HCF and LCM Exercises with Word Problems and Solutions

HCF (Highest Common Factor) is the largest number that divides two or more numbers exactly, while LCM (Least Common Multiple) is the smallest number that is a common multiple of two or more numbers. These concepts are essential in number theory and are widely used in simplifying fractions, solving word problems, and handling real-life mathematical situations. This section presents a structured set of HCF and LCM exercises designed to help learners understand and apply these concepts effectively through step-by-step solutions.

Table of Contents

• What Are HCF and LCM?
• Three Methods to Find HCF and LCM
• Key Formulas
• Exercise Set 1: Finding HCF (Class 5–6)
• Exercise Set 2: Finding LCM (Class 5–6)
• Exercise Set 3: The Fundamental Relationship - HCF × LCM = Product of Two Numbers
• Exercise Set 4: HCF Word Problems (Class 7–8)
• Exercise Set 5: LCM Word Problems (Class 7–8)
• Exercise Set 6: Real-Life LCM and HCF Problems (Class 8–9)
• Common Mistakes to Avoid

What Are HCF and LCM?

HCF: Highest Common Factor

The HCF (also called GCF: Greatest Common Factor or GCD: Greatest Common Divisor) of two or more numbers is the largest number that divides all of them exactly without leaving a remainder.

LCM: Least Common Multiple

The LCM of two or more numbers is the smallest positive number that is a multiple of all of them; in other words, the smallest number they all divide into evenly.

Three Methods to Find HCF and LCM

Method 1: Listing Factors / Multiples

Write out all factors (for HCF) or multiples (for LCM) and pick the right one.

HCF of 24 and 36:

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

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

Highest common factor = 12

⇒ HCF(24, 36) = 12

LCM of 6 and 8:

Multiples of 6 : 6, 12, 18, 24, 30 ...

Multiples of 8 : 8, 16, 24, 32 ...

Smallest common multiple = 24  ⇒ LCM(6, 8) = 24

 

Method 2: Prime Factorisation

Express each number as a product of prime factors, then apply these rules:

HCF = product of common prime factors, each raised to the LOWEST power

LCM = product of ALL prime factors, each raised to the HIGHEST power

Example: HCF and LCM of 36, 48, and 72

36 = 2² × 3²

48 = 2⁴ × 3

72 = 2³ × 3²

HCF = 2^(2) × 3^(1) = 2² × 3 = 12

LCM =  24×32= 16 × 9 = 144

 

Method 3: Long Division (Euclidean Algorithm)

Repeatedly divide the larger number by the smaller; the last non-zero remainder is the HCF.

HCF of 408 and 1032:

1032 ÷ 408 = 2  remainder 216

 408 ÷ 216 = 1  remainder 192

 216 ÷ 192 = 1  remainder  24

 192 ÷  24 = 8  remainder   0

HCF(408, 1032) = 24

Key Formulas 

Exercise Set 1: Finding HCF (Class 5–6)

Exercise 1.1: Find the HCF of 24 and 36

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

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

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

HCF = 12

Exercise 1.2: Find the HCF of 18 and 48

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

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

Highest common factor = 6

HCF(18, 48) = 6

Exercise 1.3: Find the HCF of 135 and 225 (prime factorisation)

Solution: 135 = 3 × 3 × 3 × 5 = 3³ × 5

225 = 3 × 3 × 5 × 5 = 3² × 5²

HCF = 3^(2) × 5^(1) = 9 × 5 = 45

Exercise 1.4: Find the HCF of 32 and 14

Solution: 32 = 2⁵

14 = 2 × 7

Common prime: 2^(1) = 2

HCF(32, 14) = 2

Exercise 1.5: Find the HCF of 80 and 90

Solution: 80 = 2⁴ × 5

90 = 2 × 3² × 5

HCF = 2^(1) × 5^(1) = 2 × 5 = 10

Exercise Set 2: Finding LCM(Class 5–6)

 Exercise 2.1: Find the LCM of 3 and 4 (listing method)

Solution: Multiples of 3: 3, 6, 9, 12, 15 ...

Multiples of 4: 4, 8, 12, 16 ...

