Checkpoint 1
Assess Your Understanding (page 149)
3.1
1. Use a calculator to divide each number by its prime factors.
a) 1260 = 2 · 2 · 3 · 3 · 5 · 7
= 22 · 32 · 5 · 7
b) 4224 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 3 · 11
= 27 · 3 · 11
c) 6120 = 2 · 2 · 2 · 3 · 3 · 5 · 17
= 23 · 32 · 5 · 17
d) 1045 = 5 · 11 · 19
e) 3024 = 2 · 2 · 2 · 2 · 3 · 3 · 3 · 7
= 24 · 33 · 7
f) 3675 = 3 · 5 · 5 · 7 · 7
= 3 · 52 · 72
2. List the factors of each number, then highlight the common factors.
a) 40 = 2 · 2 · 2 · 5
48 = 2 · 2 · 2 · 6
56 = 2 · 2 · 2 · 7
The greatest common factor is: 2 · 2 · 2 = 8
b) 84 = 2 · 2 · 3 · 7
120 = 2 · 2 · 2 · 3 · 5
144 = 2 · 2 · 2 · 2 · 3 · 3
The greatest common factor is: 2 · 2 · 3 = 12
c) 145 = 5 · 29
205 = 5 · 41
320 = 2 · 2 · 2 · 2 · 2 · 2 · 5
The greatest common factor is: 5
d) 208 = 2 · 2 · 2 · 2 · 13
368 = 2 · 2 · 2 · 2 · 23
528 = 2 · 2 · 2 · 2 · 3 · 11
The greatest common factor is: 2 · 2 · 2 · 2 = 16
e) 856 = 2 · 2 · 2 · 107
1200 = 2 · 2 · 2 · 2 · 3 · 5 · 5
1368 = 2 · 2 · 2 · 3 · 3 · 19
The greatest common factor is: 2 · 2 · 2 = 8
f) 950 = 2 · 5 · 5 · 19
1225 = 5 · 5 · 7 · 7
1550 = 2 · 5 · 5 · 31
The greatest common factor is: 5 · 5 = 25
3. Factor each number.
Highlight the greatest power of each prime factor in any list.
The least common multiple is the product of the greatest power of each prime factor.
a) 12 = 2 · 2 · 3 = 22 · 3
15 = 3 · 5
21 = 3 · 7
The least common multiple is: 22 · 3 · 5 · 7 = 420
b) 12 = 2 · 2 · 3 = 22 · 3
20 = 2 · 2 · 5 = 22 · 5
32 = 2 · 2 · 2 · 2 · 2 = 25
The least common multiple is: 25 · 3 · 5 = 480
c) 18 = 2 · 3 · 3 = 2 · 32
24 = 2 · 2 · 2 · 3 = 23 · 3
30 = 2 · 3 · 5
The least common multiple is: 23 · 32 · 5 = 360
d) 30 = 2 · 3 · 5
32 = 2 · 2 · 2 · 2 · 2 = 25
40 = 2 · 2 · 2 · 5 = 23 · 5
The least common multiple is: 25 · 3 · 5 = 480
e) 49 = 7 · 7 = 72
56 = 2 · 2 · 2 · 7 = 23 · 7
64 = 2 · 2 · 2 · 2 · 2 · 2 = 26
The least common multiple is: 26 · 72 = 3136
f) 50 = 2 · 5 · 5 = 2 · 52
55 = 5 · 11
66 = 2 · 3 · 11
The least common multiple is: 2 · 3 · 52 · 11 = 1650
4. a)
b)
c)
8
5
88 15
+
=
+
3 11 33
33
103
=
33
13 4
91 20
–
=
–
5
7
35 35
71
=
35
9
7
27
70
÷
=
÷
10
3
30
30
27
=
70
5. Find the least common multiple of 365 and 260.
365 = 5 · 73
260 = 2 · 2 · 5 · 13
The least common multiple is: 2 · 2 · 5 · 13 · 73 = 18 980
The first day on both calendars would occur again after 18 980 days, which is
18 980
years, or 52
365
years.
