HCF and LCM questions with answers and concepts

Learn HCF and LCM with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

91 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

HCF (Highest Common Factor), also known as GCD (Greatest Common Divisor), is the largest number that divides two or more numbers without leaving a remainder. LCM (Lowest Common Multiple) is the smallest number that is a multiple of two or more numbers. HCF and LCM are fundamental concepts used extensively in problems involving divisibility, synchronization of cycles, comparison of fractions, and finding common periods in real-world scenarios such as scheduling and production.

Key formulas and rules

8 rules

Product Rule: $HCF (a, b) \times LCM (a, b) = a \times b$ (valid for exactly two numbers)

HCF of two numbers: $HCF (a, b) = \frac{a \times b}{LCM ( a , b )}$

LCM of two numbers: $LCM (a, b) = \frac{a \times b}{HCF ( a , b )}$

HCF of fractions: $HCF (\frac{a}{b}, \frac{c}{d}) = \frac{HCF ( a , c )}{LCM ( b , d )}$ (fractions must be in lowest terms)

LCM of fractions: $LCM (\frac{a}{b}, \frac{c}{d}) = \frac{LCM ( a , c )}{HCF ( b , d )}$ (fractions must be in lowest terms)

Ratio property: If $HCF (a, b) = h$ , then $a = h \times m$ and $b = h \times n$ , where $g cd (m, n) = 1$

LCM from ratio: $LCM (a, b) = h \times m \times n$

Co-prime numbers: Two integers are co-prime iff $HCF (a, b) = 1$

Key concepts

Prime Factorisation Method

Express each number as a product of prime factors. For HCF, take each common prime factor with its MINIMUM power (lowest exponent) across all numbers and multiply them. For LCM, take each prime factor that appears in ANY number with its MAXIMUM power (highest exponent) and multiply them. Example: $12 = 2^{2}$ x $3^{1}$ , $18 = 2^{1}$ x $3^{2}$ . $H C F = 2^{1}$ x $3^{1} = 6$ . $L C M = 2^{2}$ x $3^{2} = 36$ .

Division Method (Euclidean Algorithm)

For finding HCF of two numbers, divide the larger number by the smaller and note the remainder. Then divide the previous divisor by this remainder. Repeat until the remainder becomes 0. The last non-zero remainder's divisor is the HCF. This method is especially efficient for large numbers where prime factorisation is impractical. Example: HCF(48, 18): $48/18 = 2$ remainder 12; $18/12 = 1$ remainder 6; $12/6 = 2$ remainder 0 -> $H C F = 6$ .

Relationship Between HCF, LCM and Product

For two positive integers a and b: HCF(a, b) x $L C M (a, b) = a$ x b. This holds true for exactly two numbers. For three or more numbers, the product relationship does not generalise simply. Note: This formula is very useful for checking work and solving problems where HCF and LCM are given to find the original numbers.

HCF and LCM of Fractions

HCF of fractions = HCF of numerators / LCM of denominators. LCM of fractions = LCM of numerators / HCF of denominators. Crucial condition: All fractions MUST be reduced to lowest terms before applying the formulas. For example, for 2/4 and 5/6: simplifying 2/4 to 1/2 gives $H C F (1, 5) / L C M (2, 6) = 1/6$ (correct). Applying the formula unreduced to 2/4 gives $H C F (2, 5) / L C M (4, 6) = 1/12$ (incorrect, because 1/12 is not the largest common divisor).

Co-prime Numbers

Two numbers are co-prime (or relatively prime) if their HCF is 1. This means they share no common factor other than 1. Note: Co-prime numbers need not themselves be prime (e.g., 8 and 9 are co-prime). If two numbers are co-prime, their LCM is simply their product. Useful property: If a and b are co-prime, then $H C F (a, b) = 1$ and $L C M (a, b) = a$ x b.

Bells and Lights Problems

Classic applications of LCM: 'Bells problem' — multiple bells tolling at regular intervals will all toll together at intervals equal to their LCM. 'Lights/Flashlights problem' — lights flashing at different intervals will flash together at the LCM of their periods. 'Work scheduling' — if workers take different times to complete a task, the next simultaneous occurrence is at the LCM of their cycles.

Finding Numbers Given HCF and LCM

If $H C F = h$ and $L C M = L$ of two numbers, and one number is given as a, then the other number = (h x L) / a. This follows directly from HCF x LCM = Product. Example: $H C F = 6$ , $L C M = 60$ , one number = 12. Other number = (6 x 60) / 12 = 30. Check: $H C F (12, 30) = 6$ , $L C M (12, 30) = 60$

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Find the HCF and LCM of 12, 15, and 18 using prime factorisation.

Solution

Step 1: Prime factorise each number.

12 = 2^{2} \times 3^{1}

15 = 3^{1} \times 5^{1}

18 = 2^{1} \times 3^{2}

Step 2: Calculate HCF (lowest power of common prime factors).

