What are prime numbers?

Numbers greater than 1 with exactly two factors — 1 and itself. Examples: 2, 3, 5, 7, 11.

How do you find the HCF of two numbers?

List all factors of each number and find the greatest common factor, or use the prime factorisation method.

What are the divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, 11?

2: last digit even; 3: sum of digits divisible by 3; 4: last two digits divisible by 4; 5: last digit 0 or 5; 6: divisible by both 2 and 3; 8: last three digits divisible by 8; 9: sum of digits divisible by 9; 10: last digit 0; 11: alternating sum of digits divisible by 11.

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NCERT Solutions for Class 6 Maths Chapter 3 — Playing with Numbers

Playing with numbers is about understanding how numbers relate to each other through factors, multiples, primes, and divisibility. These concepts form the backbone of arithmetic and number theory. Every solution below is worked out step by step with clear reasoning.

Key Concepts

Concept	What It Means	Example
Factors	Numbers that divide a given number exactly (leaving no remainder)	Factors of 12: 1, 2, 3, 4, 6, 12
Multiples	Products of a number with any whole number	Multiples of 4: 4, 8, 12, 16, 20, ...
Prime Numbers	Numbers greater than 1 with exactly two factors — 1 and itself	2, 3, 5, 7, 11, 13, 17, 19, 23, ...
Composite Numbers	Numbers greater than 1 with more than two factors	4, 6, 8, 9, 10, 12, ...
Prime Factorisation	Expressing a number as a product of its prime factors	60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
Common Factors	Factors shared by two or more numbers	Common factors of 12 and 18: 1, 2, 3, 6
Common Multiples	Multiples shared by two or more numbers	Common multiples of 4 and 6: 12, 24, 36, ...
HCF (Highest Common Factor)	The greatest number that divides two or more numbers exactly	HCF of 12 and 18 = 6
LCM (Lowest Common Multiple)	The smallest number that is a multiple of two or more numbers	LCM of 4 and 6 = 12
Divisibility Rules	Quick tests to check if a number is divisible by another	A number divisible by 2 ends in 0, 2, 4, 6, or 8

Exercise 3.1 — Factors and Multiples

Question 1: Write all the factors of the following numbers

Answer

(a) 24

Solution

Step 1: Find pairs of numbers whose product is 24:
1 × 24 = 24
2 × 12 = 24
3 × 8 = 24
4 × 6 = 24
Step 2: List all factors in order: 1, 2, 3, 4, 6, 8, 12, 24

AnswerFactors of 24 are 1, 2, 3, 4, 6, 8, 12, 24

(b) 36

Solution

Step 1: Find pairs:
1 × 36 = 36
2 × 18 = 36
3 × 12 = 36
4 × 9 = 36
6 × 6 = 36
Step 2: List all factors: 1, 2, 3, 4, 6, 9, 12, 18, 36

AnswerFactors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36

(c) 48

Solution

Step 1: Find pairs:
1 × 48 = 48
2 × 24 = 48
3 × 16 = 48
4 × 12 = 48
6 × 8 = 48
Step 2: List all factors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

AnswerFactors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Question 2: Write the first five multiples of

Answer

(a) 5

Solution

Step 1: Multiply 5 by 1, 2, 3, 4, 5:
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25

AnswerFirst five multiples of 5 are 5, 10, 15, 20, 25

(b) 8

Solution

8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40

AnswerFirst five multiples of 8 are 8, 16, 24, 32, 40

(c) 12

Solution

12 × 1 = 12, 12 × 2 = 24, 12 × 3 = 36, 12 × 4 = 48, 12 × 5 = 60

AnswerFirst five multiples of 12 are 12, 24, 36, 48, 60

Question 3: State whether the following are true or false

Answer

(a) 5 is a factor of 35. → True. 35 ÷ 5 = 7 with no remainder.

(b) 7 is a factor of 42. → True. 42 ÷ 7 = 6 with no remainder.

