Factors Questions with Answers and PDF

Factors Questions is a helpful topic for students who want to understand maths in a simple way. Factors Questions show us how to find the numbers that divide another number exactly without leaving any remainder. This topic is important because it builds a strong maths base and improves problem solving skills. Students can practise different questions to see how factors are used in real examples.

Factors Questions with Solutions

Question 1: Find all factors of each number:

a) 24

Check divisibility systematically:

24 ÷ 1  = 24 

24 ÷ 2  = 12 

24 ÷ 3  = 8  

24 ÷ 4  = 6  

24 ÷ 5  = 4.8 

24 ÷ 6  = 4 

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Total: 8 factors

b) 36

36 ÷ 1  = 36

36 ÷ 2  = 18 

36 ÷ 3  = 12 

36 ÷ 4  = 9  

36 ÷ 6  = 6  

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Total: 9 factors 

c) 48

Pairs: (1,48), (2,24), (3,16), (4,12), (6,8)

Factors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Total: 10 factors

Question 2: Count factors without listing them all:

Using prime factorisation method:

If n = p^a × q^b × r^c

Number of factors = (a+1)(b+1)(c+1)

Examples:

a) 72 = 2³ × 3²

   Factors = (3+1)(2+1) = 4 × 3 = 12

b) 100 = 2² × 5²

   Factors = (2+1)(2+1) = 3 × 3 = 9

c) 120 = 2³ × 3 × 5

   Factors = (3+1)(1+1)(1+1) = 4×2×2 = 16

Question 3: Find even and odd factors of 60.

60 = 2² × 3 × 5

All factors: 1,2,3,4,5,6,10,12,15,20,30,60

Odd factors (no 2 in them):

60 = 3 × 5 

Odd factors: 1, 3, 5, 15

Count = (1+1)(1+1) = 4

Even factors = Total − Odd = 12 − 4 = 8

Even factors: 2,4,6,10,12,20,30,60

Question 4: Find prime factorisation:

a) 180

180 ÷ 2 = 90

90  ÷ 2 = 45

45  ÷ 3 = 15

15  ÷ 3 = 5

5   ÷ 5 = 1

180 = 2² × 3² × 5

Answer: 2² × 3² × 5

b) 252

252 ÷ 2 = 126

126 ÷ 2 = 63

63  ÷ 3 = 21

21  ÷ 3 = 7

7   ÷ 7 = 1

252 = 2² × 3² × 7

Question 5: Draw factor tree for 90.

FACTOR TREE FOR 90:

           90

          /   \

        2   45

            /  \

          5     9

               /  \

              3   3

90 = 2 × 5 × 3 × 3

90 = 2 × 3² × 5

Circled prime factors: 2, 5, 3, 3

Question 6: Factor tree for 48:

FACTOR TREE FOR 48:

           48

          /  \

         2   24

            /  \

           2   12

              /  \

             2    6

                 / \

                2   3

48 = 2 × 2 × 2 × 2 × 3

48 = 2⁴ × 3

Question 7: Express in prime factor form:

a) 150 = 2 × 3 × 5²

b) 196 = 2² × 7²

c) 360 = 2³ × 3² × 5

d) 504 = 2³ × 3² × 7

Question 8: Find common factors of 18 and 24.

Factors of 18: 1, 2, 3, 6, 9, 18

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Common Factors: 1, 2, 3, 6

   18 only    COMMON    24 only

  {9, 18}   {1,2,3,6}  {4,8,12,24}

Answer: Common factors = 1, 2, 3, 6

Question 9: Find GCF (HCF) of 36 and 48.

Method 1: List factors

Factors of 36: 1,2,3,4,6,9,12,18,36

Factors of 48: 1,2,3,4,6,8,12,16,24,48

Common: 1,2,3,4,6,12

GCF = 12

Method 2: Prime factorisation

36 = 2² × 3²

48 = 2⁴ × 3

GCF = 2² × 3 = 4 × 3 = 12

Answer: GCF = 12

Question 10: Find GCF of 72, 90, and 108.

