Factors and Divisors
Key Concepts
Solved Examples: 41
Practice Questions: 5
Solved Examples
Question 1: Find all the factors of 126.
Find the factor pairs of 126:
1 × 126 = 126
2 × 63 = 126
3 × 42 = 126
6 × 21 = 126
7 × 18 = 126
9 × 14 = 126
Therefore, the factors of 126 are: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126
1 × 126 = 126
2 × 63 = 126
3 × 42 = 126
6 × 21 = 126
7 × 18 = 126
9 × 14 = 126
Therefore, the factors of 126 are: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126
Question 2: Express 540 as the product of its prime factors.
Divide the number successively by the smallest possible prime numbers.
540 ÷ 2 = 270
270 ÷ 2 = 135
135 ÷ 3 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Therefore, 540 = 2² × 3³ × 5
540 ÷ 2 = 270
270 ÷ 2 = 135
135 ÷ 3 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Therefore, 540 = 2² × 3³ × 5
Question 3: Find the total number of factors of 360.
First, express 360 as the product of its prime factors.
360 = 2³ × 3² × 5¹
The number of factors is given by:
(3 + 1)(2 + 1)(1 + 1)
= 4 × 3 × 2
= 24
360 = 2³ × 3² × 5¹
The number of factors is given by:
(3 + 1)(2 + 1)(1 + 1)
= 4 × 3 × 2
= 24
Question 4: The prime factorisation of a number is 2⁴ × 3² × 5³. Find the total number of factors of the number.
Using the formula,
Number of factors
= (4 + 1)(2 + 1)(3 + 1)
= 5 × 3 × 4
= 60
Number of factors
= (4 + 1)(2 + 1)(3 + 1)
= 5 × 3 × 4
= 60
Question 5: Find the smallest positive integer that has exactly 12 factors.
Let the required number be of the form:
N = pᵃ × qᵇ × rᶜ, where p, q, r are prime numbers.
The total number of factors of N is given by:
(a + 1)(b + 1)(c + 1)
Since the number has exactly 12 factors,
(a + 1)(b + 1)(c + 1) = 12
The possible factorizations of 12 are:
12
6 × 2
4 × 3
3 × 2 × 2
The corresponding forms of the number are:
p¹¹ → Smallest possible number = 2¹¹ = 2048
p⁵q → Smallest possible number = 2⁵ × 3 = 96
p³q² → Smallest possible number = 2³ × 3² = 72
p²qr → Smallest possible number = 2² × 3 × 5 = 60
Among these, the smallest number is 60.
N = pᵃ × qᵇ × rᶜ, where p, q, r are prime numbers.
The total number of factors of N is given by:
(a + 1)(b + 1)(c + 1)
Since the number has exactly 12 factors,
(a + 1)(b + 1)(c + 1) = 12
The possible factorizations of 12 are:
12
6 × 2
4 × 3
3 × 2 × 2
The corresponding forms of the number are:
p¹¹ → Smallest possible number = 2¹¹ = 2048
p⁵q → Smallest possible number = 2⁵ × 3 = 96
p³q² → Smallest possible number = 2³ × 3² = 72
p²qr → Smallest possible number = 2² × 3 × 5 = 60
Among these, the smallest number is 60.
Question 6: Find the sum of all the factors of 360.
First, express 360 as the product of its prime factors.
360 = 2³ × 3² × 5
The sum of all factors is given by:
(1 + 2 + 2² + 2³)(1 + 3 + 3²)(1 + 5)
= (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5)
= 15 × 13 × 6
= 1170
360 = 2³ × 3² × 5
The sum of all factors is given by:
(1 + 2 + 2² + 2³)(1 + 3 + 3²)(1 + 5)
= (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5)
= 15 × 13 × 6
= 1170
Question 7: Find the product of all the factors of 12.
Prime factorisation: 12 = 2² × 3¹
Number of factors
= (2 + 1)(1 + 1)
= 3 × 2
= 6
The product of all the factors is given by:
Product of Factors = N^(Number of Factors ÷ 2)
Therefore, product of Factors
= 12^(6 ÷ 2)
= 12³
= 1728
Number of factors
= (2 + 1)(1 + 1)
= 3 × 2
= 6
The product of all the factors is given by:
Product of Factors = N^(Number of Factors ÷ 2)
Therefore, product of Factors
= 12^(6 ÷ 2)
= 12³
= 1728
Question 8: Find the number of odd factors of 360.
Prime factorisation: 360 = 2³ × 3² × 5¹
The factor 2³ contributes only to the even factors.
To obtain the odd factors, ignore the power of 2.
Therefore, odd factors are obtained from
3² × 5¹
Number of odd factors
= (2 + 1)(1 + 1)
= 3 × 2
= 6
The factor 2³ contributes only to the even factors.
To obtain the odd factors, ignore the power of 2.
Therefore, odd factors are obtained from
3² × 5¹
Number of odd factors
= (2 + 1)(1 + 1)
= 3 × 2
= 6
Question 9: Find the number of even factors of 540.
Prime factorisation: 540 = 2² × 3³ × 5¹
Total number of factors
= (2 + 1)(3 + 1)(1 + 1)
= 3 × 4 × 2
= 24
Number of odd factors
Ignore the power of 2.
