HCF and LCM Questions for Placement 2026 (with Solutions)

Last Updated: March 2026

Introduction to HCF and LCM

Highest Common Factor (HCF) and Lowest Common Multiple (LCM) are fundamental concepts in number theory that appear frequently in placement exams. These concepts test your understanding of factors, multiples, and relationships between numbers. Mastery of HCF and LCM shortcuts can significantly improve your speed in quantitative aptitude sections.

Why This Topic is Important

HCF and LCM questions assess:

Understanding of number relationships
Ability to factorize quickly
Application of formulas in word problems
Logical reasoning with real-world scenarios
Speed calculation techniques

Companies That Ask HCF and LCM Questions (with Frequency)

Company	Frequency	Difficulty Level
TCS	Very High	Easy to Moderate
Infosys	Very High	Easy to Moderate
Wipro	High	Easy
Cognizant	High	Easy to Moderate
Accenture	High	Easy
Capgemini	High	Moderate
IBM	Moderate	Moderate
Tech Mahindra	High	Easy to Moderate
HCL	Moderate	Easy
LTI Mindtree	Moderate	Moderate

KEY FORMULAS / CONCEPTS

╔══════════════════════════════════════════════════════════════════╗
║                    HCF AND LCM FORMULA SHEET                     ║
╠══════════════════════════════════════════════════════════════════╣
║                                                                  ║
║  DEFINITIONS                                                     ║
║  ────────────────────────────────────────────────────────────   ║
║  HCF (GCD): Largest number that divides all given numbers       ║
║  LCM: Smallest number divisible by all given numbers            ║
║                                                                  ║
║  KEY RELATIONSHIP                                                ║
║  ────────────────────────────────────────────────────────────   ║
║  For two numbers a and b:                                       ║
║  HCF(a,b) × LCM(a,b) = a × b                                    ║
║                                                                  ║
║  IMPORTANT FORMULAS                                              ║
║  ────────────────────────────────────────────────────────────   ║
║  HCF of fractions = HCF of numerators / LCM of denominators     ║
║  LCM of fractions = LCM of numerators / HCF of denominators     ║
║                                                                  ║
║  HCF of (a-b, b) when a > b = HCF(a, b)                         ║
║  HCF(a^n - 1, a^m - 1) = a^HCF(n,m) - 1                         ║
║                                                                  ║
║  LCM of first n natural numbers                                  ║
║  = Product of highest powers of primes ≤ n                      ║
║                                                                  ║
║  FOR TWO NUMBERS                                                 ║
║  ────────────────────────────────────────────────────────────   ║
║  If HCF = h, then numbers are ha and hb where HCF(a,b) = 1      ║
║  LCM = hab                                                      ║
║                                                                  ║
║  PRODUCT OF TWO NUMBERS                                          ║
║  ────────────────────────────────────────────────────────────   ║
║  If product = P, HCF = H, LCM = L, then P = H × L               ║
║                                                                  ║
╚══════════════════════════════════════════════════════════════════╝

30 Practice Questions with Step-by-Step Solutions

Question 1

Find the HCF of 144 and 180.

Solution: 144 = 2⁴ × 3² 180 = 2² × 3² × 5 HCF = 2² × 3² = 36

Question 2

Find the LCM of 12, 18, and 30.

Solution: 12 = 2² × 3 18 = 2 × 3² 30 = 2 × 3 × 5 LCM = 2² × 3² × 5 = 180

Question 3

The HCF of two numbers is 12 and their LCM is 360. If one number is 72, find the other.

Solution: Product = HCF × LCM = 12 × 360 = 4320 Other number = 4320/72 = 60

Question 4

Find the greatest number that divides 43, 91, and 183 leaving the same remainder in each case.

Solution: Required number = HCF of (91-43), (183-91), (183-43) = HCF of 48, 92, 140 48 = 2⁴ × 3 92 = 2² × 23 140 = 2² × 5 × 7 HCF = 2² = 4

Question 5

Find the smallest number which when divided by 12, 15, 18, and 27 leaves remainder 8 in each case.

Solution: LCM of 12, 15, 18, 27 = 540 Number = 540 + 8 = 548

Question 6

The ratio of two numbers is 3:4 and their HCF is 5. Find their LCM.

Solution: Numbers are 3×5 = 15 and 4×5 = 20 LCM = 60 (Or: LCM = 3×4×5 = 60)

Question 7

Find the HCF of 2^6 - 1 and 2^8 - 1.

Solution: Using property: HCF(2^m - 1, 2^n - 1) = 2^HCF(m,n) - 1 HCF(6, 8) = 2 Answer = 2² - 1 = 3

Question 8

Three bells toll at intervals of 12, 15, and 18 minutes. If they toll together at 8 AM, when will they next toll together?

Solution: LCM of 12, 15, 18 = 180 minutes = 3 hours Next together at 11 AM

Question 9

Find the largest 4-digit number divisible by 12, 15, and 20.