(c) 9 is a multiple of 27. → False. 27 is a multiple of 9, not the other way around. 9 × 3 = 27, so 27 is a multiple of 9.

(d) 24 is a multiple of 6. → True. 6 × 4 = 24.

Question 4: Find the common factors of

Answer

(a) 20 and 28

Solution

Step 1: Factors of 20: 1, 2, 4, 5, 10, 20
Step 2: Factors of 28: 1, 2, 4, 7, 14, 28
Step 3: Common factors: 1, 2, 4

AnswerCommon factors of 20 and 28 are 1, 2, 4

(b) 15 and 25

Solution

Step 1: Factors of 15: 1, 3, 5, 15
Step 2: Factors of 25: 1, 5, 25
Step 3: Common factors: 1, 5

AnswerCommon factors of 15 and 25 are 1, 5

(c) 35 and 50

Solution

Step 1: Factors of 35: 1, 5, 7, 35
Step 2: Factors of 50: 1, 2, 5, 10, 25, 50
Step 3: Common factors: 1, 5

AnswerCommon factors of 35 and 50 are 1, 5

Question 5: Find the first three common multiples of

Answer

(a) 4 and 6

Solution

Step 1: Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, ...
Step 2: Multiples of 6: 6, 12, 18, 24, 30, 36, ...
Step 3: Common multiples: 12, 24, 36, ...

AnswerFirst three common multiples are 12, 24, 36

(b) 3 and 5

Solution

Step 1: Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...
Step 2: Multiples of 5: 5, 10, 15, 20, 25, 30, 35, ...
Step 3: Common multiples: 15, 30, 45, ...

AnswerFirst three common multiples are 15, 30, 45

Exercise 3.2 — Prime and Composite Numbers

Question 1: Which of the following numbers are prime? Which are composite?

Answer

2, 5, 8, 11, 14, 17, 20, 23, 27, 31, 36, 41

Solution

Number	Factors	Prime or Composite?
2	1, 2	Prime (exactly 2 factors)
5	1, 5	Prime (exactly 2 factors)
8	1, 2, 4, 8	Composite (4 factors)
11	1, 11	Prime (exactly 2 factors)
14	1, 2, 7, 14	Composite (4 factors)
17	1, 17	Prime (exactly 2 factors)
20	1, 2, 4, 5, 10, 20	Composite (6 factors)
23	1, 23	Prime (exactly 2 factors)
27	1, 3, 9, 27	Composite (4 factors)
31	1, 31	Prime (exactly 2 factors)
36	1, 2, 3, 4, 6, 9, 12, 18, 36	Composite (9 factors)
41	1, 41	Prime (exactly 2 factors)

Prime numbers: 2, 5, 11, 17, 23, 31, 41

Composite numbers: 8, 14, 20, 27, 36

Question 2: Express the following as the sum of two odd primes

Answer

(a) 14

Solution

Step 1: Odd prime numbers less than 14: 3, 5, 7, 11, 13
Step 2: Try combinations: 3 + 11 = 14 ✓

Answer14 = 3 + 11 (also 7 + 7 = 14)

(b) 20

Solution

Step 1: Odd primes less than 20: 3, 5, 7, 11, 13, 17, 19
Step 2: Try combinations: 3 + 17 = 20 ✓

Answer20 = 3 + 17 (also 7 + 13 = 20)

(c) 36

Solution

Step 1: Try combinations of odd primes: 5 + 31 = 36 ✓

Answer36 = 5 + 31 (also 7 + 29 = 36, 13 + 23 = 36)

Question 3: Express the following numbers as the sum of three odd primes

Answer

(a) 21

Solution

Step 1: We need three odd primes that add to 21.
Step 2: Try 3 + 5 + 13 = 21 ✓

Answer21 = 3 + 5 + 13

(b) 31

Solution

Step 1: Try 3 + 5 + 23 = 31 ✓

Answer31 = 3 + 5 + 23 (also 5 + 7 + 19 = 31)

Question 4: Write five pairs of twin primes (twin primes differ by 2)

Solution

Twin primes are pairs of primes that differ by 2.