= (3 + 1)(1 + 1)
= 4 × 2
= 8
Therefore,
Number of even factors = Total factors − Odd factors
= 24 − 8
= 16
Total number of factors
= (2 + 1)(3 + 1)(1 + 1)
= 3 × 4 × 2
= 24
Number of odd factors
Ignore the power of 2.
= (3 + 1)(1 + 1)
= 4 × 2
= 8
Therefore,
Number of even factors = Total factors − Odd factors
= 24 − 8
= 16
Question 10: Find the number of odd and even factors of 756.
Prime factorisation:
756 = 2² × 3³ × 7¹
Total number of factors
= (2 + 1)(3 + 1)(1 + 1)
= 3 × 4 × 2
= 24
Number of odd factors
Ignore the power of 2.
= (3 + 1)(1 + 1)
= 4 × 2
= 8
Number of even factors
= 24 − 8
= 16
Answer: Odd factors = 8 and Even factors = 16
756 = 2² × 3³ × 7¹
Total number of factors
= (2 + 1)(3 + 1)(1 + 1)
= 3 × 4 × 2
= 24
Number of odd factors
Ignore the power of 2.
= (3 + 1)(1 + 1)
= 4 × 2
= 8
Number of even factors
= 24 − 8
= 16
Answer: Odd factors = 8 and Even factors = 16
Question 11: Find the number of distinct prime factors of 420.
Prime factorisation of 420
420 = 2² × 3 × 5 × 7
The distinct prime factors are 2, 3, 5 and 7
Therefore, the number of distinct prime factors is 4
420 = 2² × 3 × 5 × 7
The distinct prime factors are 2, 3, 5 and 7
Therefore, the number of distinct prime factors is 4
Question 12: Find the greatest prime factor of 1386.
Prime factorisation:
1386 ÷ 2 = 693
693 ÷ 3 = 231
231 ÷ 3 = 77
77 ÷ 7 = 11
11 ÷ 11 = 1
Therefore,
1386 = 2 × 3² × 7 × 11
The greatest prime factor is 11.
1386 ÷ 2 = 693
693 ÷ 3 = 231
231 ÷ 3 = 77
77 ÷ 7 = 11
11 ÷ 11 = 1
Therefore,
1386 = 2 × 3² × 7 × 11
The greatest prime factor is 11.
Question 13: Find the least composite factor of 990.
Prime factorisation of 990
990 = 2 × 3² × 5 × 11
The factors of 990 include
1, 2, 3, 5, 6, …
Here
1 is neither prime nor composite.
2, 3 and 5 are prime numbers.
6 is composite.
Therefore, the least composite factor of 990 is 6
990 = 2 × 3² × 5 × 11
The factors of 990 include
1, 2, 3, 5, 6, …
Here
1 is neither prime nor composite.
2, 3 and 5 are prime numbers.
6 is composite.
Therefore, the least composite factor of 990 is 6
Question 14: Find the least number by which 360 must be multiplied to make it a perfect square.
Prime factorisation:
360 = 2³ × 3² × 5¹
For a perfect square, every prime factor must have an even exponent.
The exponents of 2 and 5 are odd.
Multiply by 2 × 5 = 10.
Now,
360 × 10
= 2⁴ × 3² × 5²
which is a perfect square.
Answer: 10
360 = 2³ × 3² × 5¹
For a perfect square, every prime factor must have an even exponent.
The exponents of 2 and 5 are odd.
Multiply by 2 × 5 = 10.
Now,
360 × 10
= 2⁴ × 3² × 5²
which is a perfect square.
Answer: 10
Question 15: Find the least number by which 1080 must be divided to make it a perfect cube.
Prime factorisation:
1080 = 2³ × 3³ × 5¹
For a perfect cube, every exponent must be a multiple of 3.
The exponent of 5 is 1.
Divide by 5.
Then,
1080 ÷ 5
= 2³ × 3³
= 216
= 6³
Answer: 5
1080 = 2³ × 3³ × 5¹
For a perfect cube, every exponent must be a multiple of 3.
The exponent of 5 is 1.
Divide by 5.
Then,
1080 ÷ 5
= 2³ × 3³
= 216
= 6³
Answer: 5
Question 16: Find the largest perfect square that divides 1764.
Prime factorisation:
1764 = 2² × 3² × 7²
Every exponent is even.
Hence, 1764 itself is a perfect square.
Answer: 1764
1764 = 2² × 3² × 7²
Every exponent is even.
Hence, 1764 itself is a perfect square.
Answer: 1764
Question 17: Find the largest perfect cube that divides 4320.
Prime factorisation:
4320 = 2⁵ × 3³ × 5¹
For a perfect cube, each exponent must be a multiple of 3.
Take the greatest multiples of 3 not exceeding each exponent.
2⁵ → 2³
3³ → 3³
5¹ → 5⁰
Largest perfect cube
= 2³ × 3³
= 8 × 27
= 216
Answer: 216
4320 = 2⁵ × 3³ × 5¹
For a perfect cube, each exponent must be a multiple of 3.