HCF (12, 15, 18) = 3^{1} = 3

Step 3: Calculate LCM (highest power of all appearing prime factors).

LCM (12, 15, 18) = 2^{2} \times 3^{2} \times 5^{1} = 4 \times 9 \times 5 = 180

Answer: $HCF = 3$ and $LCM = 180$ .

Example 2

Two numbers are co-prime and their product is 105. Find their LCM.

Solution

Step 1: Apply the fundamental HCF-LCM product identity.

HCF (a, b) \times LCM (a, b) = a \times b

Step 2: By definition of co-prime numbers, $HCF (a, b) = 1$ .

1 \times LCM (a, b) = 105 ⟹ LCM (a, b) = 105

Answer: 105.

Example 3

Find the smallest number which when divided by 12, 16, 24, and 36 leaves a remainder of 7 in each case.

Solution

Step 1: Find the LCM of the divisors.

12 = 2^{2} \times 3, 16 = 2^{4}, 24 = 2^{3} \times 3, 36 = 2^{2} \times 3^{2}

LCM (12, 16, 24, 36) = 2^{4} \times 3^{2} = 16 \times 9 = 144

Step 2: Add the required constant remainder.

N = LCM + r = 144 + 7 = 151

Answer: 151.

Example 4

Find the greatest number that will divide 43, 91, and 183 so as to leave the same remainder in each case.

Solution

Step 1: Compute pairwise differences.

91 - 43 = 48

183 - 91 = 92

183 - 43 = 140

Step 2: Find the HCF of the pairwise differences.

48 = 2^{4} \times 3

92 = 2^{2} \times 23

140 = 2^{2} \times 5 \times 7

HCF (48, 92, 140) = 2^{2} = 4

Answer: 4.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

The LCM of two numbers is 48. The numbers are in the ratio 2:3. Find the numbers.

6 and 9
8 and 12
16 and 24
12 and 18

Answer: C. 16 and 24

Explanation

Step 1: Let the numbers be 2x and 3x
Step 2: $L C M = 6 x$ (since 2 and 3 are co-prime)
Step 3: $6 x = 48$ , so $x = 8$
Step 4: Numbers are $2 \times 8 = 16$ and $3 \times 8 = 24$
Answer: 16 and 24

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

The product of two numbers is 4107. If their HCF is 37, find the greater number.

Answer: A. 111

Explanation

Step 1: Let numbers be 37a and 37b where a and b are co-prime
Step 2: Product = 37a x $37 b = 1369 ab = 4107$
Step 3: $ab = 4107/1369 = 3$
Step 4: Since a and b are co-prime, and $ab = 3$ , possible pairs: (1,3) or (3,1)
Step 5: Numbers are $37 \times 1 = 37$ and $37 \times 3 = 111$
Step 6: Greater number = 111
Answer: 111

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds. In 30 minutes, how many times do they toll together?

Answer: A. 16

Explanation

Step 1: Find LCM of 2, 4, 6, 8, 10, 12
Step 2: $L C M = 2^{3}$ x 3 x $5 = 120$ seconds = 2 minutes
Step 3: They toll together at 0, 2, 4, 6... minutes
Step 4: In 30 minutes: $30/2 = 15$ intervals
Step 5: Number of $t im es = 15 + 1$ (including start at 0) = 16
Answer: 16

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always verify: HCF is always <= each number, and LCM is always >= each number. If your HCF exceeds any of the numbers, something is wrong.
Memorise: HCF x LCM = Product (only valid for exactly two numbers). For three or more numbers, use prime factorisation directly.
The Euclidean Division Method (repeated division) is faster than prime factorisation for large numbers — use it in exams to save time.
When numbers are co-prime, their LCM equals their product. This shortcut frequently appears in competitive exams.
For fraction problems, remember: HCF of fractions = HCF(numerators) / LCM(denominators); LCM of fractions = LCM(numerators) / HCF(denominators). Always reduce all fractions to their lowest terms first before applying these formulas.
In 'find both numbers' problems given HCF and LCM, write numbers as HCF x m and HCF x n where m and n are co-prime, then use the LCM to find mn.

Ready to test your understanding?

Work through 91 questions with explanations after each answer.