(3, 5): Both prime, difference = 2 ✓
(5, 7): Both prime, difference = 2 ✓
(11, 13): Both prime, difference = 2 ✓
(17, 19): Both prime, difference = 2 ✓
(29, 31): Both prime, difference = 2 ✓

Question 5: Write the prime factorisation of the following numbers

Answer

(a) 36

Solution

Step 1: Divide by smallest prime factor:
36 ÷ 2 = 18
18 ÷ 2 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1
Step 2: So, 36 = 2 × 2 × 3 × 3 = 2² × 3²

Answer36 = 2² × 3²

(b) 60

Solution

Step 1: Divide by smallest prime factor:
60 ÷ 2 = 30
30 ÷ 2 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Step 2: So, 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

Answer60 = 2² × 3 × 5

(c) 84

Solution

Step 1: Divide by smallest prime factor:
84 ÷ 2 = 42
42 ÷ 2 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
Step 2: So, 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7

Answer84 = 2² × 3 × 7

(d) 96

Solution

Step 1: Divide by smallest prime factor:
96 ÷ 2 = 48
48 ÷ 2 = 24
24 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 3 = 1
Step 2: So, 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

Answer96 = 2⁵ × 3

Exercise 3.3 — Divisibility Rules

Divisibility Rules Summary

Divisible by	Rule
2	Last digit is 0, 2, 4, 6, or 8 (even)
3	Sum of digits is divisible by 3
4	Last two digits form a number divisible by 4
5	Last digit is 0 or 5
6	Divisible by both 2 and 3
8	Last three digits form a number divisible by 8
9	Sum of digits is divisible by 9
10	Last digit is 0
11	Alternating sum of digits is divisible by 11 (or equals 0)

Question 1: Test the divisibility of the following by 2, 3, 4, 5, 6, 8, 9, 10, 11

Answer

(a) 54672

Solution

By 2: Last digit = 2 (even) → Divisible ✓
By 3: Sum of digits = 5+4+6+7+2 = 24; 24 ÷ 3 = 8 → Divisible ✓
By 4: Last two digits = 72; 72 ÷ 4 = 18 → Divisible ✓
By 5: Last digit = 2 (not 0 or 5) → Not divisible ✗
By 6: Divisible by both 2 and 3 → Divisible ✓
By 8: Last three digits = 672; 672 ÷ 8 = 84 → Divisible ✓
By 9: Sum of digits = 24; 24 ÷ 9 ≠ whole number → Not divisible ✗
By 10: Last digit = 2 (not 0) → Not divisible ✗
By 11: Alternating sum = 5−4+6−7+2 = 2; 2 not divisible by 11 → Not divisible ✗

(b) 137269

Solution

By 2: Last digit = 9 (odd) → Not divisible ✗
By 3: Sum = 1+3+7+2+6+9 = 28; 28 ÷ 3 ≠ whole number → Not divisible ✗
By 4: Last two digits = 69; 69 ÷ 4 ≠ whole number → Not divisible ✗
By 5: Last digit = 9 (not 0 or 5) → Not divisible ✗
By 6: Not divisible by 2 → Not divisible ✗
By 8: Last three digits = 269; 269 ÷ 8 ≠ whole number → Not divisible ✗
By 9: Sum = 28; 28 ÷ 9 ≠ whole number → Not divisible ✗
By 10: Last digit = 9 (not 0) → Not divisible ✗
By 11: Alternating sum = 1−3+7−2+6−9 = 0; 0 is divisible by 11 → Divisible ✓

Question 2: Using divisibility test, determine which of the following are divisible by 4 and by 8