Smallest common = 12

LCM(3, 4) = 12

Exercise 2.2: Find the  LCM of 4 and 12 (prime factorisation)

Solution: 4  = 2²

12 = 2² × 3

LCM = 2^(2) × 3^(1) = 4 × 3 = 12

Exercise 2.3:  Find the LCM of 54 and 60

Solution: 54 = 2 × 3³

60 = 2² × 3 × 5

LCM = 2^(2) × 3^(3) × 5^(1)

 LCM  = 4 × 27 × 5 = 540

Exercise 2.4: Find the LCM of 12, 18, and 24

Solution: 12 = 2² × 3

18 = 2 × 3²

24 = 2³ × 3

LCM = 2^(max 3) × 3^(max 2) = 8 × 9 = 72

Exercise 2.5:  Find the LCM of 7, 12, and 21

Solution: 7  = 7

12 = 2² × 3

21 = 3 × 7

LCM = 2² × 3 × 7 = 4 × 3 × 7 = 84

Exercise Set 3: The Fundamental Relationship - HCF × LCM = Product of Two Numbers 

For any two numbers, the product of their HCF and LCM is equal to the product of the numbers:

HCF(a, b) × LCM(a, b) = a × b

Exercise 3.1: The HCF of two numbers is 12 and their product is 2,160. Find their LCM.

Solution: HCF × LCM = Product of numbers

12 × LCM  = 2,160

LCM = 2,160 ÷ 12 = 180

Exercise 3.2: Two numbers have LCM 360 and a product of 4,320. Find their HCF.

Solution: HCF × LCM = Product

HCF × 360 = 4,320

HCF = 4,320 ÷ 360 = 12

One valid pair is 120 and 36  (HCF = 12, LCM = 360, 120 × 36 = 4,320).

Exercise Set 4: HCF Word Problems (Class 7–8)

Exercise 4.1: Find the greatest number dividing 72, 96, and 120 with the same remainder

Solution: Differences: 96 − 72 = 24,  120 − 96 = 24,  120 − 72 = 48

HCF(24, 24, 48) = 24

The greatest number dividing 72, 96, and 120 with the same remainder is 24.

HCF(24, 36) = 12

Exercise 4.2: Find the largest 3-digit number divisible by HCF of 24 and 36

Solution: Largest 3-digit multiple of 12: 999 ÷ 12 = 83.25  ⇒   83 × 12 = 996

The largest 3-digit number divisible by HCF of 24 and 36 is 996

Exercise 4.3: Find the greatest number dividing 3959, 4082, and 4164 leaving the same remainder

Solution: Differences:

4082 − 3959 = 123

4164 − 4082 =  82

4164 − 3959 = 205

HCF(123, 82, 205):

123 = 3 × 41

 82 = 2 × 41

205 = 5 × 41

HCF = 41

⇒ the greatest number dividing 3959, 4082, and 4164 leaving the same remainder is 41

Exercise 4.4: Find two numbers whose sum is 1001 and HCF is 7

Solution: Let numbers be 7a and 7b  (where a and b are co-prime)

7a + 7b = 1001  ⇒   a + b = 143

Co-prime pairs adding to 143: 72 + 71 ⇒  gcd(72, 71) = 1 

⇒ Numbers = 7 × 72 and 7 × 71 = 504 and 497

Exercise 4.5: A room is 4 m 37 cm long and 3 m 23 cm wide. Find the minimum number of identical square slabs needed to tile it exactly.

Solution: Length = 437 cm,  Width = 323 cm

Side of largest square slab = HCF(437, 323)

437 = 19 × 23

323 = 17 × 19

HCF = 19 cm

Number of slabs = (437 × 323) ÷ (19 × 19)

 = 141,251 ÷ 361 = 391

The minimum number of identical square slabs needed to tile it exactly is 391 slabs.

Exercise Set 5: LCM Word Problems (Class 7–8)

Exercise 5.1: Find the smallest number divisible by 6, 8, 9, and 12

Solution:  6 = 2 × 3

 8 = 2³

 9 = 3²

12 = 2² × 3

LCM = 2³ × 3² = 8 × 9 = 72

The smallest number divisible by 6, 8, 9, and 12 is 72

Exercise 5.2: Find the smallest 4-digit number divisible by 12, 15, and 18

Solution: 12 = 2² × 3

15 = 3 × 5

18 = 2 × 3²

LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180

Smallest 4-digit multiple of 180:

1000 ÷ 180 = 5.55  ⇒   6 × 180 = 1080

The smallest 4-digit number divisible by 12, 15, and 18 is 1080

Exercise 5.3: Find the smallest number leaving a remainder of 3 when divided by 5, 8, and 12

Solution: Required number = LCM(5, 8, 12) + 3

 5 = 5

 8 = 2³

12 = 2² × 3

LCM = 2³ × 3 × 5 = 120

The smallest number leaving a remainder of 3 when divided by 5, 8, and 12 is 120 + 3 = 123

Exercise Set 6: Real-Life LCM and HCF Problems (Class 8–9)

Exercise 6.1: Four bells ring at intervals of 6, 8, 12, and 18 minutes. They ring together at 12 noon. How many times do they ring together in the next 6 hours (excluding noon)?

