📑 Topics on this page (Jump to Topic):
📌 1. Basics & Methods
- HCF (Highest Common Factor) = the largest number that divides all the given numbers exactly.
- LCM (Least Common Multiple) = the smallest number divisible by all the given numbers.
- Prime factorisation: HCF = lowest powers of common primes; LCM = highest powers of all primes.
- Division (Euclid) method: divide the larger by the smaller, then the divisor by the remainder, until the remainder is 0. The last divisor = HCF.
✏️ Example: 36 = 2²×3², 60 = 2²×3×5 → HCF = 2²×3 = 12, LCM = 2²×3²×5 = 180.
📌 2. Key Formulas
📘 Formulas:
- For two numbers: HCF × LCM = product of the numbers (not valid for three numbers)
- HCF of fractions = HCF of numerators / LCM of denominators
- LCM of fractions = LCM of numerators / HCF of denominators
- If the HCF is h, write the numbers as hx and hy where x, y are co-prime
- The HCF is always a factor of the LCM
⚠️ Exam Traps:
- If the HCF does not divide the LCM, such a pair is impossible (e.g. HCF 12, LCM 160).
- LCM of co-prime numbers = their product.
📌 3. Remainder & Applied Problems
| Situation | Answer |
|---|---|
| Largest number dividing a, b, c leaving remainder r each time | HCF(a−r, b−r, c−r) |
| Largest number dividing a, b, c leaving remainders p, q, s | HCF(a−p, b−q, c−s) |
| Largest number giving the same remainder with a, b, c | HCF(|a−b|, |b−c|, |c−a|) |
| Smallest number leaving remainder r when divided by x, y, z | LCM(x, y, z) + r |
| Smallest number where (divisor − remainder) = k is the same for all | LCM(x, y, z) − k |
| When bells/lights coincide again | LCM |
| Minimum number of square tiles | Area ÷ (HCF)² |
💡 Shortcut Trick: "Largest/maximum size/equal pieces" → HCF; "smallest/together again/minimum" → LCM.
📌 4. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. The HCF of two numbers is 4 and their LCM is 80. If one number is 16, the other is: Practice
Correct answer: (B)
Solution: Other number = HCF × LCM ÷ first = 4 × 80 ÷ 16 = 20.
Solution: Other number = HCF × LCM ÷ first = 4 × 80 ÷ 16 = 20.
Q2. The HCF of 144 and 72 is: Practice
Correct answer: (C)
Solution: Taking the lowest powers of common primes, HCF(144, 72) = 72.
Solution: Taking the lowest powers of common primes, HCF(144, 72) = 72.
Q3. The LCM of 6/7 and 3/5 is: Practice
Correct answer: (D)
Solution: LCM = LCM(numerators)/HCF(denominators) = 6/1.
Solution: LCM = LCM(numerators)/HCF(denominators) = 6/1.
Q4. Find the largest number that divides 115, 310 and 401 leaving remainder 11 in each case. Practice
Correct answer: (A)
Solution: HCF(115−11, 310−11, 401−11) = HCF(104, 299, 390) = 13.
Solution: HCF(115−11, 310−11, 401−11) = HCF(104, 299, 390) = 13.
Q5. The smallest number which leaves remainder 3 when divided by 4, 6 and 8 is: Practice
Correct answer: (B)
Solution: LCM(4, 6, 8) = 24; number = 24 + 3 = 27.
Solution: LCM(4, 6, 8) = 24; number = 24 + 3 = 27.
Q6. Three bells ring at intervals of 6, 8 and 12 minutes. If they ring together at 9:00, after how many minutes will they next ring together? Practice
Correct answer: (C)
Solution: LCM(6, 8, 12) = 24 minutes.
Solution: LCM(6, 8, 12) = 24 minutes.
Q7. A floor of 225 cm × 360 cm is to be paved with identical square tiles without cutting. What is the minimum number of tiles needed? Practice
Correct answer: (D)
Solution: Tile side = HCF(225, 360) = 45 cm. Tiles = (225×360)/(45×45) = 5 × 8 = 40.
Solution: Tile side = HCF(225, 360) = 45 cm. Tiles = (225×360)/(45×45) = 5 × 8 = 40.