Answer

(a) 572

Solution

By 4: Last two digits = 72; 72 ÷ 4 = 18 → Divisible by 4 ✓
By 8: Last three digits = 572; 572 ÷ 8 = 71.5 → Not divisible by 8 ✗

(b) 726352

Solution

By 4: Last two digits = 52; 52 ÷ 4 = 13 → Divisible by 4 ✓
By 8: Last three digits = 352; 352 ÷ 8 = 44 → Divisible by 8 ✓

(c) 12159

Solution

By 4: Last two digits = 59; 59 ÷ 4 ≠ whole number → Not divisible by 4 ✗
By 8: Last three digits = 159; 159 ÷ 8 ≠ whole number → Not divisible by 8 ✗

Question 3: Using divisibility test, determine which of the following are divisible by 11

Answer

(a) 7134

Solution

Step 1: Alternating sum: 7 − 1 + 3 − 4 = 5
Step 2: 5 is not divisible by 11.

AnswerNot divisible by 11.

(b) 901351

Solution

Step 1: Alternating sum: 9 − 0 + 1 − 3 + 5 − 1 = 11
Step 2: 11 is divisible by 11.

AnswerDivisible by 11. ✓

(c) 10827

Solution

Step 1: Alternating sum: 1 − 0 + 8 − 2 + 7 = 14
Step 2: 14 is not divisible by 11.

AnswerNot divisible by 11.

Exercise 3.4 — HCF and LCM

Question 1: Find the HCF of the following by listing all factors

Answer

(a) 18 and 24

Solution

Step 1: Factors of 18: 1, 2, 3, 6, 9, 18
Step 2: Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Step 3: Common factors: 1, 2, 3, 6
Step 4: Highest common factor = 6

AnswerHCF of 18 and 24 = 6

(b) 30 and 45

Solution

Step 1: Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
Step 2: Factors of 45: 1, 3, 5, 9, 15, 45
Step 3: Common factors: 1, 3, 5, 15
Step 4: Highest common factor = 15

AnswerHCF of 30 and 45 = 15

(c) 36 and 84

Solution

Step 1: Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Step 2: Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Step 3: Common factors: 1, 2, 3, 4, 6, 12
Step 4: Highest common factor = 12

AnswerHCF of 36 and 84 = 12

Question 2: Find the HCF using prime factorisation method

Answer

(a) 48 and 72

Solution

Step 1: Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3
Step 2: Prime factorisation of 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
Step 3: Common prime factors: 2³ × 3 = 8 × 3 = 24

AnswerHCF of 48 and 72 = 24

(b) 56 and 98

Solution

Step 1: Prime factorisation of 56 = 2 × 2 × 2 × 7 = 2³ × 7
Step 2: Prime factorisation of 98 = 2 × 7 × 7 = 2 × 7²
Step 3: Common prime factors: 2¹ × 7¹ = 2 × 7 = 14

AnswerHCF of 56 and 98 = 14

(c) 120 and 144

Solution

Step 1: Prime factorisation of 120 = 2³ × 3 × 5
Step 2: Prime factorisation of 144 = 2⁴ × 3²
Step 3: Common prime factors: 2³ × 3¹ = 8 × 3 = 24

AnswerHCF of 120 and 144 = 24

Question 3: Find the LCM of the following by listing multiples

Answer

(a) 6 and 8

Solution

Step 1: Multiples of 6: 6, 12, 18, 24, 30, 36, ...
Step 2: Multiples of 8: 8, 16, 24, 32, 40, ...
Step 3: Smallest common multiple = 24

AnswerLCM of 6 and 8 = 24

(b) 12 and 18

Solution

Step 1: Multiples of 12: 12, 24, 36, 48, 60, ...
Step 2: Multiples of 18: 18, 36, 54, 72, ...
Step 3: Smallest common multiple = 36

AnswerLCM of 12 and 18 = 36

(c) 15 and 25

Solution

Step 1: Multiples of 15: 15, 30, 45, 60, 75, ...
Step 2: Multiples of 25: 25, 50, 75, 100, ...
Step 3: Smallest common multiple = 75

