 21
7) 7
1
2)360
2)180
2) 90
3) 45
3) 15
5) 5
1
756 = 22 x 33 x 7
360 =23x 32 x 5
The HCF is the largest number that both numbers can be divided by.
Both numbers have 2 as a prime factor.
Yes. Both have (2 x 2) as a common factor.
No. They do not both have (2 x 2 x 2) as a common factor.
Yes. Both have 3 as a common factor.
Yes. Both have (3 x 3) as a common factor.
No. They do not both have (3 x 3 x 3) as a common factor.
No. Both numbers have no more prime factors in common.
So the highest multiples of prime numbers that both numbers have in
common are 2 x 2 and 3 x 3.
HCF = 22 x 32
= 36
The LCM is the smallest number that is a multiple of both 756 and 360,
so it must have all the prime factors (including repeats) for 756 and 360.
The factor 2 must be repeated 3 times in the LCM.
The factor 3 must be repeated 3 times in the LCM.
The factor 5 must be repeated once in the LCM.
The other prime factor that either 756 or 360 have is 7.
LCM = 23 x 33 x 5 x 7
= 7560
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HCF = 2 x 2 x 3 x 3
= 36
2) 756
2) 378
3) 189
3) 63
21
360
180
90
30
10
LCM = 2 x 2 x 3 x 3 x 21 x 10
= 36 x 21 x 10
= 7560
The LCM must be divisible by the HCF because the HCF is a factor of
both numbers and both numbers are factors of the LCM.
Use the method above to find the HCF and LCM of 1650 and 1155.
HCF = 3 x 5 x 11
= 165
3) 1650
5) 550
11) 110
10
1155
385
77
7
LCM = 3 x 5 x 11 x 10 x 7
= 11 550
Finding the HCF and LCM of more than two large numbers
The three numbers below have already been expressed as products of
their prime factors.
550 550 = 2 x 5 x 5 x 7 x 11 x 11 x 13
115 500 = 2 x 2 x 3 x 5 x 5 x 5 x 7 x 11
193 050 = 2 x 3 x 3 x 3 x 5 x 5 x 11 x 13
HCF = 2 x 5 x 5 x 11
= 2 x 52 x 11
LCM = 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 7 x 11 x 11 x 13
= 22 x 33 x 53 x 7 x 112 x 13
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Finding roots
1) 784
3
2) 3375
2)784
2)392
2)196
2) 98
7) 49
7) 7
1
3)3375
3)1125
3) 375
5) 125
5) 25
5)
5
1
3375 = 33 x 53
=3x5
3
784 = 28
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3
3
784 = 24 x 72
= 22 x 7
3375 = 15
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