Solution: LCM of 12, 15, 20 = 60 Largest 4-digit number = 9999 9999 = 60 × 166 + 39 Required number = 9999 - 39 = 9960

Question 10

Two numbers are in ratio 4:5 and their LCM is 180. Find the numbers.

Solution: Let numbers be 4x and 5x LCM = 20x = 180 x = 9 Numbers are 36 and 45

Question 11

Find the HCF of 1/2, 2/3, and 3/4.

Solution: HCF of fractions = HCF of numerators / LCM of denominators = HCF(1,2,3) / LCM(2,3,4) = 1/12

Question 12

Find the LCM of 1/3, 2/9, and 5/6.

Solution: LCM of fractions = LCM of numerators / HCF of denominators = LCM(1,2,5) / HCF(3,9,6) = 10/3

Question 13

The sum of two numbers is 60 and their HCF is 10. How many such pairs exist?

Solution: Let numbers be 10a and 10b where HCF(a,b) = 1 10a + 10b = 60 → a + b = 6 Pairs with HCF 1: (1,5) only Answer = 1 pair (10, 50)

Question 14

Find the least number which when divided by 6, 8, 9, and 12 leaves remainder 1, but is divisible by 13.

Solution: LCM of 6, 8, 9, 12 = 72 Number = 72k + 1 Need 72k + 1 ≡ 0 (mod 13) 72 ≡ 7 (mod 13), so 7k + 1 ≡ 0 → 7k ≡ 12 ≡ 25 ≡ ... → k ≡ 4 (mod 13) k = 4: Number = 72×4 + 1 = 289

Question 15

Four runners complete a circular track in 24, 36, 48, and 64 seconds. When will they meet at the starting point?

Solution: LCM of 24, 36, 48, 64 = LCM(2³×3, 2²×3², 2⁴×3, 2⁶) = 2⁶ × 3² = 64 × 9 = 576 seconds = 9 minutes 36 seconds

Question 16

The product of two numbers is 2160 and their HCF is 12. Find their LCM.

Solution: LCM = Product/HCF = 2160/12 = 180

Question 17

Find the greatest number that will divide 1657 and 2037 leaving remainders 6 and 5 respectively.

Solution: Number divides (1657-6) = 1651 and (2037-5) = 2032 HCF of 1651 and 2032 2032 - 1651 = 381 1651 = 4×381 + 127 381 = 3×127 HCF = 127

Question 18

Three numbers are in ratio 2:3:4 and their LCM is 240. Find their HCF.

Solution: Let numbers be 2x, 3x, 4x LCM = 12x = 240 x = 20 = HCF

Question 19

Find the smallest number which is exactly divisible by 12, 15, 20, and 27.

Solution: LCM of 12, 15, 20, 27 12 = 2² × 3 15 = 3 × 5 20 = 2² × 5 27 = 3³ LCM = 2² × 3³ × 5 = 540

Question 20

The HCF and LCM of two numbers are 18 and 3780 respectively. If one number is 270, find the other.

Solution: Other = (HCF × LCM)/First = (18 × 3780)/270 = 252

Question 21

Find the HCF of 4^5 - 1 and 4^7 - 1.

Solution: HCF(4^5 - 1, 4^7 - 1) = 4^HCF(5,7) - 1 = 4^1 - 1 = 3

Question 22

A number leaves remainder 7 when divided by 12, remainder 10 when divided by 15, and remainder 13 when divided by 18. Find the smallest such number.

Solution: Notice: 12-7 = 15-10 = 18-13 = 5 Number + 5 is divisible by 12, 15, 18 LCM of 12, 15, 18 = 180 Number = 180 - 5 = 175

Question 23

The HCF of three numbers is 12 and their LCM is 2400. If two numbers are 48 and 72, find the third number.

Solution: For three numbers: HCF × LCM is not necessarily equal to product But: If HCF of all three is 12, each number = 12 × something 48 = 12 × 4, 72 = 12 × 6, let third = 12 × k LCM(4, 6, k) × 12 = 2400 LCM(4, 6, k) = 200 = 2³ × 5² 4 = 2², 6 = 2 × 3 So k must have 2³ × 5² and no 3 k = 8 × 25 = 200 Third number = 12 × 200 / GCD adjustment... = 200 (or verify: 12 × 25 = 300 also works) Actually: Third = 300 (LCM(48,72,300) = 2400, HCF = 12 ✓)

Question 24

Find the least number which when divided by 5, 6, 7, 8 leaves remainder 3, but when divided by 9 leaves remainder 0.

Solution: LCM of 5, 6, 7, 8 = 840 Number = 840k + 3 Need 840k + 3 ≡ 0 (mod 9) 840 ≡ 3 (mod 9), so 3k + 3 ≡ 0 → k ≡ 2 (mod 3) k = 2: Number = 840×2 + 3 = 1683

Question 25

Two numbers differ by 24 and their LCM is 180. How many such pairs exist?