Q8. The smallest number which when divided by 9, 10, 15 leaves remainders 2, 3, 8 respectively is: Practice
Correct answer: (A)
Solution: Each time (divisor − remainder) = 7. Number = LCM − 7 = 90 − 7 = 83.
Solution: Each time (divisor − remainder) = 7. Number = LCM − 7 = 90 − 7 = 83.
Q9. Two numbers are in the ratio 4 : 5 and their LCM is 100. Their sum is: Practice
Correct answer: (B)
Solution: Numbers are 4k, 5k; LCM = 20k = 100 → k = 5. Sum = 9k = 45.
Solution: Numbers are 4k, 5k; LCM = 20k = 100 → k = 5. Sum = 9k = 45.
Q10. What is the largest number which divides 65, 142 and 175 leaving the same remainder? Practice
Correct answer: (C)
Solution: HCF(142−65, 175−142, 175−65) = HCF(77, 33, 110) = 11.
Solution: HCF(142−65, 175−142, 175−65) = HCF(77, 33, 110) = 11.
Q11. The largest 4-digit number divisible by each of 15, 25 and 30 is: Practice
Correct answer: (D)
Solution: LCM = 150; 9999 ÷ 150 leaves 99; 9999 − 99 = 9900.
Solution: LCM = 150; 9999 ÷ 150 leaves 99; 9999 − 99 = 9900.
Q12. How many pairs of numbers are possible whose HCF is 16 and LCM is 1680? Practice
Correct answer: (A)
Solution: Numbers 16x, 16y; xy = 1680/16 = 105, with x, y co-prime. Pairs: (16,1680), (48,560), (80,336), (112,240) → 4.
Solution: Numbers 16x, 16y; xy = 1680/16 = 105, with x, y co-prime. Pairs: (16,1680), (48,560), (80,336), (112,240) → 4.
Q13. The HCF of two numbers is 16 and their LCM is 240. If one number is 48, the other is: Practice
Correct answer: (B)
Solution: Other number = HCF × LCM ÷ first = 16 × 240 ÷ 48 = 80.
Solution: Other number = HCF × LCM ÷ first = 16 × 240 ÷ 48 = 80.
Q14. The HCF of 72 and 36 is: Practice
Correct answer: (C)
Solution: Taking the lowest powers of common primes, HCF(72, 36) = 36.
Solution: Taking the lowest powers of common primes, HCF(72, 36) = 36.
Q15. The LCM of 9/10 and 8/15 is: Practice
Correct answer: (D)
Solution: LCM = LCM(numerators)/HCF(denominators) = 72/5.
Solution: LCM = LCM(numerators)/HCF(denominators) = 72/5.
Q16. Find the largest number that divides 860, 934 and 1156 leaving remainder 9 in each case. Practice
Correct answer: (A)
Solution: HCF(860−9, 934−9, 1156−9) = HCF(851, 925, 1147) = 37.
Solution: HCF(860−9, 934−9, 1156−9) = HCF(851, 925, 1147) = 37.
Q17. The smallest number which leaves remainder 4 when divided by 5, 6 and 8 is: Practice
Correct answer: (B)
Solution: LCM(5, 6, 8) = 120; number = 120 + 4 = 124.
Solution: LCM(5, 6, 8) = 120; number = 120 + 4 = 124.
Q18. Three bells ring at intervals of 4, 6 and 9 minutes. If they ring together at 9:00, after how many minutes will they next ring together? Practice
Correct answer: (C)
Solution: LCM(4, 6, 9) = 36 minutes.
Solution: LCM(4, 6, 9) = 36 minutes.
Q19. A floor of 180 cm × 240 cm is to be paved with identical square tiles without cutting. What is the minimum number of tiles needed? Practice
Correct answer: (D)
Solution: Tile side = HCF(180, 240) = 60 cm. Tiles = (180×240)/(60×60) = 3 × 4 = 12.
Solution: Tile side = HCF(180, 240) = 60 cm. Tiles = (180×240)/(60×60) = 3 × 4 = 12.
Q20. The smallest number which when divided by 9, 10, 15 leaves remainders 4, 5, 10 respectively is: Practice
Correct answer: (A)
Solution: Each time (divisor − remainder) = 5. Number = LCM − 5 = 90 − 5 = 85.
Solution: Each time (divisor − remainder) = 5. Number = LCM − 5 = 90 − 5 = 85.