Solution:  6 = 2 × 3

 8 = 2³

12 = 2² × 3

18 = 2 × 3²

LCM = 2³ × 3² = 72 minutes

6 hours = 360 minutes

Number of times = 360 ÷ 72 = 5 times

5 times the bells will ring together in next 6 hours.

Exercise 6.2: Bus A departs every 20 minutes and Bus B every 45 minutes. Both depart together at 6:00 AM. When is the next time they depart together?

Solution: 20 = 2² × 5

45 = 3² × 5

LCM(20, 45) = 2² × 3² × 5 = 180 minutes = 3 hours

6:00 AM + 3 hours = 9:00 AM

Both buses will next depart together at 9:00 AM

Exercise 6.3: Three swimmers start simultaneously. Swimmer A completes a lap every 12 seconds, B every 15 seconds, and C every 18 seconds. When are they all at the starting point together again?

Solution: 12 = 2² × 3

15 = 3 × 5

18 = 2 × 3²

LCM = 2² × 3² × 5 = 180 seconds = 3 minutes

All three are at the start together after 3 minutes

Exercise 6.4: Three ropes are 36 m, 48 m, and 60 m long. Find the greatest length that can cut all three into equal pieces with no waste.

Solution: 36 = 2² × 3²

48 = 2⁴ × 3

60 = 2² × 3 × 5

HCF = 2² × 3 = 12

Greatest piece = 12 m
Cutting all three into equal pieces of 12 m will get 3 pieces from first rope, 4 from second, 5 from third.

Exercise 6.5: A teacher has 40 pens and 60 pencils. She distributes them equally among students so each student gets the same number of pens and pencils, with nothing left over. Find the maximum number of students.

Solution: HCF(40, 60):

40 = 2³ × 5

60 = 2² × 3 × 5

HCF = 2² × 5 = 20

The maximum number of students is 20 students

Each student gets 2 pens and 3 pencils

Exercise 6.6: A shopkeeper has 420 laddoos, 490 barfis, and 350 peda. He wants to pack them into identical packets (same number of each type per packet, nothing left over). Find the maximum number of packets.

Solution: HCF(420, 490, 350):

420 = 2² × 3 × 5 × 7

490 = 2 × 5 × 7²

350 = 2 × 5² × 7

HCF = 2 × 5 × 7 = 70

The maximum number of packets is 70 packets

Each packet contains 6 laddoos, 7 barfis, 5 pedas.

Exercise 6.7: A courtyard is 18 m 72 cm long and 13 m 20 cm wide. Find the largest square tile that can pave it exactly and the total number of tiles needed.

Solution:  Length = 1872 cm,  Width = 1320 cm

HCF(1872, 1320):

1872 = 2⁴ × 3² × 13

1320 = 2³ × 3 × 5 × 11

HCF = 2³ × 3 = 24 cm

Number of tiles = (1872 × 1320) ÷ (24 × 24)

 = 2,470,960 ÷ 576 = 4,290

Therefore, Tile size = 24 cm × 24 cm, and number of tiles = 4,290

Common Mistakes to Avoid

• Using HCF when LCM is needed (or vice versa). Rule of thumb: HCF for splitting and dividing; LCM for combining and meeting.

• Applying HCF × LCM = product to three or more numbers. This formula works only for exactly two numbers. Don't use it for three.

• Wrong powers in prime factorisation. For HCF take the minimum power; for LCM take the maximum power. Mixing these up is the most common error in this topic.

• Not reducing the differences first in ‘same remainder’ problems. When a number divides several values leaving the same remainder, find the HCF of the differences, not the original numbers.

• Confusing co-prime with prime. 'Co-prime' means two numbers share no common factor other than 1 (HCF = 1). The numbers themselves do not have to be prime; 8 and 15 are co-prime, but neither is a prime number.

• Forgetting to add the remainder back. In problems like ‘find the smallest number divisible by 8, 9, 10 leaving remainder 7’, students often stop at the LCM and forget to add 7.