Start practice

Next topic: Decimal Fractions

HCF and LCM questions with answers and concepts

Learn HCF and LCM with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

91 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Key formulas and rules

8 rules

Product Rule: $HCF (a, b) \times LCM (a, b) = a \times b$ (valid for exactly two numbers)

HCF of two numbers: $HCF (a, b) = \frac{a \times b}{LCM ( a , b )}$

LCM of two numbers: $LCM (a, b) = \frac{a \times b}{HCF ( a , b )}$

HCF of fractions: $HCF (\frac{a}{b}, \frac{c}{d}) = \frac{HCF ( a , c )}{LCM ( b , d )}$ (fractions must be in lowest terms)

LCM of fractions: $LCM (\frac{a}{b}, \frac{c}{d}) = \frac{LCM ( a , c )}{HCF ( b , d )}$ (fractions must be in lowest terms)

Ratio property: If $HCF (a, b) = h$ , then $a = h \times m$ and $b = h \times n$ , where $g cd (m, n) = 1$

LCM from ratio: $LCM (a, b) = h \times m \times n$

Co-prime numbers: Two integers are co-prime iff $HCF (a, b) = 1$

Key concepts

Prime Factorisation Method

Division Method (Euclidean Algorithm)

Relationship Between HCF, LCM and Product

HCF and LCM of Fractions

Co-prime Numbers

Bells and Lights Problems

Finding Numbers Given HCF and LCM

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Find the HCF and LCM of 12, 15, and 18 using prime factorisation.

Solution

Step 1: Prime factorise each number.

12 = 2^{2} \times 3^{1}

15 = 3^{1} \times 5^{1}

18 = 2^{1} \times 3^{2}

Step 2: Calculate HCF (lowest power of common prime factors).

HCF (12, 15, 18) = 3^{1} = 3

Step 3: Calculate LCM (highest power of all appearing prime factors).

LCM (12, 15, 18) = 2^{2} \times 3^{2} \times 5^{1} = 4 \times 9 \times 5 = 180

Answer: $HCF = 3$ and $LCM = 180$ .

Example 2

Two numbers are co-prime and their product is 105. Find their LCM.

Solution

Step 1: Apply the fundamental HCF-LCM product identity.

HCF (a, b) \times LCM (a, b) = a \times b

Step 2: By definition of co-prime numbers, $HCF (a, b) = 1$ .

1 \times LCM (a, b) = 105 ⟹ LCM (a, b) = 105

Answer: 105.

Example 3

Find the smallest number which when divided by 12, 16, 24, and 36 leaves a remainder of 7 in each case.

Solution

Step 1: Find the LCM of the divisors.

12 = 2^{2} \times 3, 16 = 2^{4}, 24 = 2^{3} \times 3, 36 = 2^{2} \times 3^{2}

LCM (12, 16, 24, 36) = 2^{4} \times 3^{2} = 16 \times 9 = 144

Step 2: Add the required constant remainder.

N = LCM + r = 144 + 7 = 151

Answer: 151.

Example 4

Find the greatest number that will divide 43, 91, and 183 so as to leave the same remainder in each case.

Solution

Step 1: Compute pairwise differences.

91 - 43 = 48

183 - 91 = 92

183 - 43 = 140

Step 2: Find the HCF of the pairwise differences.

48 = 2^{4} \times 3

92 = 2^{2} \times 23

140 = 2^{2} \times 5 \times 7

HCF (48, 92, 140) = 2^{2} = 4

Answer: 4.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

The LCM of two numbers is 48. The numbers are in the ratio 2:3. Find the numbers.

6 and 9
8 and 12
16 and 24
12 and 18

Answer: C. 16 and 24

Explanation

Step 1: Let the numbers be 2x and 3x
Step 2: $L C M = 6 x$ (since 2 and 3 are co-prime)
Step 3: $6 x = 48$ , so $x = 8$
Step 4: Numbers are $2 \times 8 = 16$ and $3 \times 8 = 24$
Answer: 16 and 24

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

The product of two numbers is 4107. If their HCF is 37, find the greater number.

Answer: A. 111

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds. In 30 minutes, how many times do they toll together?

Answer: A. 16

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always verify: HCF is always <= each number, and LCM is always >= each number. If your HCF exceeds any of the numbers, something is wrong.
Memorise: HCF x LCM = Product (only valid for exactly two numbers). For three or more numbers, use prime factorisation directly.
The Euclidean Division Method (repeated division) is faster than prime factorisation for large numbers — use it in exams to save time.
When numbers are co-prime, their LCM equals their product. This shortcut frequently appears in competitive exams.
For fraction problems, remember: HCF of fractions = HCF(numerators) / LCM(denominators); LCM of fractions = LCM(numerators) / HCF(denominators). Always reduce all fractions to their lowest terms first before applying these formulas.
In 'find both numbers' problems given HCF and LCM, write numbers as HCF x m and HCF x n where m and n are co-prime, then use the LCM to find mn.

Ready to test your understanding?

Work through 91 questions with explanations after each answer.

Start practice

Next topic: Decimal Fractions

Key formulas and rules

Key concepts

Prime Factorisation Method

Division Method (Euclidean Algorithm)

Relationship Between HCF, LCM and Product

HCF and LCM of Fractions

Co-prime Numbers

Bells and Lights Problems

Finding Numbers Given HCF and LCM

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Prime Factorisation Method

Division Method (Euclidean Algorithm)

Relationship Between HCF, LCM and Product

HCF and LCM of Fractions

Co-prime Numbers

Bells and Lights Problems

Finding Numbers Given HCF and LCM

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.