Take the greatest multiples of 3 not exceeding each exponent.
2⁵ → 2³
3³ → 3³
5¹ → 5⁰
Largest perfect cube
= 2³ × 3³
= 8 × 27
= 216
Answer: 216
Question 18: Find the smallest perfect square that is divisible by 540.
Prime factorisation:
540 = 2² × 3³ × 5¹
For a perfect square, all exponents must be even.
Multiply by 3 × 5 = 15.
Then,
540 × 15
= 2² × 3⁴ × 5²
= 8100
Answer: 8100
540 = 2² × 3³ × 5¹
For a perfect square, all exponents must be even.
Multiply by 3 × 5 = 15.
Then,
540 × 15
= 2² × 3⁴ × 5²
= 8100
Answer: 8100
Question 19: Find the smallest perfect cube that is divisible by 756.
Prime factorisation:
756 = 2² × 3³ × 7¹
For a perfect cube, every exponent must be a multiple of 3.
Multiply by:
2¹ × 7²
= 2 × 49
= 98
Verification:
756 × 98
= 2³ × 3³ × 7³
= (2 × 3 × 7)³
= 42³
Hence, the required least multiplier is 98.
Answer: 98
756 = 2² × 3³ × 7¹
For a perfect cube, every exponent must be a multiple of 3.
Multiply by:
2¹ × 7²
= 2 × 49
= 98
Verification:
756 × 98
= 2³ × 3³ × 7³
= (2 × 3 × 7)³
= 42³
Hence, the required least multiplier is 98.
Answer: 98
Question 20: Find the greatest perfect cube that divides 9072.
Prime factorisation:
9072 = 2⁴ × 3⁴ × 7¹
For a perfect cube, each exponent must be the greatest multiple of 3 not exceeding the given exponent.
Therefore,
2⁴ → 2³
3⁴ → 3³
7¹ → 7⁰
Hence, the greatest perfect cube is
2³ × 3³
= 8 × 27
= 216
Answer: 216
9072 = 2⁴ × 3⁴ × 7¹
For a perfect cube, each exponent must be the greatest multiple of 3 not exceeding the given exponent.
Therefore,
2⁴ → 2³
3⁴ → 3³
7¹ → 7⁰
Hence, the greatest perfect cube is
2³ × 3³
= 8 × 27
= 216
Answer: 216
Question 21: Find the number of factors of 1080 that are exactly divisible by 6.
Prime factorisation 1080 = 2³ × 3³ × 5
A factor of 1080 has the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 3
0 ≤ b ≤ 3
0 ≤ c ≤ 1
For a factor to be divisible by 6 = 2 × 3, it must contain at least one factor of 2 and at least one factor of 3.
Therefore
a = 1, 2 or 3 → 3 choices
b = 1, 2 or 3 → 3 choices
c = 0 or 1 → 2 choices
Hence, the number of factors divisible by 6 is 3 × 3 × 2 = 18
A factor of 1080 has the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 3
0 ≤ b ≤ 3
0 ≤ c ≤ 1
For a factor to be divisible by 6 = 2 × 3, it must contain at least one factor of 2 and at least one factor of 3.
Therefore
a = 1, 2 or 3 → 3 choices
b = 1, 2 or 3 → 3 choices
c = 0 or 1 → 2 choices
Hence, the number of factors divisible by 6 is 3 × 3 × 2 = 18
Question 22: Find the number of factors of 240 that are strictly greater than 10.
Prime factorisation 240 = 2⁴ × 3 × 5
Therefore, the total number of factors is
(4 + 1)(1 + 1)(1 + 1)
= 5 × 2 × 2
= 20
Now list the factors of 240 in ascending order
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240
The factors strictly greater than 10 are
12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240
So there are 12 such factors.
Therefore, the total number of factors is
(4 + 1)(1 + 1)(1 + 1)
= 5 × 2 × 2
= 20
Now list the factors of 240 in ascending order
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240
The factors strictly greater than 10 are
12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240
So there are 12 such factors.
Question 23: Find the number of factors of 10800 that are multiples of 15 but are not even numbers (odd multiples of 15).
Prime factorisation of 10,800 = 2⁵ × 3³ × 5²
A factor of 10,800 has the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 5
0 ≤ b ≤ 3
0 ≤ c ≤ 2
For a factor to be a multiple of 15,
15 = 3 × 5
Therefore, the factor must contain at least one factor of 3 → b = 1, 2 or 3 at least one factor of 5 → c = 1 or 2
The factor must not be even, so a = 0
Therefore, the number of possible factors is 1 × 3 × 2 = 6
The six factors are 15, 45, 135, 75, 225, 675
A factor of 10,800 has the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 5
0 ≤ b ≤ 3
0 ≤ c ≤ 2
For a factor to be a multiple of 15,
15 = 3 × 5
Therefore, the factor must contain at least one factor of 3 → b = 1, 2 or 3 at least one factor of 5 → c = 1 or 2
The factor must not be even, so a = 0
Therefore, the number of possible factors is 1 × 3 × 2 = 6
The six factors are 15, 45, 135, 75, 225, 675
Question 24: Find how many factors of 3600 are perfect cubes.