3.2
6. I could list the prime factors, then arrange them in two equal groups, or I could estimate then use
guess and check. I will use prime factors.
a) 400 = 2 · 2 · 2 · 2 · 5 · 5
= (2 · 2 · 5)(2 · 2 · 5)
= 20 · 20
400 = 20
b)
784 = 2 · 2 · 2 · 2 · 7 · 7
= (2 · 2 · 7)(2 · 2 · 7)
= 28 · 28
784 = 28
c)
576 = 2 · 2 · 2 · 2 · 2 · 2 · 3 · 3
= (2 · 2 · 2 · 3)(2 · 2 · 2 · 3)
= 24 · 24
576 = 24
d)
1089 = 3 · 3 · 11 · 11
= (3 · 11)(3 · 11)
= 33 · 33
1089 = 33
e)
1521 = 3 · 3 · 13 · 13
= (3 · 13)(3 · 13)
= 39 · 39
1521 = 39
f)
3025 = 5 · 5 · 11 · 11
= (5 · 11)(5 · 11)
= 55 · 55
3025 = 55
7. I could list the prime factors, then arrange them in three equal groups, or I could estimate then use
guess and check. I will use prime factors.
a) 1728 = 2 · 2 · 2 · 2 · 2 · 2 · 3 · 3 · 3
= (2 · 2 · 3)(2 · 2 · 3)(2 · 2 · 3)
= 12 · 12 · 12
3
1728 = 12
b) 3375 = 3 · 3 · 3 · 5 · 5 · 5
= (3 · 5)(3 · 5)(3 · 5)
= 15 · 15 · 15
3
3375 = 15
c) 8000 = 2 · 2 · 2 · 2 · 2 · 2 · 5 · 5 · 5
= (2 · 2 · 5)(2 · 2 · 5)(2 · 2 · 5)
= 20 · 20 · 20
3
8000 = 20
d) 5832 = 2 · 2 · 2 · 3 · 3 · 3 · 3 · 3 · 3
= (2 · 3 · 3)(2 · 3 · 3)(2 · 3 · 3)
= 18 · 18 · 18
3
5832 = 18
e) 10 648 = 2 · 2 · 2 · 11 · 11 · 11
= (2 · 11)(2 · 11)(2 · 11)
= 22 · 22 · 22
3 10 648 = 22
f) 9261 = 3 · 3 · 3 · 7 · 7 · 7
= (3 · 7)(3 · 7)(3 · 7)
= 21 · 21 · 21
3
9261 = 21
8. Use a calculator to write each number as a product of its prime factors. If the factors can be arranged
in 2 equal groups, then the number is a perfect square. If the factors can be arranged in 3 equal
groups, then the number is a perfect cube.
a) 2808 = 2 · 2 · 2 · 3 · 3 · 3 · 13
The factors cannot be arranged in 2 or 3 equal groups, so 2808 is neither a perfect square nor a
perfect cube.
b) 3136 = 2 · 2 · 2 · 2 · 2 · 2 · 7 · 7
= (2 · 2 · 2 · 7)(2 · 2 · 2 · 7)
The factors can be arranged in 2 equal groups, so 3136 is a perfect square.
c) 4096 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
= (2 · 2 · 2 · 2 · 2 · 2)(2 · 2 · 2 · 2 · 2 · 2)
= (2 · 2 · 2 · 2)(2 · 2 · 2 · 2)(2 · 2 · 2 · 2)
The factors can be arranged in 2 equal groups and 3 equal groups, so 4096 is both a perfect square
and a perfect cube.
d) 4624 = 2 · 2 · 2 · 2 · 17 · 17
= (2 · 2 · 17)(2 · 2 · 17)
The factors can be arranged in 2 equal groups, so 4624 is a perfect square.
e) 5832 = 2 · 2 · 2 · 3 · 3 · 3 · 3 · 3 · 3
= (2 · 3 · 3)(2 · 3 · 3)(2 · 3 · 3)
The factors can be arranged in 3 equal groups, so 5832 is a perfect cube.
f) 9270 = 2 · 3 · 3 · 5 · 103
The factors cannot be arranged in 2 or 3 equal groups, so 9270 is neither a perfect square nor a
perfect cube.
9. Use estimation or guess and check to determine the perfect square and perfect cube closest to the first
number in each given pair, then calculate the squares and cubes of all whole numbers until the second
number in each pair is reached or exceeded.
a) 400 – 500
202 = 400
212 = 441
222 = 484
232 = 529
3
3
7 = 343
8 = 512
Between 400 and 500, the perfect squares are 441 and 484; there are no perfect cubes.
b) 900 – 1000
302 = 900
312 = 961
322 = 1024
3
3
9 = 729
10 = 1000
Between 900 and 1000, the perfect square is 961; there are no perfect cubes.
c) 1100 – 1175
332 = 1089
342 = 1156
352 = 1225
3
3
10 = 1000
11 = 1331
Between 1100 and 1175, the perfect square is 1156; there are no perfect cubes.
10. The edge length of the cube, in metres, is 3 2197 .
2197 = 13 · 13 · 13
So, 3 2197 = 13
The surface area of one face, in square metres, is: 132 = 169
The surface area of the cube, in square metres, is: 6(169) = 1014
One can of paint covers about 40 m2.
1014
So, the number of cans of paint needed is:
= 25.35
40
Twenty-six cans of paint are needed.