72  = 2³ × 3²

90  = 2  × 3² × 5

108 = 2² × 3³

GCF = lowest power of common primes

= 2¹ × 3² = 2 × 9 = 18

Answer: GCF = 18

Question 11: Which number has more factors 48 or 60?

48 = 2⁴ × 3

Factors = (4+1)(1+1) = 5×2 = 10

60 = 2² × 3 × 5

Factors = (2+1)(1+1)(1+1) = 3×2×2 = 12

60 has more factors (12 > 10)

Answer: 60 has more factors

Question 12: A teacher wants to arrange 48 students into equal rows. What are the possible arrangements?

Find all factors of 48:

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Possible arrangements (rows × students per row):

1 × 48

2 × 24

3 × 16

4 × 12

6 × 8

8 × 6

(and their reverses)

Answer: 10 different arrangements possible

Question 13: A ribbon of 72 cm needs to be cut into equal pieces. What lengths are possible with no wastage?

All factors of 72:

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Answer: The ribbon can be cut into pieces of

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, or 72 cm

Question 14: I am a two-digit number. I have exactly 3 factors. What could I be?

Numbers with exactly 3 factors must be:

p² (square of a prime), because:

Factors of p² are: 1, p, p²

Two-digit perfect squares of primes:

2² = 4 

3² = 9 

5² = 25 

7² = 49 

11² = 121 

Answer: 25 or 49

Question 15: What is the smallest number with exactly 5 factors?

5 factors means: (a+1) = 5 → a = 4

So n = p⁴ (prime to the power 4)

Smallest: 2⁴ = 16

Factors of 16: 1, 2, 4, 8, 16 

Answer: 16

Factors MCQs

Q1: Which is NOT a factor of 30?

a) 6    b) 8    c) 10    d) 15

Answer: b) 8

(30 ÷ 8 = 3.75, not exact)

Q2: How many factors does 25 have?

Factors: 1, 5, 25

a) 2    b) 3    c) 4    d) 5

Answer: b) 3

Q3: Which number has the most factors?

a) 12    b) 15    c) 17    d) 11

Factors: 12→6, 15→4, 17→2, 11→2

Answer: a) 12

Q4: The GCF of 12 and 18 is:

a) 3    b) 6    c) 9    d) 12

Answer: b) 6

Q5: How many factors does 2³ × 5² have?

(3+1)(2+1) = 4 × 3 = 12

a) 8    b) 10    c) 12    d) 6

Answer: c) 12

Q6: Which of these is a prime factorisation of 84?

a) 2×6×7    b) 4×21    c) 2²×3×7    d) 2×42

Answer: c) 2² × 3 × 7

Q7: Common factors of 24 and 36 are:

a) 1,2,3,6      b) 1,2,3,4,6,12

c) 1,2,3,6,12   d) 1,2,4,6

Factors of 24: 1,2,3,4,6,8,12,24

Factors of 36: 1,2,3,4,6,9,12,18,36

Common: 1,2,3,4,6,12

Answer: b) 1,2,3,4,6,12

Q8: If a number has 4 factors and is less than 20, how many such numbers exist?

4 factors means: p³ or p×q

p³: 2³ = 8, 3³ = 27(too big)

p×q: 2 × 3 = 6, 2 × 5 = 10, 2 × 7 = 14, 3 × 5 = 15, 2 × 11 = 22(no), 3 × 7 = 21(no)

     = 6, 10, 14, 15

Total: 8, 6, 10, 14, 15 

a) 4    b) 5    c) 6    d) 3

Answer: b) 5

Q9: The number of common factors of 2³×3² and 2²×3³ is:

GCF = 2² × 3² = 36

Factors of 36 = (2+1)(2+1) = 9

a) 6    b) 8    c) 9    d) 12

Answer: c) 9

Download PDF - Factors Questions

Related Topics:

• Class 5 Maths Factors
• Factors and Multiples
• Factors and Multiples for Class 5