Prime factorisation 3600 = 2⁴ × 3² × 5²
Any factor of 3600 is of the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 4
0 ≤ b ≤ 2
0 ≤ c ≤ 2
For the factor to be a perfect cube, every exponent must be a multiple of 3.
For a, the possible values are 0, 3 → 2 choices
For b, the only possible value is 0 → 1 choice
For c, the only possible value is 0 → 1 choice
Therefore, the number of perfect-cube factors is 2 × 1 × 1 = 2
The two perfect-cube factors are 1 and 2³ = 8
Any factor of 3600 is of the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 4
0 ≤ b ≤ 2
0 ≤ c ≤ 2
For the factor to be a perfect cube, every exponent must be a multiple of 3.
For a, the possible values are 0, 3 → 2 choices
For b, the only possible value is 0 → 1 choice
For c, the only possible value is 0 → 1 choice
Therefore, the number of perfect-cube factors is 2 × 1 × 1 = 2
The two perfect-cube factors are 1 and 2³ = 8
Question 25: Find the number of factors of 4320 that end with a digit of 0.
Prime factorisation 4320 = 2⁵ × 3³ × 5
A factor of 4320 has the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 5
0 ≤ b ≤ 3
0 ≤c ≤ 1
For a factor to end in 0, it must be divisible by 10 = 2 × 5.
Therefore, the factor must contain
At least one factor of 2 → a = 1, 2, 3, 4 or 5 → 5 choices
The factor 5 → c = 1 → 1 choice
Any allowed power of 3 → b = 0, 1, 2 or 3 → 4 choices
Therefore, the number of factors ending in 0 is 5 × 4 × 1 = 20
A factor of 4320 has the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 5
0 ≤ b ≤ 3
0 ≤c ≤ 1
For a factor to end in 0, it must be divisible by 10 = 2 × 5.
Therefore, the factor must contain
At least one factor of 2 → a = 1, 2, 3, 4 or 5 → 5 choices
The factor 5 → c = 1 → 1 choice
Any allowed power of 3 → b = 0, 1, 2 or 3 → 4 choices
Therefore, the number of factors ending in 0 is 5 × 4 × 1 = 20
Question 26: Find the number of factors of 1620 that are multiples of 9 but not multiples of 27.
Prime factorisation of 1620 = 2² × 3⁴ × 5
Any factor of 1620 is of the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 2
0 ≤ b ≤ 4
0 ≤ c ≤ 1
For a factor to be a multiple of 9 = 3² b ≥ 2
For it not to be a multiple of 27 = 3³ b < 3 Therefore b = 2 So there is 1 choice for the power of 3. For the power of 2 a = 0, 1, 2 → 3 choices For the power of 5 c = 0, 1 → 2 choices Therefore, the number of required factors is: 3 × 1 × 2 = 6[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row admin_label="SE_Question" _builder_version="4.27.6" _module_preset="default" custom_margin="-50px||||false|false" global_colors_info="{}"][et_pb_column type="4_4" _builder_version="4.27.6" _module_preset="default" global_colors_info="{}"][et_pb_text _builder_version="4.27.6" _module_preset="default" text_font="|300|||||||" text_font_size="18px" custom_padding="0.5%||0.5%|1%|false|false" text_font_size_tablet="17px" text_font_size_phone="16px" text_font_size_last_edited="on|phone" border_radii="on|3px|3px|3px|3px" border_width_all="1px" border_color_all="#0d9488" border_width_all_tablet="1px" border_width_all_phone="2px" border_width_all_last_edited="on|phone" global_colors_info="{}"]Question 27: Find the number of composite factors of 840.
Any factor of 1620 is of the form 2ᵃ × 3ᵇ × 5ᶜ
where:
0 ≤ a ≤ 2
0 ≤ b ≤ 4
0 ≤ c ≤ 1
For a factor to be a multiple of 9 = 3² b ≥ 2
For it not to be a multiple of 27 = 3³ b < 3 Therefore b = 2 So there is 1 choice for the power of 3. For the power of 2 a = 0, 1, 2 → 3 choices For the power of 5 c = 0, 1 → 2 choices Therefore, the number of required factors is: 3 × 1 × 2 = 6[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row admin_label="SE_Question" _builder_version="4.27.6" _module_preset="default" custom_margin="-50px||||false|false" global_colors_info="{}"][et_pb_column type="4_4" _builder_version="4.27.6" _module_preset="default" global_colors_info="{}"][et_pb_text _builder_version="4.27.6" _module_preset="default" text_font="|300|||||||" text_font_size="18px" custom_padding="0.5%||0.5%|1%|false|false" text_font_size_tablet="17px" text_font_size_phone="16px" text_font_size_last_edited="on|phone" border_radii="on|3px|3px|3px|3px" border_width_all="1px" border_color_all="#0d9488" border_width_all_tablet="1px" border_width_all_phone="2px" border_width_all_last_edited="on|phone" global_colors_info="{}"]Question 27: Find the number of composite factors of 840.