Solution: Let numbers be ha and hb where HCF = h, HCF(a,b) = 1 h(b-a) = 24 and hab = 180 Possible h values (divisors of both): 1, 2, 3, 4, 6, 12 For h = 12: b-a = 2, ab = 15 → (3,5) ✓ → Numbers: 36, 60 For h = 6: b-a = 4, ab = 30 → (5,6) - diff is 1, not 4; no solution For h = 4: b-a = 6, ab = 45 → (5,9)? diff 4, no; (3,15)? HCF not 1 For h = 3: b-a = 8, ab = 60 → (5,12) - diff 7, no; (6,10) - HCF not 1 For h = 2: b-a = 12, ab = 90 → No valid coprime pair For h = 1: b-a = 24, ab = 180 → (6,30) HCF not 1; (5,36)? diff 31 Only 1 pair: (36, 60)

Question 26

Find the greatest number which divides 400, 542, and 764 leaving remainders 22, 26, and 34 respectively.

Solution: Number divides: 400-22=378, 542-26=516, 764-34=730 HCF of 378, 516, 730 378 = 2 × 3³ × 7 516 = 2² × 3 × 43 730 = 2 × 5 × 73 HCF = 2

Question 27

The product of HCF and LCM of two numbers is 720. If one number is 36 more than the other, find the numbers.

Solution: Let numbers be x and x+36 x(x+36) = 720 x² + 36x - 720 = 0 (x+60)(x-12) = 0 x = 12 Numbers are 12 and 48

Question 28

Find the smallest number of 5 digits divisible by 12, 15, and 18.

Solution: LCM of 12, 15, 18 = 180 Smallest 5-digit number = 10000 10000 = 180 × 55 + 100 Required = 10000 + (180-100) = 10080

Question 29

The HCF of two numbers is 8 and their LCM is 96. If one number is 32, find how many values the other number can take.

Solution: Other = (8 × 96)/32 = 24 Since 24 < 32, and numbers are fixed, only 1 value (Alternatively, if we considered different arrangements, still only one distinct other number)

Question 30

Find the HCF of 2^100 - 1 and 2^120 - 1.

Solution: HCF(2^100 - 1, 2^120 - 1) = 2^HCF(100,120) - 1 = 2^20 - 1 = 1048575

SHORTCUTS & TRICKS

Trick 1: Same Remainder Problems

If a number leaves same remainder r when dividing a, b, c: Number = HCF(a-r, b-r, c-r)

Trick 2: Common Difference Pattern

If a number leaves remainders where (divisor - remainder) is constant: Number = LCM - constant

Trick 3: Product Relationship

For two numbers: HCF × LCM = Product of numbers Always true, very useful!

Trick 4: Ratio Method

If numbers are in ratio a:b and HCF is h: Numbers are ah and bh, LCM = abh

Trick 5: Fraction HCF/LCM

HCF of fractions = HCF of numerators / LCM of denominators LCM of fractions = LCM of numerators / HCF of denominators

Trick 6: 2^n - 1 Property

HCF(2^a - 1, 2^b - 1) = 2^HCF(a,b) - 1 Very useful for competitive exams!

Trick 7: Meeting Problems

For cyclic events, LCM gives when they synchronize again.

Common Mistakes to Avoid

Confusing HCF and LCM: HCF is always ≤ smallest number, LCM is always ≥ largest number.
Product Formula Misuse: HCF × LCM = Product only works for TWO numbers, not three or more.
Prime Factorization Errors: Double-check your factorization, especially for larger numbers.
Remainder Sign: When dealing with negative remainders, convert to positive by adding divisor.
Fraction Confusion: Remember HCF of fractions uses LCM in denominator (opposite of what you might expect).
Same Remainder vs Different Remainders: Read carefully! Same remainder → HCF, specific pattern → LCM.
Multiple Answers: Some problems have multiple valid pairs - check if question asks for count or specific pair.

5 Frequently Asked Questions

Q1: How do I quickly find HCF of large numbers? A: Use Euclidean algorithm: HCF(a,b) = HCF(b, a mod b). Repeat until remainder is 0.

Q2: Can HCF be greater than the smallest number? A: No! HCF is always less than or equal to the smallest number in the set.

Q3: What's the fastest way to find LCM? A: Prime factorization method is usually fastest for placement exams. For two numbers, you can also use: LCM = (a × b)/HCF.

Q4: How do I identify which formula to use? A: Read the problem carefully:

"Greatest number dividing" → HCF
"Smallest number divisible by" → LCM
Same remainder → HCF of differences
Specific remainders → LCM adjustment

Q5: Are HCF and LCM questions calculation-heavy? A: They can be, but with shortcuts and prime factorization, most can be solved in 30-45 seconds.

Practice these 30 questions thoroughly to master HCF and LCM for your placement exams!