AnswerLCM of 15 and 25 = 75

Question 4: Find the LCM using prime factorisation method

Answer

(a) 12 and 20

Solution

Step 1: 12 = 2² × 3; 20 = 2² × 5
Step 2: LCM = highest power of each prime factor = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60

AnswerLCM of 12 and 20 = 60

(b) 24 and 36

Solution

Step 1: 24 = 2³ × 3; 36 = 2² × 3²
Step 2: LCM = 2³ × 3² = 8 × 9 = 72

AnswerLCM of 24 and 36 = 72

(c) 18 and 42

Solution

Step 1: 18 = 2 × 3²; 42 = 2 × 3 × 7
Step 2: LCM = 2 × 3² × 7 = 2 × 9 × 7 = 126

AnswerLCM of 18 and 42 = 126

Question 5: Verify: HCF × LCM = Product of the two numbers

Answer

(a) For 12 and 20

Solution

Step 1: HCF(12, 20) = 4
Step 2: LCM(12, 20) = 60
Step 3: HCF × LCM = 4 × 60 = 240
Step 4: Product = 12 × 20 = 240
Step 5: 240 = 240 → Verified ✓

(b) For 18 and 42

Solution

Step 1: HCF(18, 42) = 6
Step 2: LCM(18, 42) = 126
Step 3: HCF × LCM = 6 × 126 = 756
Step 4: Product = 18 × 42 = 756
Step 5: 756 = 756 → Verified ✓

Exercise 3.5 — Additional Divisibility Rules and Word Problems

Question 1: If a number is divisible by 9, is it necessarily divisible by 3?

Solution

Step 1: If a number is divisible by 9, then the sum of its digits is divisible by 9.
Step 2: If the sum is divisible by 9, it must also be divisible by 3 (because 9 = 3 × 3).
Step 3: Therefore, the number is also divisible by 3.

AnswerYes, every number divisible by 9 is also divisible by 3.

Example: 54 is divisible by 9 (5 + 4 = 9), and 54 is also divisible by 3 (5 + 4 = 9, 9 ÷ 3 = 3).

Question 2: If a number is divisible by 3, is it necessarily divisible by 9?

Solution

Step 1: Not necessarily. Divisibility by 3 does not guarantee divisibility by 9.

AnswerNo. Example: 12 is divisible by 3 (1 + 2 = 3), but 12 is NOT divisible by 9 (1 + 2 = 3, 3 ÷ 9 ≠ whole number).

Question 3: A number is divisible by both 5 and 12. By which other number will it always be divisible?

Solution

Step 1: If a number is divisible by both 5 and 12, it must be divisible by the LCM of 5 and 12.
Step 2: LCM(5, 12) = 5 × 12 = 60 (since 5 and 12 have no common factor other than 1)
Step 3: The number will always be divisible by 60, and therefore also by all factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

AnswerThe number will always be divisible by 60 (and also by 2, 3, 4, 6, 10, 15, 20, 30).

Question 4: The product of two numbers is 180. If their HCF is 3, find their LCM.

Solution

Step 1: We know that HCF × LCM = Product of the two numbers
Step 2: 3 × LCM = 180
Step 3: LCM = 180 ÷ 3 = 60

AnswerLCM = 60

Question 5: The HCF of two numbers is 12 and their LCM is 144. If one of the numbers is 36, find the other number.

Solution

Step 1: HCF × LCM = Product of the two numbers
Step 2: 12 × 144 = First number × Second number
Step 3: 1728 = 36 × Second number
Step 4: Second number = 1728 ÷ 36 = 48

AnswerThe other number is 48

Question 6: Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a container that can measure the petrol of both tankers exactly.