Prime factorisation 840 = 2³ × 3 × 5 × 7
The total number of factors is
(3 + 1)(1 + 1)(1 + 1)(1 + 1)
= 4 × 2 × 2 × 2
= 32
Among these factors
1 is neither prime nor composite.
The distinct prime factors are 2, 3, 5 and 7, so there are 4 prime factors.
Therefore, the number of composite factors is
32 − 4 − 1
= 27
The total number of factors is
(3 + 1)(1 + 1)(1 + 1)(1 + 1)
= 4 × 2 × 2 × 2
= 32
Among these factors
1 is neither prime nor composite.
The distinct prime factors are 2, 3, 5 and 7, so there are 4 prime factors.
Therefore, the number of composite factors is
32 − 4 − 1
= 27
Question 28: Find the total number of factors of 1200 that are greater than 5 but less than 100.
Prime factorisation 1200 = 2⁴ × 3 × 5²
The total number of factors is
(4 + 1)(1 + 1)(2 + 1)
= 5 × 2 × 3
= 30
Now, identify the factors outside the required range. Factors less than or equal to 5 are
1, 2, 3, 4, 5
Number of such factors = 5
Factors greater than or equal to 100 are 100, 120, 150, 200, 240, 300, 400, 600, 1200
Number of such factors = 9
Therefore, the number of factors outside the required range is 5 + 9 = 14
Hence, the number of factors greater than 5 but less than 100 is 30 − 14 = 16
The total number of factors is
(4 + 1)(1 + 1)(2 + 1)
= 5 × 2 × 3
= 30
Now, identify the factors outside the required range. Factors less than or equal to 5 are
1, 2, 3, 4, 5
Number of such factors = 5
Factors greater than or equal to 100 are 100, 120, 150, 200, 240, 300, 400, 600, 1200
Number of such factors = 9
Therefore, the number of factors outside the required range is 5 + 9 = 14
Hence, the number of factors greater than 5 but less than 100 is 30 − 14 = 16
Question 29: Find how many natural numbers less than or equal to 120 are coprime to 120.
Prime factorisation 120 = 2³ × 3 × 5
A number is coprime to 120 if it has no common factor greater than 1 with 120.
Therefore, it must not be divisible by 2, 3 or 5.
From 1 to 120:
Numbers divisible by 2 are 120 ÷ 2 = 60
Numbers divisible by 3 are 120 ÷ 3 = 40
Numbers divisible by 5 are 120 ÷ 5 = 24
Now apply inclusion-exclusion.
Numbers divisible by both 2 and 3 are 120 ÷ 6 = 20
Numbers divisible by both 2 and 5 are 120 ÷ 10 = 12
Numbers divisible by both 3 and 5 are 120 ÷ 15 = 8
Numbers divisible by 2, 3 and 5 are 120 ÷ 30 = 4
Therefore, the number of integers from 1 to 120 having at least one common prime factor with 120 is
60 + 40 + 24 − 20 − 12 − 8 + 4 = 88
Hence, the number coprime to 120 is 120 − 88 = 32
A number is coprime to 120 if it has no common factor greater than 1 with 120.
Therefore, it must not be divisible by 2, 3 or 5.
From 1 to 120:
Numbers divisible by 2 are 120 ÷ 2 = 60
Numbers divisible by 3 are 120 ÷ 3 = 40
Numbers divisible by 5 are 120 ÷ 5 = 24
Now apply inclusion-exclusion.
Numbers divisible by both 2 and 3 are 120 ÷ 6 = 20
Numbers divisible by both 2 and 5 are 120 ÷ 10 = 12
Numbers divisible by both 3 and 5 are 120 ÷ 15 = 8
Numbers divisible by 2, 3 and 5 are 120 ÷ 30 = 4
Therefore, the number of integers from 1 to 120 having at least one common prime factor with 120 is
60 + 40 + 24 − 20 − 12 − 8 + 4 = 88
Hence, the number coprime to 120 is 120 − 88 = 32
Question 30: Find the total number of positive integers less than 90 that share a common factor greater than 1 with 90 (numbers that are not coprime to 90).
Prime factorisation 90 = 2 × 3² × 5
A positive integer shares a common factor greater than 1 with 90 if it is divisible by at least one of the prime factors 2, 3 or 5.
From 1 to 89
Numbers divisible by 2 are 89 ÷ 2 = 44
Numbers divisible by 3 are 89 ÷ 3 = 29
Numbers divisible by 5 are 89 ÷ 5 = 17
Now subtract the overlaps.
Numbers divisible by both 2 and 3 = 89 ÷ 6 = 14
Numbers divisible by both 2 and 5 = 89 ÷ 10 = 8
Numbers divisible by both 3 and 5 = 89 ÷ 15 = 5
Numbers divisible by 2, 3 and 5 = 89 ÷ 30 = 2
Using inclusion-exclusion:
44 + 29 + 17 − 14 − 8 − 5 + 2 = 65
Therefore, the total number of positive integers less than 90 that share a common factor greater than 1 with 90 is
65
A positive integer shares a common factor greater than 1 with 90 if it is divisible by at least one of the prime factors 2, 3 or 5.
