1. Definitions
- HCF — largest number that divides all. LCM — smallest number divisible by all.
- For two numbers: HCF × LCM = product. HCF always divides LCM.
2. Fractions
HCF = HCF of numerators / LCM of denominators. LCM = LCM of numerators / HCF of denominators.
3. Problem types
| Asked | Use |
|---|---|
| Greatest number dividing a, b, c exactly | HCF |
| Greatest number leaving remainders r₁, r₂ | HCF(a − r₁, b − r₂) |
| Greatest number leaving the same remainder | HCF of differences |
| Least number divisible by a, b, c | LCM |
| Least number leaving remainder r in each | LCM + r |
| Bells ring together / runners meet | LCM of intervals |
| Largest tile, container, rope piece | HCF |
| Least number of square tiles | Area ÷ (HCF)² |
4. Solved examples
(ASO 2021) HCF of 1/5, 2/7, 3/11 = 1/385.
(ASO 2015) 90 and 144 students → books = LCM = 720.
Bells toll at 6, 8, 12 min; together again after LCM 24 min.
HCF 12, LCM 360, one number 72 → other = 4320/72 = 60.
Greatest number dividing 43, 91, 183 with the same remainder → HCF(48, 92, 140) = 4.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. The HCF of 60, 36 and 24 is:
(a) 36 (b) 6 (c) 12 (d) 24
Answer: (c) 12
Solution: Factorise: all three are multiples of 12, and 5, 3, 2 share no factor. HCF = 12.
Example 2. The LCM of 9, 12 and 15 is:
(a) 90 (b) 108 (c) 180 (d) 360
Answer: (c) 180
Solution: Take the highest power of each prime: LCM = 180.
Example 3. The LCM of 12, 15 and 20 is:
(a) 120 (b) 180 (c) 60 (d) 30
Answer: (c) 60
Solution: Take the highest power of each prime: LCM = 60.
Example 4. The HCF of two numbers is 16 and their LCM is 336. If one number is 48, the other is:
(a) 64 (b) 112 (c) 128 (d) 21
Answer: (b) 112
Solution: HCF × LCM = product of the numbers. 16 × 336 = 48 × N → N = 112.
ASO exam level
Example 5. Two numbers are in the ratio 5 : 7 and their HCF is 8. The numbers are:
(a) 13 and 15 (b) 48 and 64 (c) 40 and 56 (d) 280 and 8
Answer: (c) 40 and 56
Solution: Numbers = 5×8 and 7×8 = 40 and 56.
Example 6. Three bells ring at intervals of 12, 18 and 24 minutes. They ring together now. After how long will they next ring together?
(a) 0 h 36 min (b) 1 h 12 min (c) 54 min (d) 2 hours
Answer: (b) 1 h 12 min
Solution: Answer = LCM(12, 18, 24) = 72 min = 1 h 12 min.
Example 7. Three containers hold 621, 529 and 368 litres of milk. What is the largest measure that can measure all three exactly?
(a) 19 litres (b) 46 litres (c) 2 litres (d) 23 litres
Answer: (d) 23 litres
Solution: Largest measure = HCF(621, 529, 368) = 23 litres.
Example 8. Three containers hold 208, 299 and 247 litres of milk. What is the largest measure that can measure all three exactly?
(a) 2 litres (b) 9 litres (c) 26 litres (d) 13 litres
Answer: (d) 13 litres
Solution: Largest measure = HCF(208, 299, 247) = 13 litres.
Example 9. The least number which, when divided by 4, 6 and 10, leaves a remainder of 1 in each case is:
(a) 59 (b) 121 (c) 61 (d) 60
Answer: (c) 61
Solution: LCM = 60. Required number = 60 + 1 = 61.
Example 10. The HCF of 1/5, 2/7, 3/11 is:
(a) 385/1 (b) 6/1 (c) 1/385 (d) 1/770
Answer: (c) 1/385
Solution: HCF of fractions = HCF of numerators / LCM of denominators = 1/385.
Example 11. The smallest 3-digit number exactly divisible by both 6 and 12 is:
(a) 106 (b) 108 (c) 96 (d) 120
Answer: (b) 108
Solution: LCM(6, 12) = 12. Smallest multiple of 12 that has 3 digits = 108.
Advanced level
Example 12. The least number which, when divided by 5, 6 and 7, leaves remainders 2, 3 and 4 respectively is:
(a) 210 (b) 213 (c) 207 (d) 206
Answer: (c) 207
Solution: In each case divisor − remainder = 3. So the number = LCM − 3 = 210 − 3 = 207.
Example 13. The least number which, when divided by 5, 6 and 7, leaves remainders 1, 2 and 3 respectively is:
(a) 210 (b) 206 (c) 205 (d) 214
Answer: (b) 206
Solution: In each case divisor − remainder = 4. So the number = LCM − 4 = 210 − 4 = 206.
Example 14. The greatest number that divides 214, 594 and 291 leaving remainders 5, 5 and 6 respectively is:
(a) 18 (b) 24 (c) 19 (d) 38
Answer: (c) 19
Solution: Subtract the remainders: 209, 589, 285. Their HCF = 19.
Topic-wise practice for HCF and LCM (Quant): Only LCM, only HCF, HCF × LCM = product, bells and remainder types. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- LCM of numbers
- HCF of numbers
- HCF × LCM = product of two numbers
- HCF and LCM of fractions and decimals
- Least number which leaves a given remainder
- Greatest number which divides leaving remainders
- Bells ringing together, runners meeting
- Largest tile or container size