Solution

Step 1: We need the largest container that can measure both exactly. This is the HCF of 850 and 680.
Step 2: Prime factorisation of 850 = 2 × 5² × 17
Step 3: Prime factorisation of 680 = 2³ × 5 × 17
Step 4: Common prime factors = 2 × 5 × 17 = 170

AnswerThe maximum capacity of the container is 170 litres.

Question 7: Three bells ring at intervals of 12 minutes, 15 minutes, and 18 minutes respectively. If they ring together at 8:00 AM, when will they next ring together?

Solution

Step 1: We need the LCM of 12, 15, and 18.
Step 2: 12 = 2² × 3; 15 = 3 × 5; 18 = 2 × 3²
Step 3: LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes
Step 4: 180 minutes = 3 hours
Step 5: They will ring together at 8:00 AM + 3 hours = 11:00 AM

AnswerThe three bells will next ring together at 11:00 AM.

Question 8: The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm respectively. Find the longest tape that can measure the dimensions exactly.

Solution

Step 1: We need the HCF of 825, 675, and 450.
Step 2: Prime factorisation:
825 = 3 × 5² × 11
675 = 3³ × 5²
450 = 2 × 3² × 5²
Step 3: Common prime factors = 3 × 5² = 3 × 25 = 75

AnswerThe longest tape is 75 cm.

Common Mistakes

1. Confusing factors and multiples: A factor divides into a number (e.g., 3 is a factor of 12). A multiple is the product of a number (e.g., 12 is a multiple of 3). Factors are always less than or equal to the number; multiples are always greater than or equal to the number.

2. Forgetting that 1 is neither prime nor composite... wait, 1 is NOT prime (it has only one factor) and NOT composite (it doesn't have more than two factors). 1 is a special number — it is neither prime nor composite. Also, 2 is the only even prime number.

3. Incorrect prime factorisation: Always divide by the smallest prime factor first and continue until you reach 1. Do not skip steps or use composite numbers as divisors.

4. HCF vs LCM confusion in word problems: HCF is for "greatest/longest/largest" that divides exactly (sharing, distributing equally). LCM is for "smallest/least" common time or quantity (events happening together, meeting again at the same time).

5. Wrong divisibility test for 11: The rule for 11 is the alternating sum of digits (subtract and add alternately). Some students simply add all digits, which is the rule for 3 and 9, not 11.

6. Assuming divisibility by 4 implies divisibility by 2 or vice versa incorrectly: Every number divisible by 4 is divisible by 2, but NOT every number divisible by 2 is divisible by 4. For example, 6 is divisible by 2 but not by 4.

Exam Checklist

[ ] I can find all factors of a number by listing factor pairs systematically.
[ ] I can list multiples of a number and identify common multiples.
[ ] I can distinguish prime numbers (exactly 2 factors) from composite numbers (more than 2 factors).
[ ] I can express a number as a product of its prime factors using the division method.
[ ] I can apply divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, and 11 correctly.
[ ] I can find the HCF of two or more numbers using both the listing method and the prime factorisation method.
[ ] I can find the LCM of two or more numbers using both the listing method and the prime factorisation method.
[ ] I can verify that HCF × LCM = Product of two numbers for any pair.
[ ] I can identify whether a word problem requires HCF (sharing/distributing equally) or LCM (events occurring together).
[ ] I know that 1 is neither prime nor composite, and 2 is the only even prime number.

Frequently Asked Questions

A factor divides a number exactly; a multiple is the product of a number with any whole number.

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers | KalviGavani

NCERT Solutions for Class 6 Maths Chapter 3 — Playing with Numbers

Key Concepts

Exercise 3.1 — Factors and Multiples

Exercise 3.2 — Prime and Composite Numbers

Exercise 3.3 — Divisibility Rules

Divisibility Rules Summary

Exercise 3.4 — HCF and LCM

Exercise 3.5 — Additional Divisibility Rules and Word Problems

Common Mistakes

Exam Checklist

Frequently Asked Questions