From 1 to 89
Numbers divisible by 2 are 89 ÷ 2 = 44
Numbers divisible by 3 are 89 ÷ 3 = 29
Numbers divisible by 5 are 89 ÷ 5 = 17
Now subtract the overlaps.
Numbers divisible by both 2 and 3 = 89 ÷ 6 = 14
Numbers divisible by both 2 and 5 = 89 ÷ 10 = 8
Numbers divisible by both 3 and 5 = 89 ÷ 15 = 5
Numbers divisible by 2, 3 and 5 = 89 ÷ 30 = 2
Using inclusion-exclusion:
44 + 29 + 17 − 14 − 8 − 5 + 2 = 65
Therefore, the total number of positive integers less than 90 that share a common factor greater than 1 with 90 is
65
Question 31: Find the sum of the reciprocals of all the factors of 180.
Prime factorisation 180 = 2² × 3² × 5
The sum of all factors of 180 is:
(1 + 2 + 2²)(1 + 3 + 3²)(1 + 5)
= (1 + 2 + 4)(1 + 3 + 9)(1 + 5)
= 7 × 13 × 6
= 546
Now, for every factor d of 180, the number 180 ÷ d is also a factor of 180.
Therefore, Σ(1/d) = Σ(d/180)
So Sum of reciprocals of the factors
= (Sum of all factors) ÷ 180
= 546 ÷ 180
= 91/30
The sum of all factors of 180 is:
(1 + 2 + 2²)(1 + 3 + 3²)(1 + 5)
= (1 + 2 + 4)(1 + 3 + 9)(1 + 5)
= 7 × 13 × 6
= 546
Now, for every factor d of 180, the number 180 ÷ d is also a factor of 180.
Therefore, Σ(1/d) = Σ(d/180)
So Sum of reciprocals of the factors
= (Sum of all factors) ÷ 180
= 546 ÷ 180
= 91/30
Question 31: Find the sum of all the even factors of 240.
Prime factorisation 240 = 2⁴ × 3 × 5
An even factor must contain at least one factor of 2. Therefore, the exponent of 2 can be
1, 2, 3 or 4
The exponents of 3 and 5 can be 0 or 1
So the sum of all even factors is
(2 + 2² + 2³ + 2⁴)(1 + 3)(1 + 5)
= (2 + 4 + 8 + 16)(4)(6)
= 30 × 4 × 6
= 720
An even factor must contain at least one factor of 2. Therefore, the exponent of 2 can be
1, 2, 3 or 4
The exponents of 3 and 5 can be 0 or 1
So the sum of all even factors is
(2 + 2² + 2³ + 2⁴)(1 + 3)(1 + 5)
= (2 + 4 + 8 + 16)(4)(6)
= 30 × 4 × 6
= 720
Question 32: Find the sum of all the odd factors of 540.
Prime factorisation 540 = 2² × 3³ × 5
Odd factors cannot contain any factor of 2. Therefore, we ignore 2² and consider
3³ × 5
The sum of all odd factors is
(1 + 3 + 3² + 3³)(1 + 5)
= (1 + 3 + 9 + 27)(6)
= 40 × 6
= 240
Odd factors cannot contain any factor of 2. Therefore, we ignore 2² and consider
3³ × 5
The sum of all odd factors is
(1 + 3 + 3² + 3³)(1 + 5)
= (1 + 3 + 9 + 27)(6)
= 40 × 6
= 240
Question 33: Find the sum of all the factors of 360 that are exactly divisible by 10.
Prime factorisation 360 = 2³ × 3² × 5
A factor of 360 has the form 2ᵃ × 3ᵇ × 5ᶜ
For a factor to be divisible by 10 = 2 × 5, it must contain at least one factor of 2 and one factor of 5.
Therefore
a = 1, 2 or 3
b = 0, 1 or 2
c = 1
Hence, the required sum is
(2 + 2² + 2³)(1 + 3 + 3²)(5)
= (2 + 4 + 8)(1 + 3 + 9)(5)
= 14 × 13 × 5
= 910
A factor of 360 has the form 2ᵃ × 3ᵇ × 5ᶜ
For a factor to be divisible by 10 = 2 × 5, it must contain at least one factor of 2 and one factor of 5.
Therefore
a = 1, 2 or 3
b = 0, 1 or 2
c = 1
Hence, the required sum is
(2 + 2² + 2³)(1 + 3 + 3²)(5)
= (2 + 4 + 8)(1 + 3 + 9)(5)
= 14 × 13 × 5
= 910
Question 34: Find the sum of all the factors of 120 that are perfect squares.
Prime factorisation 120 = 2³ × 3 × 5
A factor of 120 is a perfect square only when all its prime exponents are even.
For the power of 2: 0 or 2 → 2 choices
For the powers of 3 and 5: 0 only → 1 choice each
Therefore, the perfect-square factors are: 1 and 2² = 4
Their sum is 1 + 4 = 5
A factor of 120 is a perfect square only when all its prime exponents are even.
For the power of 2: 0 or 2 → 2 choices
For the powers of 3 and 5: 0 only → 1 choice each
Therefore, the perfect-square factors are: 1 and 2² = 4
Their sum is 1 + 4 = 5
Question 35: Find the sum of all the factors of 720 that are multiples of 12.
Prime factorisation 720 = 2⁴ × 3² × 5
A factor of 720 has the form 2ᵃ × 3ᵇ × 5ᶜ
For a factor to be a multiple of 12 = 2² × 3
we need
a = 2, 3 or 4
b = 1 or 2
c = 0 or 1
Therefore, the sum of all such factors is (2² + 2³ + 2⁴)(3 + 3²)(1 + 5)
= (4 + 8 + 16)(3 + 9)(6)
= 28 × 12 × 6
= 2,016
A factor of 720 has the form 2ᵃ × 3ᵇ × 5ᶜ
For a factor to be a multiple of 12 = 2² × 3
we need
a = 2, 3 or 4
b = 1 or 2
c = 0 or 1
Therefore, the sum of all such factors is (2² + 2³ + 2⁴)(3 + 3²)(1 + 5)
= (4 + 8 + 16)(3 + 9)(6)
= 28 × 12 × 6
= 2,016
Question 36: Find the sum of all the factors of 1000 except the number 1 and the number 1000 itself (the sum of proper non-trivial divisors).
Prime factorisation 1000 = 2³ × 5³
The sum of all factors is
(1 + 2 + 2² + 2³)(1 + 5 + 5² + 5³)
= (1 + 2 + 4 + 8)(1 + 5 + 25 + 125)
= 15 × 156
= 2,340
The proper non-trivial divisors exclude 1 and the number itself, 1000.
Therefore sum of proper non-trivial divisors
= 2,340 − 1 − 1,000
= 1,339
The sum of all factors is
(1 + 2 + 2² + 2³)(1 + 5 + 5² + 5³)
= (1 + 2 + 4 + 8)(1 + 5 + 25 + 125)
= 15 × 156
= 2,340
The proper non-trivial divisors exclude 1 and the number itself, 1000.
Therefore sum of proper non-trivial divisors
= 2,340 − 1 − 1,000
= 1,339
Question 37: Find the sum of all the factors of 144 that are composite numbers.
Prime factorisation 144 = 2⁴ × 3²
The sum of all factors is
(1 + 2 + 2² + 2³ + 2⁴)(1 + 3 + 3²)
= (1 + 2 + 4 + 8 + 16)(1 + 3 + 9)
= 31 × 13
= 403
Now, among the factors of 144
1 is neither prime nor composite.
The prime factors are 2 and 3.
Therefore, the sum of the composite factors is
403 − 1 − 2 − 3
= 397
The sum of all factors is
(1 + 2 + 2² + 2³ + 2⁴)(1 + 3 + 3²)
= (1 + 2 + 4 + 8 + 16)(1 + 3 + 9)
= 31 × 13
= 403
Now, among the factors of 144
1 is neither prime nor composite.
The prime factors are 2 and 3.
Therefore, the sum of the composite factors is
403 − 1 − 2 − 3
= 397
Question 38: Find the least number by which 72 must be multiplied to make it a perfect square and a perfect cube simultaneously (a perfect 6th power).
Prime factorisation 72 = 2³ × 3²
For a number to be both a perfect square and a perfect cube, all prime exponents must be multiples of 6.
For 2³, we need to reach the next multiple of 6
2³ → 2⁶
So multiply by 2³.
For 3², we need
3² → 3⁶
So multiply by 3⁴.
Therefore, the least required multiplier is
2³ × 3⁴
= 8 × 81
= 648
Verification
72 × 648 = 2⁶ × 3⁶
= (2 × 3)⁶
= 6⁶
Therefore, the result is both a perfect square and a perfect cube.
For a number to be both a perfect square and a perfect cube, all prime exponents must be multiples of 6.
For 2³, we need to reach the next multiple of 6
2³ → 2⁶
So multiply by 2³.
For 3², we need
3² → 3⁶
So multiply by 3⁴.
Therefore, the least required multiplier is
2³ × 3⁴
= 8 × 81
= 648
Verification
72 × 648 = 2⁶ × 3⁶
= (2 × 3)⁶
= 6⁶
Therefore, the result is both a perfect square and a perfect cube.
Question 39: Find how many factors of 1,000,000 (one million) are perfect squares.
Prime factorisation 1,000,000 = 10⁶ = 2⁶ × 5⁶
Any factor of 1,000,000 is of the form
2ᵃ × 5ᵇ
where 0 ≤ a ≤ 6
and 0 ≤ b ≤ 6
For a factor to be a perfect square, both exponents must be even.
For a, the possible values are
0, 2, 4, 6 → 4 choices
For b, the possible values are
0, 2, 4, 6 → 4 choices
Therefore, the number of perfect-square factors is
4 × 4 = 16
Any factor of 1,000,000 is of the form
2ᵃ × 5ᵇ
where 0 ≤ a ≤ 6
and 0 ≤ b ≤ 6
For a factor to be a perfect square, both exponents must be even.
For a, the possible values are
0, 2, 4, 6 → 4 choices
For b, the possible values are
0, 2, 4, 6 → 4 choices
Therefore, the number of perfect-square factors is
4 × 4 = 16
Question 40: Find the least number by which 250 must be multiplied to make it a perfect 4th power.
Prime factorisation 250 = 2 × 5³
For a number to be a perfect fourth power, the exponent of every prime factor must be a multiple of 4.
For 2¹, we need 2¹ → 2⁴
So we need 2³.
For 5³, we need 5³ → 5⁴
So we need 5¹.
Therefore, the least number required is
2³ × 5
= 8 × 5
= 40
Verification
250 × 40 = 10,000
= 10⁴
For a number to be a perfect fourth power, the exponent of every prime factor must be a multiple of 4.
For 2¹, we need 2¹ → 2⁴
So we need 2³.
For 5³, we need 5³ → 5⁴
So we need 5¹.
Therefore, the least number required is
2³ × 5
= 8 × 5
= 40
Verification
250 × 40 = 10,000
= 10⁴
Question 41: Find the smallest perfect square that is divisible by both 24 and 45.
Prime factorisation of
24 = 2³ × 3 and
45 = 3² × 5
Therefore, their LCM is
= 2³ × 3² × 5
= 360
So the required number must be a multiple of 360.
Prime factorisation
360 = 2³ × 3² × 5
For a perfect square, every prime exponent must be even.
The exponents of 2 and 5 are odd, so multiply by
2 × 5 = 10
Therefore 360 × 10 = 3600 and
3600 = 60²
Hence, the smallest perfect square divisible by both 24 and 45 is 3600
24 = 2³ × 3 and
45 = 3² × 5
Therefore, their LCM is
= 2³ × 3² × 5
= 360
So the required number must be a multiple of 360.
Prime factorisation
360 = 2³ × 3² × 5
For a perfect square, every prime exponent must be even.
The exponents of 2 and 5 are odd, so multiply by
2 × 5 = 10
Therefore 360 × 10 = 3600 and
3600 = 60²
Hence, the smallest perfect square divisible by both 24 and 45 is 3600
What are Factors & Divisors?
Factors (also called divisors) are numbers that divide a given number exactly, leaving no remainder. In other words, if a number can be divided by another number without leaving a remainder, then that number is called a factor or divisor of the given number.
For example, the factors of 12 are 1, 2, 3, 4, 6, and 12, because each of these numbers divides 12 exactly.
Every positive integer has at least two factors: 1 and the number itself. A prime number has exactly two factors, whereas a composite number has more than two factors.
Prime factorisation is a useful method for finding the factors of a number. By expressing a number as the product of its prime factors, we can determine the number of factors, sum of factors, product of factors, and distinguish between even, odd, and prime factors.
A clear understanding of factors and divisors is essential for solving many problems involving HCF & LCM, divisibility, prime numbers, number of divisors, and other number system concepts. It also forms the basis of many questions asked in quantitative aptitude and logical reasoning.
Summary of Divisibility Rules
| Concept | Rule / Formula | Key Point |
|---|---|---|
| Factor / Divisor | Divides a number exactly | Leaves a remainder of 0. |
| Prime Number | Exactly 2 factors | The factors are 1 and the number itself. |
| Composite Number | More than 2 factors | Can be expressed as a product of smaller integers. |
| Number of Factors | If N = pᵃ × qᵇ × rᶜ, then Number of Factors = (a + 1)(b + 1)(c + 1) | Obtain the exponents from the prime factorisation. |
| Sum of Factors | (1 + p + p² + … + pᵃ)(1 + q + q² + … + qᵇ)… | Multiply the sums corresponding to each prime factor. |
| Product of Factors | N^(Number of Factors ÷ 2) | Applicable for every positive integer. |
| Perfect Square | Odd number of factors | All other numbers have an even number of factors. |
| Every positive integer has at least two factors: 1 and the number itself. | ||
Common Mistakes
- Confusing factors with multiples: A factor divides a number exactly, whereas a multiple is obtained by multiplying the number by an integer.
- Missing factor pairs: When listing factors manually, always find factors in pairs to avoid omitting any.
- Incorrect prime factorisation: An error in prime factorisation leads to incorrect answers for the number, sum, and product of factors.
- Using the wrong exponents while finding the number of factors: Remember that if N = pᵃ × qᵇ × rᶜ, then the number of factors is (a + 1)(b + 1)(c + 1).
- Forgetting to include 1 and the number itself as factors: Every positive integer has at least these two factors.
- Confusing distinct prime factors with total prime factors: Count only the different prime numbers unless the question states otherwise.
- Including even factors while counting odd factors: Ignore the power of 2 when finding the number of odd factors.
- Not making all exponents even for a perfect square: Every prime factor must have an even exponent.
- Not making all exponents multiples of 3 for a perfect cube: Every prime factor must have an exponent that is a multiple of 3.
- Stopping after finding one possible answer: In questions involving the smallest number with a given number of factors, consider all possible exponent combinations before choosing the smallest number.
Practice Questions
Question 1: Find the number of even factors and odd factors of 540.
Question 2: Find the smallest positive integer having exactly 18 factors.
Question 3: Find the least number by which 840 must be multiplied to make it a perfect square.
Question 4: Find the greatest perfect square that divides 2520.
Question 5: Find the smallest perfect cube that is divisible by 540.
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